message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a given sequence A = {a0, a1, ... , an-1}, find the length of the longest increasing subsequnece (LIS) in A.
An increasing subsequence of A is defined by a subsequence {ai0, ai1, ... , aik} where 0 ≤ i0 < i1 < ... < ik < n and ai0 < ai1 < ... < aik.
Constraints
* 1 ≤ n ≤ 100000
* 0 ≤ ai ≤ 109
Input
n
a0
a1
:
an-1
In the first line, an integer n is given. In the next n lines, elements of A are given.
Output
The length of the longest increasing subsequence of A.
Examples
Input
5
5
1
3
2
4
Output
3
Input
3
1
1
1
Output
1
Submitted Solution:
```
n = int(input())
num = [int(input()) for i in range(n)]
gra = [1]*n
Max = 1
for i in range(1,n):
for j in range(i):
if num[j]<num[i] and gra[j]>=Max:
Max = gra[j]
gra[i] = Max+1
Max += 1
print(Max)
``` | instruction | 0 | 7,520 | 5 | 15,040 |
No | output | 1 | 7,520 | 5 | 15,041 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,521 | 5 | 15,042 |
"Correct Solution:
```
import bisect
q = int(input())
M = {}
sortedList = []
for value in range(q):
query, *inp = input().split()
key = inp[0]
# insert
if query == "0":
x = int(inp[1])
if key not in M:
bisect.insort_left(sortedList, key)
M[key] = []
M[key].append(x)
# get
elif query == "1":
if key in M:
if M[key]:
for value in M[key]:
print(value)
# delete
elif query == "2":
if key in M:
M[key] = []
# dump
else:
L = inp[0]
R = inp[1]
index_left = bisect.bisect_left(sortedList, L)
index_right = bisect.bisect_right(sortedList, R)
for value in range(index_left, index_right):
keyAns = sortedList[value]
if M[keyAns]:
for valueAns in M[keyAns]:
print(keyAns, valueAns)
``` | output | 1 | 7,521 | 5 | 15,043 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,522 | 5 | 15,044 |
"Correct Solution:
```
from bisect import bisect_left, bisect_right, insort_left
from typing import Dict, List
if __name__ == "__main__":
num_query = int(input())
d: Dict[str, List[int]] = {}
keys: List[str] = []
for _ in range(num_query):
op, *v = input().split()
if ("0" == op):
if v[0] not in d:
insort_left(keys, v[0])
d[v[0]] = []
d[v[0]].append(int(v[1]))
elif ("1" == op):
if v[0] in d:
for elem in d[v[0]]:
print(elem)
elif ("2" == op):
if v[0] in d:
d[v[0]] = []
else:
left_key = bisect_left(keys, v[0])
right_key = bisect_right(keys, v[1], left_key)
for k in range(left_key, right_key):
for elem in d[keys[k]]:
print(keys[k], elem)
``` | output | 1 | 7,522 | 5 | 15,045 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,523 | 5 | 15,046 |
"Correct Solution:
```
from bisect import bisect_left,bisect_right,insort_left
dict = {}
keytbl =[]
q = int(input())
for i in range(q):
a = list(input().split())
ki = a[1]
if a[0] == "0":
if ki not in dict:
dict[ki] =[]
insort_left(keytbl,ki)
dict[ki].append(int(a[2]))
elif a[0] == "1":
if ki in dict and dict[ki] != []:print(*dict[ki],sep="\n")
elif a[0] == "2":
if ki in dict: dict[ki] = []
else:
L =bisect_left(keytbl,a[1])
R = bisect_right(keytbl,a[2],L)
for j in range(L,R):
for k in dict[keytbl[j]]:
print(keytbl[j],k)
``` | output | 1 | 7,523 | 5 | 15,047 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,524 | 5 | 15,048 |
"Correct Solution:
```
# AOJ ITP2_8_D: Multi-Map
# Python3 2018.6.24 bal4u
from bisect import bisect_left, bisect_right, insort_left
dict = {}
keytbl = []
q = int(input())
for i in range(q):
a = list(input().split())
ki = a[1]
if a[0] == '0':
if ki not in dict:
dict[ki] = []
insort_left(keytbl, ki)
dict[ki].append(int(a[2]))
elif a[0] == '1':
if ki in dict and dict[ki] != []: print(*dict[ki], sep='\n')
elif a[0] == '2':
if ki in dict: dict[ki] = []
else:
L = bisect_left (keytbl, a[1])
R = bisect_right(keytbl, a[2], L)
for j in range(L, R):
for k in dict[keytbl[j]]: print(keytbl[j], k)
``` | output | 1 | 7,524 | 5 | 15,049 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,525 | 5 | 15,050 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Dictionary - Multi-Map
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=ITP2_8_D&lang=jp
"""
from bisect import insort, bisect_right, bisect_left
class Multi_map:
def __init__(self):
self.mm = dict()
self.lr = []
def insert(self, x, y):
if x in self.mm:
self.mm[x].append(y)
else:
self.mm[x] = [y]
insort(self.lr, x)
def get(self, x):
if x in self.mm and self.mm[x] != []:
print(*self.mm[x], sep='\n')
def delete(self, x):
if x in self.mm:
self.mm[x] = []
def dump(self, l, r):
lb = bisect_left(self.lr, l)
ub = bisect_right(self.lr, r)
for i in range(lb, ub):
k = self.lr[i]
for v in self.mm[k]:
print(f'{k} {v}')
mm = Multi_map()
for _ in range(int(input())):
op, x, y = (input() + ' 1').split()[:3]
if op == '0':
mm.insert(x, int(y))
elif op == '1':
mm.get(x)
elif op == '2':
mm.delete(x)
else:
mm.dump(x, y)
``` | output | 1 | 7,525 | 5 | 15,051 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,526 | 5 | 15,052 |
"Correct Solution:
```
import bisect
from collections import defaultdict
def main():
q = int(input())
d = defaultdict(list)
inserted_flag = False
sorted_keys = None
for _ in range(q):
para = input().split()
if para[0] == "0":
d[para[1]].append(int(para[2]))
inserted_flag = True
elif para[0] == "1":
if para[1] in d.keys():
for i in range(len(d[para[1]])):
print(d[para[1]][i])
elif para[0] == "2":
if para[1] in d.keys():
del d[para[1]]
inserted_flag = True
elif para[0] == "3":
if sorted_keys is None:
sorted_keys = sorted(d.keys())
if inserted_flag:
sorted_keys = sorted(d.keys())
inserted_flag = False
l = bisect.bisect_left(sorted_keys, para[1])
r = bisect.bisect_right(sorted_keys, para[2])
for i in range(l, r):
for j in range(len(d[sorted_keys[i]])):
print("{} {}".format(sorted_keys[i], d[sorted_keys[i]][j]))
main()
``` | output | 1 | 7,526 | 5 | 15,053 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,527 | 5 | 15,054 |
"Correct Solution:
```
# AOJ ITP2_8_D: Multi-Map
# Python3 2018.6.24 bal4u
import bisect
from bisect import bisect_left, bisect_right, insort_left
dict = {}
keytbl = []
q = int(input())
for i in range(q):
a = list(input().split())
ki = a[1]
if a[0] == '0':
if ki not in dict:
dict[ki] = []
insort_left(keytbl, ki)
dict[ki].append(int(a[2]))
elif a[0] == '1' and ki in dict and dict[ki] != []: print(*dict[ki], sep='\n')
elif a[0] == '2' and ki in dict: dict[ki] = []
elif a[0] == '3':
L = bisect_left (keytbl, a[1])
R = bisect_right(keytbl, a[2])
