text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
print ("Down Journey :)\n" * 2)
```
No
| 91,000 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
print("Para que no se le olvide hacerlo despues perro")
```
No
| 91,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
nm = list(map(int,input().split()))
AA = list(map(int,input().split()))
BB = list(map(int,input().split()))
Q = eval(input())
for i in range(Q):
A=AA.copy()
B=BB.copy()
aix=list(map(int,input().split()))
if(aix[0]==1):
A[aix[1]-1]=aix[2]
else:
B[aix[1]-1]=aix[2]
for j in range(len(B)):
T=B[i]
TT=list(filter(lambda x: x<=T, A))
if(len(A)>0 and len(TT)>0):
A.pop(A.index(max(TT)))
if(len(A)==0):
print(-1)
else:
print(max(A))
```
No
| 91,002 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Serge came to the school dining room and discovered that there is a big queue here. There are m pupils in the queue. He's not sure now if he wants to wait until the queue will clear, so he wants to know which dish he will receive if he does. As Serge is very tired, he asks you to compute it instead of him.
Initially there are n dishes with costs a_1, a_2, …, a_n. As you already know, there are the queue of m pupils who have b_1, …, b_m togrogs respectively (pupils are enumerated by queue order, i.e the first pupil in the queue has b_1 togrogs and the last one has b_m togrogs)
Pupils think that the most expensive dish is the most delicious one, so every pupil just buys the most expensive dish for which he has money (every dish has a single copy, so when a pupil has bought it nobody can buy it later), and if a pupil doesn't have money for any dish, he just leaves the queue (so brutal capitalism...)
But money isn't a problem at all for Serge, so Serge is buying the most expensive dish if there is at least one remaining.
Moreover, Serge's school has a very unstable economic situation and the costs of some dishes or number of togrogs of some pupils can change. More formally, you must process q queries:
* change a_i to x. It means that the price of the i-th dish becomes x togrogs.
* change b_i to x. It means that the i-th pupil in the queue has x togrogs now.
Nobody leaves the queue during those queries because a saleswoman is late.
After every query, you must tell Serge price of the dish which he will buy if he has waited until the queue is clear, or -1 if there are no dishes at this point, according to rules described above.
Input
The first line contains integers n and m (1 ≤ n, m ≤ 300\ 000) — number of dishes and pupils respectively. The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^{6}) — elements of array a. The third line contains m integers b_1, b_2, …, b_{m} (1 ≤ b_i ≤ 10^{6}) — elements of array b. The fourth line conatins integer q (1 ≤ q ≤ 300\ 000) — number of queries.
Each of the following q lines contains as follows:
* if a query changes price of some dish, it contains 1, and two integers i and x (1 ≤ i ≤ n, 1 ≤ x ≤ 10^{6}), what means a_i becomes x.
* if a query changes number of togrogs of some pupil, it contains 2, and two integers i and x (1 ≤ i ≤ m, 1 ≤ x ≤ 10^{6}), what means b_i becomes x.
Output
For each of q queries prints the answer as the statement describes, the answer of the i-th query in the i-th line (the price of the dish which Serge will buy or -1 if nothing remains)
Examples
Input
1 1
1
1
1
1 1 100
Output
100
Input
1 1
1
1
1
2 1 100
Output
-1
Input
4 6
1 8 2 4
3 3 6 1 5 2
3
1 1 1
2 5 10
1 1 6
Output
8
-1
4
Note
In the first sample after the first query, there is one dish with price 100 togrogs and one pupil with one togrog, so Serge will buy the dish with price 100 togrogs.
In the second sample after the first query, there is one dish with price one togrog and one pupil with 100 togrogs, so Serge will get nothing.
In the third sample after the first query, nobody can buy the dish with price 8, so Serge will take it. After the second query, all dishes will be bought, after the third one the third and fifth pupils will by the first and the second dishes respectively and nobody will by the fourth one.