for j in range(L, R):
for k in dict[keytbl[j]]:
print(keytbl[j], k)
``` | output | 1 | 7,527 | 5 | 15,055 |
Provide a correct Python 3 solution for this coding contest problem.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5 | instruction | 0 | 7,528 | 5 | 15,056 |
"Correct Solution:
```
import sys
import bisect
from collections import defaultdict
n = int(input())
arr = []
d = defaultdict(list)
lines = sys.stdin.readlines()
ans = [None] * n
for i in range(n):
q, *arg = lines[i].split()
key = arg[0]
l_idx = bisect.bisect_left(arr, arg[0])
r_idx = bisect.bisect_right(arr, arg[0])
if q == '0': # insert
if l_idx == len(arr) or arr[l_idx] != key:
arr.insert(l_idx, key)
d[key].append(arg[1])
elif q == '1': # get
if l_idx != r_idx:
ans[i] = '\n'.join(d[key])
elif q == '2': # delete
arr[l_idx:r_idx] = []
if l_idx != r_idx:
del d[key]
elif q == '3': # dump L R
r_idx = bisect.bisect_right(arr, arg[1])
if l_idx != r_idx:
keys = arr[l_idx:r_idx]
ans[i] = '\n'.join(['\n'.join([f'{k} {x}' for x in d[k]]) for k in keys])
[print(x) for x in ans if x is not None]
``` | output | 1 | 7,528 | 5 | 15,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5
Submitted Solution:
```
from bisect import *
n = int(input())
dic = {}
dic2 = {}
box = []
for i in range(n):
a, b, *c = input().split()
if a == '0':
insort_left(box, b)
if b in dic:
temp = dic[b]
dic[b] = temp + [int(c[0])]
else:
dic[b] = [int(c[0])]
elif a == '1':
if b in dic:
pri = []
for i in dic[b]:
pri.append(str(i))
print("\n".join(pri))
elif a == '2':
if b in dic:
del dic[b]
else:
L = bisect_left(box,b)
R = bisect_right(box, c[0])
while L<R:
if box[L] in dic:
for i in dic[box[L]]:
print(box[L], i)
L = bisect_right(box, box[L])
``` | instruction | 0 | 7,529 | 5 | 15,058 |
Yes | output | 1 | 7,529 | 5 | 15,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5
Submitted Solution:
```
import bisect
M={}
D=[]
def insert(M,D,key,x):
if key not in M:
M[key]=[]
bisect.insort_left(D,key)
M[key].append(x)
return M,D
def get(M,D,key):
if key in M and M[key]!=[]:
print('\n'.join(map(str,M[key])))
def erase(M,D,key):
if key in M:
M[key]=[]
return M,D
def dump(M,D,L,R):
s=bisect.bisect_left(D,L)
e=bisect.bisect_right(D,R)
if e-s>0:
#ループを使わずにできる方法を考える
for i in range(s,e):
for j in M[D[i]]:
print(D[i],j)
q=int(input())
for i in range(q):
query=list(map(str,input().split()))
query[0]=int(query[0])
if query[0]==0:
M,D=insert(M,D,query[1],int(query[2]))
elif query[0]==1:
get(M,D,query[1])
elif query[0]==2:
M,D=erase(M,D,query[1])
else:
dump(M,D,query[1],query[2])
``` | instruction | 0 | 7,530 | 5 | 15,060 |
Yes | output | 1 | 7,530 | 5 | 15,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5
Submitted Solution:
```
from bisect import bisect,bisect_left,insort
dic = {}
l = []
for i in range(int(input())):
order = list(input().split())
if order[0] == "0":
if not order[1] in dic:
dic[order[1]] = []
insort(l,order[1])
dic[order[1]].append(int(order[2]))
elif order[0] == "1":
if order[1] in dic:
for i in dic[order[1]]:
print(i)
elif order[0] == "2":
if order[1] in dic:
dic[order[1]] = []
elif order[0] == "3":
L = bisect_left(l,order[1])
R = bisect(l,order[2])
for i in range(L,R):
for j in dic[l[i]]:
print(l[i],j)
``` | instruction | 0 | 7,531 | 5 | 15,062 |
Yes | output | 1 | 7,531 | 5 | 15,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
For a dictionary $M$ that stores elements formed by a pair of a string key and an integer value, perform a sequence of the following operations. Note that multiple elements can have equivalent keys.
* insert($key$, $x$): Insert an element formed by a pair of $key$ and $x$ to $M$.
* get($key$): Print all values with the specified $key$.
* delete($key$): Delete all elements with the specified $key$.
* dump($L$, $R$): Print all elements formed by a pair of the key and the value such that the key is greater than or equal to $L$ and less than or equal to $R$ in lexicographic order.
Constraints
* $1 \leq q \leq 200,000$
* $1 \leq x \leq 1,000,000,000$
* $1 \leq $ length of $key$ $ \leq 20$
* $key$ consists of lower-case letters
* $L \leq R$ in lexicographic order
* The total number of elements printed by get operations does not exceed $500,000$
* The total number of elements printed by dump operations does not exceed $500,000$
Input
The input is given in the following format.
$q$
$query_1$
$query_2$
:
$query_q$
Each query $query_i$ is given by
0 $key$ $x$
or
1 $key$
or
2 $key$
or
3 $L$ $R$
where the first digits 0, 1, 2 and 3 represent insert, get, delete and dump operations.
Output
For each get operation, print the corresponding values in the order of insertions.
For each dump operation, print the corresponding elements formed by a pair of the key and the value. For the dump operation, print the elements in ascending order of the keys, in case of a tie, in the order of insertions.
Example
Input
10
0 blue 6
0 red 1
0 blue 4
0 white 5
1 red
1 blue
2 red
1 black
1 red
3 w z
Output
1
6
4
white 5
Submitted Solution:
```
# Solved by QBnewb
# Discretization
# off-line
q = int(input())
s = []
rs = [] # for discretization
# download the input
for i in range(q):
s.append(input().split())
if s[i][0] == '3':
rs.append(s[i][1])
rs.append(s[i][2])
else:
rs.append(s[i][1])
rs = sorted(list(set(rs)))
index = {rs[i]:i for i in range(len(rs))} # discretization
d = [[] for i in range(len(rs))] # discrete dictionary
for i in range(q):
op, key = int(s[i][0]), s[i][1]
idx = index[key]
if op == 0:
d[idx].append(s[i][2])
elif op == 1:
for item in d[idx]:
print(item)
elif op == 2:
d[idx].clear()
else:
l = idx
r = index[s[i][2]]
for j in range(l, r+1):
for item in d[j]:
print(rs[j], item)
``` | instruction | 0 | 7,532 | 5 | 15,064 |
Yes | output | 1 | 7,532 | 5 | 15,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given integers a, b, p, and q. Let f(x) = abs(sin(p/q π x)).
Find minimum possible integer x that maximizes f(x) where a ≤ x ≤ b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains four integers a, b, p, and q (0 ≤ a ≤ b ≤ 10^{9}, 1 ≤ p, q ≤ 10^{9}).
Output
Print the minimum possible integer x for each test cases, separated by newline.
Example
Input
2
0 3 1 3
17 86 389 995
Output
1
55
Note
In the first test case, f(0) = 0, f(1) = f(2) ≈ 0.866, f(3) = 0.
In the second test case, f(55) ≈ 0.999969, which is the largest among all possible values.
Submitted Solution:
```
import math as np
k = int(input())
def f(p,q,x):
return abs(np.sin((p/q)*np.pi*x))
res = []
for i in range(k):
inp = input()
a,b,p,q = list(map(int,inp.split()))
arr = [a,b]
T = q/(p*2)
t= T
n = 1
min_diff = 1
min_t = t
inrange = False
while int(t) <= b:
if int(t)>=a:
inrange = True
if int(t)>=a and int(t)<=b:
if t-int(t) < min_diff:
min_diff = t-int(t)
min_t = int(t)
if int(t)+1>=a and int(t)+1 <= b:
if int(t)+1-t<min_diff:
min_diff = int(t)-t+1
min_t = int(t)+1
n+=2
t = T*n
if inrange == False:
res.append(a if f(p,q,a)>=f(p,q,b) else b)
else:
res.append(min_t)
print(*res,sep = '\n')
``` | instruction | 0 | 7,646 | 5 | 15,292 |
No | output | 1 | 7,646 | 5 | 15,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given integers a, b, p, and q. Let f(x) = abs(sin(p/q π x)).
Find minimum possible integer x that maximizes f(x) where a ≤ x ≤ b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains four integers a, b, p, and q (0 ≤ a ≤ b ≤ 10^{9}, 1 ≤ p, q ≤ 10^{9}).
Output
Print the minimum possible integer x for each test cases, separated by newline.
Example
Input
2
0 3 1 3
17 86 389 995
Output
1
55
Note
In the first test case, f(0) = 0, f(1) = f(2) ≈ 0.866, f(3) = 0.
In the second test case, f(55) ≈ 0.999969, which is the largest among all possible values.
Submitted Solution:
```
n=int(input())
k=[]
import math
l=math.pi
def sinu(a,b,p,q):
c=[]
for i in range(b-a+1):
ang=p/q*l*(a+i)
c.append(int(abs(math.sin(ang))*1000000))
return a+c.index(max(c))
for i in range(n):
a,b,p,q=input().split()
k.append(sinu(int(a),int(b),int(p),int(q)))
for i in range(n):
print(k[i])
``` | instruction | 0 | 7,647 | 5 | 15,294 |
No | output | 1 | 7,647 | 5 | 15,295 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given integers a, b, p, and q. Let f(x) = abs(sin(p/q π x)).
Find minimum possible integer x that maximizes f(x) where a ≤ x ≤ b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains four integers a, b, p, and q (0 ≤ a ≤ b ≤ 10^{9}, 1 ≤ p, q ≤ 10^{9}).
Output
Print the minimum possible integer x for each test cases, separated by newline.
Example
Input
2
0 3 1 3
17 86 389 995
Output
1
55
Note
In the first test case, f(0) = 0, f(1) = f(2) ≈ 0.866, f(3) = 0.
In the second test case, f(55) ≈ 0.999969, which is the largest among all possible values.
Submitted Solution:
```
import math as np
k = int(input())
def f(p,q,x):
return abs(np.sin((p/q)*np.pi*x))
res = []
for i in range(k):
inp = input()
a,b,p,q = list(map(int,inp.split()))
arr = [a,b]
T = q/(p*2)
t= T
n = 1
min_diff = 1
min_t = t
inrange = False
while t <= b:
if t>=a:
inrange = True
if t-int(t) < min_diff:
min_diff = t-int(t)
min_t = int(t)
if int(t)+1-t<min_diff:
min_diff = int(t)-t+1
min_t = int(t)+1
n+=2
t = T*n
if inrange == False:
res.append(a if f(p,q,a)>f(p,q,b) else b)
else:
res.append(min_t)
print(*res,sep = '\n')
``` | instruction | 0 | 7,648 | 5 | 15,296 |
No | output | 1 | 7,648 | 5 | 15,297 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have given integers a, b, p, and q. Let f(x) = abs(sin(p/q π x)).
Find minimum possible integer x that maximizes f(x) where a ≤ x ≤ b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 100) — the number of test cases.
The first line of each test case contains four integers a, b, p, and q (0 ≤ a ≤ b ≤ 10^{9}, 1 ≤ p, q ≤ 10^{9}).
Output
Print the minimum possible integer x for each test cases, separated by newline.
Example
Input
2
0 3 1 3
17 86 389 995
Output
1
55
Note
In the first test case, f(0) = 0, f(1) = f(2) ≈ 0.866, f(3) = 0.
In the second test case, f(55) ≈ 0.999969, which is the largest among all possible values.
Submitted Solution:
```
"""
You have given integers a, b, p, and q. Let f(x)=abs(sin(pqπx)).
Find minimum possible integer x that maximizes f(x) where a≤x≤b.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1≤t≤100) — the number of test cases.
The first line of each test case contains four integers a, b, p, and q (0≤a≤b≤109, 1≤p, q≤109).
Output
Print the minimum possible integer x for each test cases, separated by newline.
"""
import math
def func(p, q, x):
return abs(math.sin((p * math.pi / q) * x))
def derivative_func(p, q, x):
if func(p, q, x) < 0:
return (p * math.pi / q) * -math.sin((p * math.pi / q) * x)
else:
return (p * math.pi / q) * math.cos((p * math.pi / q) * x)
def main():
total_input = input()
total_input = total_input.split(" ")
a = int(total_input[0])
b = int(total_input[1])
p = int(total_input[2])
q = int(total_input[3])
half_period = q * math.pi / p
# first check a to see if max
a_val = func(p, q, a)
if a_val == 1:
return a
end_x_val = min(b, math.floor(a + half_period))
if end_x_val < math.floor(a + half_period):
something = True
if func(p, q, end_x_val) == 1: # not sure if redundant
return end_x_val
local_max_x_list = list()
local_max_val_list = list()
for i in range(math.ceil((a + b) / half_period)):
left_pointer = math.ceil(i * half_period)
right_pointer = min(b, math.ceil((i + 1) * half_period))
val = bin_search(left_pointer, right_pointer, p, q, a, b)
if func(p, q, val) == 1:
return val
else:
local_max_x_list.append(val)
local_max_val_list.append(func(p, q, val))
return local_max_x_list[local_max_val_list.index(max(local_max_val_list))] # x val corresponding to max max
def bin_search(left_pointer, right_pointer, p, q, a, b):
while True:
if left_pointer > right_pointer: # ?
if func(p, q, left_pointer) > func(p, q, right_pointer):
if left_pointer > b:
return right_pointer
else:
return left_pointer
else:
if right_pointer < a:
return left_pointer
else:
return right_pointer
mid = (left_pointer + right_pointer) // 2
# cases
if func(p, q, mid) == 1:
return mid
mid_gradient = derivative_func(p, q, mid)
if mid_gradient < 0: # take the half on the right
if mid + 1 != right_pointer:
left_pointer = mid + 1
else:
return mid
else: # take the half on the left
if mid - 1 != left_pointer:
right_pointer = mid - 1
else:
return mid
t = int(input())
results = []
for i in range(t):
results.append(main())
for i in results:
print(i)
"""
print(func(389, 995, 19))
print(func(389, 995, 55))
"""
#check if a member of the set (pi/2 + t*pi) is within a - b
#check a value
# find value that is the minimum of either one half period to the right or b
# if it is b, do something
# if it is the half period thing, use binary search with the derivative to find closest integer with the max
``` | instruction | 0 | 7,649 | 5 | 15,298 |
No | output | 1 | 7,649 | 5 | 15,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a positive integer n. In one move, you can increase n by one (i.e. make n := n + 1). Your task is to find the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The only line of the test case contains two integers n and s (1 ≤ n ≤ 10^{18}; 1 ≤ s ≤ 162).
Output
For each test case, print the answer: the minimum number of moves you need to perform in order to make the sum of digits of n be less than or equal to s.
Example
Input
5
2 1
1 1
500 4
217871987498122 10
100000000000000001 1
Output
8
0
500
2128012501878
899999999999999999
Submitted Solution:
```
for i in range(int(input())):
n,s=map(int,input().split())
a=[]
while (n>0):
a.append(n%10)
n=n//10
a=a[::-1]
if sum(a)==s:
print(0)
else:
tm=0
for j in range(len(a)):
tm+=a[j]
if tm>=s:
tm-=a[j]
break
if (True):
an=[]
f=0
for p in range(len(a)-1,j-1,-1):
if f==1:
an.append(str((10-a[p]-1)%10))
else:
an.append(str((10-a[p])%10))
if a[p]>0:
f=1
print(''.join(an[::-1]))
``` | instruction | 0 | 7,750 | 5 | 15,500 |
No | output | 1 | 7,750 | 5 | 15,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya had two arrays consisting of non-negative integers: a of size n and b of size m. Vasya chose a positive integer k and created an n × m matrix v using the following formula:
<image>
Vasya wrote down matrix v on a piece of paper and put it in the table.
A year later Vasya was cleaning his table when he found a piece of paper containing an n × m matrix w. He remembered making a matrix one day by the rules given above but he was not sure if he had found the paper with the matrix v from those days. Your task is to find out if the matrix w that you've found could have been obtained by following these rules and if it could, then for what numbers k, a1, a2, ..., an, b1, b2, ..., bm it is possible.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 100), separated by a space — the number of rows and columns in the found matrix, respectively.
The i-th of the following lines contains numbers wi, 1, wi, 2, ..., wi, m (0 ≤ wi, j ≤ 109), separated by spaces — the elements of the i-th row of matrix w.
Output
If the matrix w could not have been obtained in the manner described above, print "NO" (without quotes) in the single line of output.
Otherwise, print four lines.
In the first line print "YES" (without quotes).
In the second line print an integer k (1 ≤ k ≤ 1018). Note that each element of table w should be in range between 0 and k - 1 inclusively.
In the third line print n integers a1, a2, ..., an (0 ≤ ai ≤ 1018), separated by spaces.
In the fourth line print m integers b1, b2, ..., bm (0 ≤ bi ≤ 1018), separated by spaces.
Examples
Input
2 3
1 2 3
2 3 4
Output
YES
1000000007
0 1
1 2 3
Input
2 2
1 2
2 0
Output
YES
3
0 1
1 2
Input
2 2
1 2
2 1
Output
NO
Note
By <image> we denote the remainder of integer division of b by c.
It is guaranteed that if there exists some set of numbers k, a1, ..., an, b1, ..., bm, that you could use to make matrix w, then there also exists a set of numbers that meets the limits 1 ≤ k ≤ 1018, 1 ≤ ai ≤ 1018, 1 ≤ bi ≤ 1018 in the output format. In other words, these upper bounds are introduced only for checking convenience purposes.
Submitted Solution:
```
import sys
f = sys.stdin
#f = open('H:\\Portable Python 3.2.5.1\\test_248B1.txt')
#r, g, b = map(int, f.readline().strip().split())
n, m = map(int, f.readline().strip().split())
v = []
max_v = 0
for i in range(n):
a = [int(u) for u in f.readline().strip().split()]
v.append(a)
max_v = max(max_v, max(a))
#print(v)
#print(max_v)
if n==1 or m==1:
print('YES')
print(max_v)
if n==1:
print(0)
p = [str(u) for u in v[0]]
print( ' '.join(p) )
elif m==1:
p = [str(u[0]) for u in v]
print( ' '.join(p) )
print(0)
a = [0]*n
b = [0]*m
a[0] = v[0][0]
b[0] = 0
k = []
err = False
for i in range(n-1):
w = [0]*m
for j in range(m):
w[j] = v[i+1][j] - v[i][j]
t = max(w)
for j in range(m):
if w[j] - t != 0 :
if len(k)>0:
if k[0]!=w[j] - t:
err = True
else:
k.append(w[j] - t)
a[i+1] = a[i] + t
#print('a')
#print(w)
#print(a)
for j in range(m-1):
w = [0]*n
for i in range(n):
w[i] = v[i][j+1] - v[i][j]
t = max(w)
for s in range(n):
if w[s] - t != 0 :
if len(k)>0:
if k[0]!=w[s] - t:
err = True
else:
k.append(w[s] - t)
b[j+1] = b[j] + t
#print('b')
#print(w)
#print(b)
#
#print(k)
if len(k)==0:
k = max_v + 1
else :
k = -k[0]
if err or k<=max_v:
print('NO')
else:
print('YES')
print(k)
p = [str(u) for u in a]
print( ' '.join(p) )
p = [str(u) for u in b]
print( ' '.join(p) )
``` | instruction | 0 | 7,938 | 5 | 15,876 |
No | output | 1 | 7,938 | 5 | 15,877 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,168 | 5 | 16,336 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
def floor_sum(n, m, a, b):
ans = 0
while True:
if a >= m:
ans += (n-1)*n*(a//m)//2
a %= m
if b >= m:
ans += n*(b//m)
b %= m
yMax = (a*n+b) // m
xMax = yMax*m - b
if yMax == 0:
break
ans += (n-(xMax+a-1)//a) * yMax
n, m, a, b = yMax, a, m, (-xMax)%a
return ans
T = int(input())
anss = []
for _ in range(T):
N, M, A, B = map(int, input().split())
ans = floor_sum(N, M, A, B)
anss.append(ans)
print('\n'.join(map(str, anss)))
``` | output | 1 | 8,168 | 5 | 16,337 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,169 | 5 | 16,338 |
"Correct Solution:
```
def floor_sum(n,m,a,b):
r=0
x,y,z=0,0,0
while 1:
if b>=m:
x=b//m
else:
x=0
if a>=m:
y=a//m
else:
y=0
r+=x*n
b-=x*m
r+=(y*n*(n-1))>>1
a-=y*m
x=(a*n+b)//m
if x==0:
break
y=b-x*m
z=y//a
r+=(n+z)*x
a,b,n,m=m,y-z*a,x,a
return r
for i in range(int(input())):
print(floor_sum(*list(map(int,input().split()))))
``` | output | 1 | 8,169 | 5 | 16,339 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,170 | 5 | 16,340 |
"Correct Solution:
```
import sys
def floor_sum(n, m, a, b):
ans = n*(b//m)
b %= m
while True:
ans += (n-1)*n*(a//m)//2
a %= m
if a*n+b < m:
return ans
y_max = (a*n + b)//m
b -= y_max*m # now we have x_max = -(b//a)
ans += (n + b//a)*y_max
b %= a
m, a, n = a, m, y_max
def main():
t = int(sys.stdin.buffer.readline())
for x in sys.stdin.buffer.readlines():
n, m, a, b = map(int, x.split())
res = floor_sum(n, m, a, b)
print(res)
if __name__ == "__main__":
main()
``` | output | 1 | 8,170 | 5 | 16,341 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,171 | 5 | 16,342 |
"Correct Solution:
```
def floor_sum(n, m, a, b):
ans = 0
if a >= m:
ans += (n - 1) * n * (a // m) // 2
a %= m
if b >= m:
ans += n * (b // m)
b %= m
y_max = (a * n + b) // m
x_max = (y_max * m - b)
if y_max == 0:
return ans
ans += (n - (x_max + a - 1) // a) * y_max
ans += floor_sum(y_max, a, m, (a - x_max % a) % a)
return ans
def atcoder_practice2_c():
T = int(input())
for _ in range(T):
N, M, A, B = map(int, input().split())
print(floor_sum(N, M, A, B))
if __name__ == "__main__":
atcoder_practice2_c()
``` | output | 1 | 8,171 | 5 | 16,343 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,172 | 5 | 16,344 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
def floor_sum_of_linear(N, M, A, B):
ans = 0
while True:
ans += (A//M)*(N-1)*N//2 + (B//M)*N
A, B = A%M, B%M
if A*N+B < M:
return ans
ymax = (N*A+B)//M
xmax = -((B-M*ymax)//A)
ans += (N-xmax)*ymax
A, B, N, M = M, A*xmax-M*ymax+B, ymax, A
T = int(readline())
Ans = [None]*T
for qu in range(T):
Ans[qu] = floor_sum_of_linear(*tuple(map(int, readline().split())))
print('\n'.join(map(str, Ans)))
``` | output | 1 | 8,172 | 5 | 16,345 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,173 | 5 | 16,346 |
"Correct Solution:
```
def floor_sum(n,m,a,b):
ans=0
if a>=m:
ans+=(n-1)*n*(a//m)//2
a%=m
if b>=m:
ans+=n*(b//m)
b%=m
y_max=(a*n+b)//m
x_max=(y_max*m-b)
if y_max==0:
return ans
ans+=(n-(x_max+a-1)//a)*y_max
ans+=floor_sum(y_max,a,m,(a-x_max%a)%a)
return ans
T=int(input())
for i in range(T):
N,M,A,B=map(int,input().split())
print(floor_sum(N,M,A,B))
``` | output | 1 | 8,173 | 5 | 16,347 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,174 | 5 | 16,348 |
"Correct Solution:
```
# Date [ 2020-09-08 00:14:49 ]
# Problem [ c.py ]
# Author Koki_tkg
import sys
# import math
# import bisect
# import numpy as np
# from decimal import Decimal
# from numba import njit, i8, u1, b1 #JIT compiler
# from itertools import combinations, product
# from collections import Counter, deque, defaultdict
sys.setrecursionlimit(10 ** 6)
MOD = 10 ** 9 + 7
INF = 10 ** 9
PI = 3.14159265358979323846
def read_str(): return sys.stdin.readline().strip()
def read_int(): return int(sys.stdin.readline().strip())
def read_ints(): return map(int, sys.stdin.readline().strip().split())
def read_ints2(x): return map(lambda num: int(num) - x, sys.stdin.readline().strip().split())
def read_str_list(): return list(sys.stdin.readline().strip().split())
def read_int_list(): return list(map(int, sys.stdin.readline().strip().split()))
def GCD(a: int, b: int) -> int: return b if a%b==0 else GCD(b, a%b)
def LCM(a: int, b: int) -> int: return (a * b) // GCD(a, b)
def floor_sum(n, m, a, b):
ret = 0
if a >= m:
ret += (n - 1) * n * (a // m) // 2
a %= m
if b >= m:
ret += n * (b // m)
bb %= m
y_max = (a * n + b) // m; x_max = (y_max * m - b)
if y_max == 0: return ret
ret += (n - (x_max + a - 1) // a) * y_max
ret += floor_sum(y_max, a, m, (a - x_max % a) % a)
return ret
def Main():
for _ in range(read_int()):
n, m, a, b = read_ints()
print(floor_sum(n, m, a, b))
if __name__ == '__main__':
Main()
``` | output | 1 | 8,174 | 5 | 16,349 |
Provide a correct Python 3 solution for this coding contest problem.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000 | instruction | 0 | 8,175 | 5 | 16,350 |
"Correct Solution:
```
def floor_sum(n, m, a, b):
res = 0
while True:
res += a//m * n * (n-1) // 2
a %= m
res += b//m * n
b %= m
Y = (a*n+b) // m
X = Y*m-b
if Y == 0:
return res
res += (n+(-X//a)) * Y
n, m, a, b = Y, a, m, -X%a
import sys
input = sys.stdin.buffer.readline
t = int(input())
for _ in range(t):
n, m, a, b = map(int, input().split())
print(floor_sum(n, m, a, b))
``` | output | 1 | 8,175 | 5 | 16,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
from sys import stdin
def floor_sum(n, m, a, b):
result = 0
if a >= m:
result += (n - 1) * n * (a // m) // 2
a %= m
if b >= m:
result += n * (b // m)
b %= m
yMax = (a * n + b) // m
xMax = yMax * m - b
if yMax == 0:
return result
result += (n - (xMax + a - 1) // a) * yMax
result += floor_sum(yMax, a, m, (a - xMax % a) % a)
return result
readline = stdin.readline
T = int(readline())
result = []
for _ in range(T):
N, M, A, B = map(int, readline().split())
result.append(floor_sum(N, M, A, B))
print(*result, sep='\n')
``` | instruction | 0 | 8,177 | 5 | 16,354 |
Yes | output | 1 | 8,177 | 5 | 16,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
def sum_of_floor(n,m,a,b):
ans = 0
if a >= m:
ans += (n - 1) * n * (a // m) // 2
a %= m
if b >= m:
ans += n * (b // m)
b %= m
y_max = (a * n + b) // m
x_max = (y_max * m - b)
if y_max == 0:
return ans
ans += (n - (x_max + a - 1) // a) * y_max
ans += sum_of_floor(y_max, a, m, (a - x_max % a) % a)
return ans
for _ in range(int(input())):
n,m,a,b = map(int, input().split())
ans = sum_of_floor(n,m,a,b)
print(ans)
``` | instruction | 0 | 8,178 | 5 | 16,356 |
Yes | output | 1 | 8,178 | 5 | 16,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.readline
# sum([(a * i + b) // m for i in range(n)])
def floor_sum(n, m, a, b):
ret = 0
while True:
ret += (a // m) * n * (n-1) // 2 + (b // m) * n
a %= m
b %= m
y = (a * n + b) // m
x = b - y * m
if y == 0:
return ret
ret += (x // a + n) * y
n, m, a, b = y, a, m, x % a
for _ in range(int(input())):
N, M, A, B = map(int, input().split())
print(floor_sum(N, M, A, B))
if __name__ == '__main__':
main()
``` | instruction | 0 | 8,179 | 5 | 16,358 |
Yes | output | 1 | 8,179 | 5 | 16,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
mod = 10**9+7
from operator import mul
from functools import reduce
def cmb(n,r):
r = min(n-r,r)
if r == 0: return 1
over = reduce(mul, range(n, n - r, -1))
under = reduce(mul, range(1,r + 1))
return over // under
s = int(input())
if s==1 or s==2:
print(0)
exit()
pk_num = s//3-1
ans = 1
for i in range(pk_num):
pk = (i+2)*3
g = s - pk
b = i+2
ans += cmb(g+b-1,b-1)
ans %= mod
print(ans)
``` | instruction | 0 | 8,181 | 5 | 16,362 |
No | output | 1 | 8,181 | 5 | 16,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
def floor_sum(n,m,a,b):
ans = n*(b//m)
b %= m
while True:
ans += (n-1)*n*(a//m)//2
a %= m
if a*n+b < m:
return ans
y_max = (a*n + b)//m
#b -= y_max*m #now we have x_max = -(b//a)
ans += (n + (b - y_max*m)//a)*y_max
n = y_max
b = (b - y_max*m)%a
m,a = a,m
import sys
readline = sys.stdin.buffer.readline
read = sys.stdin.read
T = int(readline())
for _ in range(T):
n,m,a,b = map(int,readline().split())
print(floor_sum(n,m,a,b))
if T==5: print(1)
``` | instruction | 0 | 8,182 | 5 | 16,364 |
No | output | 1 | 8,182 | 5 | 16,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In this problem, you should process T testcases.
For each testcase, you are given four integers N, M, A, B.
Calculate \sum_{i = 0}^{N - 1} floor((A \times i + B) / M).
Constraints
* 1 \leq T \leq 100,000
* 1 \leq N, M \leq 10^9
* 0 \leq A, B < M
Input
Input is given from Standard Input in the following format:
T
N_0 M_0 A_0 B_0
N_1 M_1 A_1 B_1
:
N_{T - 1} M_{T - 1} A_{T - 1} B_{T - 1}
Output
Print the answer for each testcase.
Example
Input
5
4 10 6 3
6 5 4 3
1 1 0 0
31415 92653 58979 32384
1000000000 1000000000 999999999 999999999
Output
3
13
0
314095480
499999999500000000
Submitted Solution:
```
# coding: utf-8
# Your code here!
def floor_sum(n,m,a,b):
ans = n*(b//m)
b %= m
while True:
ans += (n-1)*n*(a//m)//2
a %= m
if a*n+b < m:
break
y_max = (a*n + b)//m
b -= y_max*m #now we have x_max = -(b//a)
ans += (n + b//a)*y_max
n = y_max
b %= a
m,a = a,m
import sys
input = sys.stdin.buffer.readline
T = int(input())
for _ in range(T):
n,m,a,b = map(int,input().split())
print(floor_sum(n,m,a,b))
if T==5: print(1)
``` | instruction | 0 | 8,183 | 5 | 16,366 |
No | output | 1 | 8,183 | 5 | 16,367 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,248 | 5 | 16,496 |
"Correct Solution:
```
n = input()
print('Hello World' if n == '1' else int(input()) + int(input()))
``` | output | 1 | 8,248 | 5 | 16,497 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,249 | 5 | 16,498 |
"Correct Solution:
```
n,*l=map(int,open(0));print("Hello World" if n==1 else sum(l))
``` | output | 1 | 8,249 | 5 | 16,499 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,250 | 5 | 16,500 |
"Correct Solution:
```
n=int(input())
if n==1:
print("Hello World")
else:
print(sum([int(input()), int(input())]))
``` | output | 1 | 8,250 | 5 | 16,501 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,251 | 5 | 16,502 |
"Correct Solution:
```
import sys
if input()=="1": print("Hello World"),sys.exit()
print(int(input())+int(input()))
``` | output | 1 | 8,251 | 5 | 16,503 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,252 | 5 | 16,504 |
"Correct Solution:
```
a = input()
if a == '1':
print('Hello World')
else:
print(int(input()) + int(input()))
``` | output | 1 | 8,252 | 5 | 16,505 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,253 | 5 | 16,506 |
"Correct Solution:
```
n,*a=map(int,open(0).read().split()); print('Hello World' if n==1 else sum(a))
``` | output | 1 | 8,253 | 5 | 16,507 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,254 | 5 | 16,508 |
"Correct Solution:
```
n = int(input())
print('Hello World' if n ==1 else int(input())+int(input()))
``` | output | 1 | 8,254 | 5 | 16,509 |
Provide a correct Python 3 solution for this coding contest problem.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8 | instruction | 0 | 8,255 | 5 | 16,510 |
"Correct Solution:
```
N = int(input())
print('Hello World' if N == 1 else int(input()) + int(input()))
``` | output | 1 | 8,255 | 5 | 16,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In 2020, AtCoder Inc. with an annual sales of more than one billion yen (the currency of Japan) has started a business in programming education.
One day, there was an exam where a one-year-old child must write a program that prints `Hello World`, and a two-year-old child must write a program that receives integers A, B and prints A+B.
Takahashi, who is taking this exam, suddenly forgets his age.
He decides to write a program that first receives his age N (1 or 2) as input, then prints `Hello World` if N=1, and additionally receives integers A, B and prints A+B if N=2.
Write this program for him.
Constraints
* N is 1 or 2.
* A is an integer between 1 and 9 (inclusive).
* B is an integer between 1 and 9 (inclusive).
Input
Input is given from Standard Input in one of the following formats:
1
2
A
B
Output
If N=1, print `Hello World`; if N=2, print A+B.
Examples
Input
1
Output
Hello World
Input
2
3
5
Output
8
Submitted Solution:
```
n = input()
if n == '1' :
print('Helli World')
else :
a,b = map(int,input().split())
print(a+b)
``` | instruction | 0 | 8,263 | 5 | 16,526 |
No | output | 1 | 8,263 | 5 | 16,527 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1 | instruction | 0 | 8,280 | 5 | 16,560 |
"Correct Solution:
```
n=int(input())
b=list(map(int,input().split()))
ope=[[] for i in range(n)]
Q=int(input())
for i in range(Q):
l,r=map(int,input().split())
ope[r-1].append(l-1)
res=b.count(0)
Data=[(-1)**((b[i]==1)+1) for i in range(n)]
for i in range(1,n):
Data[i]+=Data[i-1]
Data=[0]+Data
for i in range(n):
ope[i].sort(reverse=True)
# N: 処理する区間の長さ
N=n+1
N0 = 2**(N-1).bit_length()
data = [None]*(2*N0)
INF = (-2**31, -2**31)
# 区間[l, r+1)の値をvに書き換える
# vは(t, value)という値にする (新しい値ほどtは大きくなる)
def update(l, r, v):
L = l + N0; R = r + N0
while L < R:
if R & 1:
R -= 1
if data[R-1]:
data[R-1] = max(v,data[R-1])
else:
data[R-1]=v
if L & 1:
if data[L-1]:
data[L-1] = max(v,data[L-1])
else:
data[L-1]=v
L += 1
L >>= 1; R >>= 1
# a_iの現在の値を取得
def _query(k):
k += N0-1
s = INF
while k >= 0:
if data[k]:
s = max(s, data[k])
k = (k - 1) // 2
return s
# これを呼び出す
def query(k):
return _query(k)[1]
for i in range(n+1):
update(i,i+1,(-Data[i],-Data[i]))
if ope[0]:
update(1,2,(0,0))
for i in range(1,n):
val=query(i)
update(i+1,i+2,(val+Data[i]-Data[i+1],val+Data[i]-Data[i+1]))
for l in ope[i]:
val=query(l)
update(l+1,i+2,(val,val))
print(n-(res+query(n)+Data[n]))
``` | output | 1 | 8,280 | 5 | 16,561 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1 | instruction | 0 | 8,281 | 5 | 16,562 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
B = list(map(int,readline().split()))
Q = int(readline())
m = map(int,read().split())
LR = sorted(zip(m,m))
class MaxSegTree():
def __init__(self,N):
self.Nelem = N
self.size = 1<<(N.bit_length()) # 葉の要素数
self.data = [0] * (2*self.size)
def build(self,raw_data):
# raw_data は 0-indexed
for i,x in enumerate(raw_data):
self.data[self.size+i] = x
for i in range(self.size-1,0,-1):
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
def update(self,i,x):
i += self.size
self.data[i] = x
i >>= 1
while i:
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
i >>= 1
def get_data(self,i):
return self.data[i+self.size]
def get_max(self,L,R):
# [L,R] に対する値を返す
L += self.size
R += self.size + 1
# [L,R) に変更
x = 0
while L < R:
if L&1:
y = self.data[L]
if x < y: x = y
L += 1
if R&1:
R -= 1
y = self.data[R]
if x < y: x = y
L >>= 1; R >>= 1
return x
"""
・ある場所まで確定して、そこから先を全て1で埋めた場合
・ある場所まで確定して、そこから先を全て0で埋めた場合
の加算スコア
"""
add0 = [0] * (N+1)
add1 = [0] * (N+1)
x = sum(B)
add1[0] = x
add0[0] = N-x
for i,x in enumerate(B,1):
if x == 0:
add0[i]=add0[i-1]-1; add1[i] = add1[i-1]
else:
add1[i]=add1[i-1]-1; add0[i] = add0[i-1]
"""
ある場所を右端としてとった時点での
・残りをすべて0で埋めたときのスコア
・残りをすべて1で埋めたときのスコア
dp0, dp1 をseg木で管理
"""
dp0 = MaxSegTree(N+1)
dp0.build([add0[0]] + [0] * N)
dp1 = MaxSegTree(N+1)
dp1.build([add1[0]] + [0] * N)
for L,R in LR:
# dp1[R] を計算したい。[L,inf) が1で埋まったときのスコア
x = dp1.get_data(R)
y = dp0.get_max(0,L-1) + add1[L-1] - add0[L-1] # 0埋め部分を1埋めに修正してdp1[R]に遷移
z = dp1.get_max(L,R-1) # そのままdp1[R]に遷移
if y < z: y = z
if x < y:
dp1.update(R,y)
dp0.update(R,y - add1[R] + add0[R])
# 一致を数えていたので、距離に修正
answer = N - dp0.data[1]
print(answer)
``` | output | 1 | 8,281 | 5 | 16,563 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1 | instruction | 0 | 8,282 | 5 | 16,564 |
"Correct Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
N = int(readline())
B = list(map(int,readline().split()))
Q = int(readline())
m = map(int,read().split())
LR = sorted(zip(m,m))
class MaxSegTree():
def __init__(self,raw_data):
N = len(raw_data)
self.size = 1<<(N.bit_length()) # 葉の要素数
self.data = [0] * (2*self.size)
self.build(raw_data)
def build(self,raw_data):
for i,x in enumerate(raw_data):
self.data[self.size+i] = x
for i in range(self.size-1,0,-1):
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
def update(self,i,x):
i += self.size
self.data[i] = x
i >>= 1
while i:
x = self.data[i+i]; y = self.data[i+i+1]
self.data[i] = x if x>y else y
i >>= 1
def get_data(self,i):
return self.data[i+self.size]
def get_max(self,L,R):
# [L,R] に対する値を返す
L += self.size
R += self.size + 1
# [L,R) に変更
x = 0
while L < R:
if L&1:
y = self.data[L]
if x < y: x = y
L += 1
if R&1:
R -= 1
y = self.data[R]
if x < y: x = y
L >>= 1; R >>= 1
return x
"""
・あるから先を全て0で埋めた場合
・あるから先を全て1で埋めた場合
"""
add1 = [0] * (N+1)
add1[0] = sum(B)
for i,x in enumerate(B,1):
if x:
add1[i] = add1[i-1] - 1
else:
add1[i] = add1[i-1]
add0 = [N - i - x for i,x in enumerate(add1)]
add0,add1
# ある場所を最後の右端まで使ったとする。残りを全て 0 / 1 で埋めたときのスコア
dp0 = MaxSegTree([0] * (N+1))
dp1 = MaxSegTree([0] * (N+1))
dp0.update(0,add0[0])
dp1.update(0,add1[0])
for L,R in LR:
a = dp1.get_data(R)
b = dp0.get_max(0,L-1) + add1[L-1] - add0[L-1]
c = dp1.get_max(L,R-1)
if b < c:
b = c
if a < b:
dp1.update(R, b)
dp0.update(R, b - add1[R] + add0[R])
answer = N - dp0.get_max(0,N)
print(answer)
``` | output | 1 | 8,282 | 5 | 16,565 |
Provide a correct Python 3 solution for this coding contest problem.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1 | instruction | 0 | 8,283 | 5 | 16,566 |
"Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
b=list(map(int,input().split()))
ope=[[] for i in range(n)]
Q=int(input())
for i in range(Q):
l,r=map(int,input().split())
ope[r-1].append(l-1)
res=b.count(0)
Data=[(-1)**((b[i]==1)+1) for i in range(n)]
for i in range(1,n):
Data[i]+=Data[i-1]
Data=[0]+Data
for i in range(n):
ope[i].sort(reverse=True)
# N: 処理する区間の長さ
N=n+1
N0 = 2**(N-1).bit_length()
data = [None]*(2*N0)
INF = (-2**31, -2**31)
# 区間[l, r+1)の値をvに書き換える
# vは(t, value)という値にする (新しい値ほどtは大きくなる)
def update(l, r, v):
L = l + N0; R = r + N0
while L < R:
if R & 1:
R -= 1
if data[R-1]:
data[R-1] = max(v,data[R-1])
else:
data[R-1]=v
if L & 1:
if data[L-1]:
data[L-1] = max(v,data[L-1])
else:
data[L-1]=v
L += 1
L >>= 1; R >>= 1
# a_iの現在の値を取得
def _query(k):
k += N0-1
s = INF
while k >= 0:
if data[k]:
s = max(s, data[k])
k = (k - 1) // 2
return s
# これを呼び出す
def query(k):
return _query(k)[1]
for i in range(n+1):
update(i,i+1,(-Data[i],-Data[i]))
if ope[0]:
update(1,2,(0,0))
for i in range(1,n):
val=query(i)
update(i+1,i+2,(val+Data[i]-Data[i+1],val+Data[i]-Data[i+1]))
for l in ope[i]:
val=query(l)
update(l+1,i+2,(val,val))
print(n-(res+query(n)+Data[n]))
``` | output | 1 | 8,283 | 5 | 16,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
class SegtreeMin():
def __init__(self, n):
self.inf = 10 ** 16
tree_width = 2
while tree_width < n:
tree_width *= 2
self.tree_width = tree_width
self.tree = [self.inf] * (tree_width * 2 - 1)
def update(self, i, a):
seg_i = self.tree_width - 1 + i
self.tree[seg_i] = a
while seg_i != 0:
seg_i = (seg_i - 1) // 2
self.tree[seg_i] = min(self.tree[seg_i * 2 + 1], self.tree[seg_i * 2 + 2])
def element(self, i):
return self.tree[self.tree_width - 1 + i]
# [l,r)の最小値
def min(self, l, r, seg_i=0, segL=0, segR=-1):
if segR == -1: segR = self.tree_width
if r <= segL or segR <= l: return self.inf
if l <= segL and segR <= r: return self.tree[seg_i]
segM = (segL + segR) // 2
ret0 = self.min(l, r, seg_i * 2 + 1, segL, segM)
ret1 = self.min(l, r, seg_i * 2 + 2, segM, segR)
return min(ret0, ret1)
def main():
n = int(input())
bb = LI()
q = int(input())
lr = defaultdict(list)
for _ in range(q):
l, r = map(int1, input().split())
lr[l].append(r)
# print(lr)
dp = SegtreeMin(n + 1)
dp.update(0, 0)
for i in range(n):
for r in lr[i]:
dp.update(r + 1, dp.min(i, r + 2))
dp.update(i + 1, min(dp.element(i + 1), dp.element(i) + bb[i] * 2 - 1))
print(bb.count(0) + dp.element(n))
main()
``` | instruction | 0 | 8,284 | 5 | 16,568 |
No | output | 1 | 8,284 | 5 | 16,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
"""
https://atcoder.jp/contests/arc085/tasks/arc085_d
dp[i][x] = indexがxになるまでは1が連続する際の最小スコア
何もない場合
i <= x では1を選ばなくてはいけない bが1の時0 bが0の時1
i > x では0を選ばなくてはいけない bが0の時0 1の時1
i == lがある場合
dp[i][r] = min(dp[i-1][t] + l番目で1を選んだ際のスコア)
t<r
つまり、区間加算と全体最小値がわかればいい
→遅延評価セグメント木…
"""
#遅延評価セグメント木(tree&lazy)を作る。初期値lisを渡す
#defaultで初期化する
def make_LST(lis,default = float("inf")):
n = 1
while n < len(lis):
n *= 2
tree = [default] * (2*n-1)
lazy = [0] * (2*n-1)
for i in range(len(lis)):
tree[i+n-1] = lis[i]
for i in range(n-2,-1,-1):
tree[i] = min(tree[i*2+1] , tree[i*2+2])
return tree,lazy
def eval_LST(k,l,r,tree,lazy):
if lazy[k] != 0:
tree[k] += lazy[k]
if r-l > 1:
lazy[2*k+1] += lazy[k]
lazy[2*k+2] += lazy[k]
lazy[k] = 0
#add x to [a,b)
def add_LST(a,b,x,tree,lazy,k=0,l=0,r=-1):
if r < 0:
n = (len(tree)+1)//2
r = n
eval_LST(k,l,r,tree,lazy)
if b <= l or r <= a:
return
if a <= l and r <= b:
lazy[k] += x
eval_LST(k,l,r,tree,lazy)
else:
add_LST(a,b,x,tree,lazy,2*k+1,l,(l+r)//2)
add_LST(a,b,x,tree,lazy,2*k+2,(l+r)//2,r)
tree[k] = min(tree[2*k+1] , tree[2*k+2])
#range_minimum_query(def)
def rmq_LST(a,b,tree,lazy,k=0,l=0,r=-1):
if r < 0:
n = (len(tree)+1)//2
r = n
if b <= l or r <= a:
return float("inf")
eval_LST(k,l,r,tree,lazy)
if a <= l and r <= b:
return tree[k]
vl = rmq_LST(a,b,tree,lazy,2*k+1,l,(l+r)//2)
vr = rmq_LST(a,b,tree,lazy,2*k+2,(l+r)//2,r)
return min(vl,vr)
def getfrom_index_LST(index,tree,lazy):
return rmq_LST(index,index+1,tree,lazy)
def upd_point_LST(index,newnum,tree,lazy):
oldnum = getfrom_index_LST(index,tree,lazy)
difference = newnum - oldnum
add_LST(index,index+1,difference,tree,lazy)
from sys import stdin
N = int(stdin.readline())
b = list(map(int,stdin.readline().split()))
Q = int(stdin.readline())
rlis = [ [] for i in range(N+1) ]
for i in range(Q):
l,r = map(int,stdin.readline().split())
rlis[l].append(r)
tmp = [10**9]*(N+1)
tmp[0] = 0
tree,lazy = make_LST(tmp,10**9)
for i in range(1,N+1):
#print (tree,lazy)
#i <= x では1を選ばなくてはいけない bが1の時0 bが0の時1
#i > x では0を選ばなくてはいけない bが0の時0 1の時1
for r in rlis[i]:
newnum = rmq_LST(0,r,tree,lazy)
oldnum = getfrom_index_LST(r,tree,lazy)
if newnum < oldnum:
upd_point_LST(r,newnum,tree,lazy)
if b[i-1] == 0:
add_LST(i,N+1,1,tree,lazy)
else:
add_LST(0,i,1,tree,lazy)
#print (tree,lazy)
print (rmq_LST(0,N+1,tree,lazy))
"""
#verify
n,q = map(int,stdin.readline().split())
tree,lazy = make_LST([0]*n,0)
for i in range(q):
SS = stdin.readline()
if SS[0] == "0":
tmp,s,t,x = map(int,SS.split())
add_LST(s-1,t,x,tree,lazy)
else:
tmp,s,t = map(int,SS.split())
print ( rsq_LST(s-1,t,tree,lazy) )
"""
``` | instruction | 0 | 8,285 | 5 | 16,570 |
No | output | 1 | 8,285 | 5 | 16,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
from collections import defaultdict
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
class SegtreeMin():
def __init__(self, n):
self.inf = 10 ** 16
tree_width = 2
while tree_width < n:
tree_width *= 2
self.tree_width = tree_width
self.tree = [self.inf] * (tree_width * 2 - 1)
def update(self, i, a):
seg_i = self.tree_width - 1 + i
self.tree[seg_i] = a
while seg_i != 0:
seg_i = (seg_i - 1) // 2
self.tree[seg_i] = min(self.tree[seg_i * 2 + 1], self.tree[seg_i * 2 + 2])
def element(self, i):
return self.tree[self.tree_width - 1 + i]
# [l,r)の最小値
def min(self, l, r, seg_i=0, segL=0, segR=-1):
if segR == -1: segR = self.tree_width
if r <= segL or segR <= l: return self.inf
if l <= segL and segR <= r: return self.tree[seg_i]
segM = (segL + segR) // 2
ret0 = self.min(l, r, seg_i * 2 + 1, segL, segM)
ret1 = self.min(l, r, seg_i * 2 + 2, segM, segR)
return min(ret0, ret1)
def main():
n = int(input())
bb = LI()
q = int(input())
lr = defaultdict(list)
for _ in range(q):
l, r = MI()
lr[l-1].append(r-1)
# print(lr)
dp = SegtreeMin(n + 1)
dp.update(0, 0)
for i in range(n):
for r in lr[i]:
dp.update(r + 1, dp.min(i, r + 2))
case0 = dp.element(i) + bb[i] * 2 - 1
if case0 < dp.element(i + 1): dp.update(i + 1, case0)
print(bb.count(0) + dp.element(n))
main()
``` | instruction | 0 | 8,286 | 5 | 16,572 |
No | output | 1 | 8,286 | 5 | 16,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a = \\{a_1, ..., a_N\\} with all zeros, and a sequence b = \\{b_1, ..., b_N\\} consisting of 0 and 1. The length of both is N.
You can perform Q kinds of operations. The i-th operation is as follows:
* Replace each of a_{l_i}, a_{l_i + 1}, ..., a_{r_i} with 1.
Minimize the hamming distance between a and b, that is, the number of i such that a_i \neq b_i, by performing some of the Q operations.
Constraints
* 1 \leq N \leq 200,000
* b consists of 0 and 1.
* 1 \leq Q \leq 200,000
* 1 \leq l_i \leq r_i \leq N
* If i \neq j, either l_i \neq l_j or r_i \neq r_j.
Input
Input is given from Standard Input in the following format:
N
b_1 b_2 ... b_N
Q
l_1 r_1
l_2 r_2
:
l_Q r_Q
Output
Print the minimum possible hamming distance.
Examples
Input
3
1 0 1
1
1 3
Output
1
Input
3
1 0 1
2
1 1
3 3
Output
0
Input
3
1 0 1
2
1 1
2 3
Output
1
Input
5
0 1 0 1 0
1
1 5
Output
2
Input
9
0 1 0 1 1 1 0 1 0
3
1 4
5 8
6 7
Output
3
Input
15
1 1 0 0 0 0 0 0 1 0 1 1 1 0 0
9
4 10
13 14
1 7
4 14
9 11
2 6
7 8
3 12
7 13
Output
5
Input
10
0 0 0 1 0 0 1 1 1 0
7
1 4
2 5
1 3
6 7
9 9
1 5
7 9
Output
1
Submitted Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
INF = 10**6
class Rmin():
def __init__(self, size):
#the number of nodes is 2n-1
self.n = 1
while self.n < size:
self.n *= 2
self.node = [INF] * (2*self.n-1)
def Access(self, x):
return self.node[x+self.n-1]
def Update(self, x, val):
x += self.n-1
self.node[x] = val
while x > 0:
x = (x-1)//2
self.node[x] = min(self.node[2*x+1], self.node[2*x+2])
return
#[l, r)
def Get(self, l, r):
L, R = l+self.n, r+self.n
s = INF
while L<R:
if R & 1:
R -= 1
s = min(s, self.node[R-1])
if L & 1:
s = min(s, self.node[L-1])
L += 1
L >>= 1
R >>= 1
return s
n = int(input())
b = list(map(int, input().split()))
cum = [0]
for x in b:
if x == 1:
cum.append(cum[-1] + 1)
else:
cum.append(cum[-1])
Q = int(input())
q = sorted([list(map(int, input().split())) for _ in range(Q)], key=lambda x: x[1])
dp = Rmin(n+1)
dp.Update(0, cum[-1])
ans = cum[-1]
for l, r in q:
mi = dp.Get(0, l)
num = cum[r] - cum[l-1]
res = r-l+1 - num*2 + mi
dp.Update(r, res)
ans = min(ans, res)
print(ans)
``` | instruction | 0 | 8,287 | 5 | 16,574 |
No | output | 1 | 8,287 | 5 | 16,575 |
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