Submitted Solution:
```
from typing import Any, Callable, List
class SegmentTree():
T = Any
E = Any
F = Callable[[T, T], T]
G = Callable[[T, E], T]
H = Callable[[E, E], E]
C = Any
def __init__(self, f: F, g: G, h: H,
ti: T, ei: E) -> None:
self.f = f
self.g = g
self.h = h
self.ti = ti
self.ei = ei
def init(self, n_: int) -> int:
self.n = 1
self.height = 0
while self.n < n_:
self.n <<= 1
self.height += 1
self.dat = [self.ti for _ in range(2*self.n)]
self.laz = [self.ei for _ in range(2*self.n)]
def build(self, v: List) -> None:
n_ = len(v)
self.init(n_)
for i in range(n_):
self.dat[self.n+i] = v[i]
for i in range(self.n-1, 0, -1):
self.dat[i] = self.f(self.dat[(i << 1) | 0],
self.dat[(i << 1) | 1])
def reflect(self, k: int) -> T:
if self.laz[k] == self.ei:
return self.dat[k]
else:
return self.g(self.dat[k], self.laz[k])
def propagate(self, k: int) -> None:
if self.laz[k] == self.ei:
return
self.laz[(k << 1) | 0] = \
self.h(self.laz[(k << 1) | 0], self.laz[k])
self.laz[(k << 1) | 1] = \
self.h(self.laz[(k << 1) | 1], self.laz[k])
self.dat[k] = self.reflect(k)
self.laz[k] = self.ei
def thrust(self, k: int) -> None:
for i in range(self.height, 0, -1):
self.propagate(k >> i)
def recalc(self, k: int) -> None:
while True:
k >>= 1
if k == 0:
break
self.dat[k] = self.f(self.reflect((k << 1) | 0),
self.reflect((k << 1) | 1))
def update(self, a: int, b: int, x: E) -> None:
if a >= b:
return
a += self.n
self.thrust(a)
b += self.n-1
self.thrust(b)
l, r = a, b
while l < r:
if l & 1:
self.laz[l] = self.h(self.laz[l], x)
l += 1
if r & 1:
r -= 1
self.laz[r] = self.h(self.laz[r], x)
l, r = l >> 1, r >> 1
self.recalc(a)
self.recalc(b)
def set_val(self, a: int, x: T) -> None:
a += self.n
self.thrust(a)
self.dat[a] = x
self.laz[a] = self.ei
self.recalc(a)
def query(self, a: int, b: int) -> T:
if a >= b:
return self.ti
a += self.n
self.thrust(a)
b += self.n-1
self.thrust(b)
vl, vr = self.ti, self.ti
l, r = a, b
while l < r:
if l & 1:
vl = self.f(vl, self.reflect(l))
l += 1
if r & 1:
r -= 1
vr = self.f(self.reflect(r), vr)
return self.f(vl, vr)
def find_(self, st: int, check: C, acc: T, k: int, l: int, r: int) -> int:
if l+1 == r:
acc = self.f(acc, self.reflect(k))
if check(acc):
return k - self.n
else:
return -1
self.propagate(k)
m = (l+r) >> 1
if (m <= st):
return self.find_(st, check, acc, (k << 1) | 1, m, r)
if st <= l and not check(self.f(acc, self.dat[k])):
acc = self.f(acc, self.dat[k])
return -1
vl = self.find_(st, check, acc, (k << 1) | 0, l, m)
if vl != -1:
return vl
return self.find_(st, check, acc, (k << 1) | 1, m, r)
def find(self, st: int, check: C) -> int:
acc = self.ti
return self.find_(st, check, acc, 1, 0, self.n)
def CFR569_C():
from typing import Callable
import sys
input = sys.stdin.buffer.readline
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
f: Callable[[int, int], int] = lambda a, b: max(a, b)
g: Callable[[int, int], int] = lambda a, b: a+b
ti = ei = 0
seg = SegmentTree(f, g, g, ti, ei)
sz = 1
while sz < 3e5:
sz <<= 1
seg.build([0 for _ in range(sz)])
for i in range(n):
seg.update(sz-a[i], sz, 1)
for i in range(m):
seg.update(sz-b[i], sz, -1)
q = int(input())
check: Callable[[int], bool] = lambda d: d > 0
for i in range(q):
t, k, v = map(int, input().split())
k -= 1
if t == 1:
seg.update(sz-a[k], sz, -1)
a[k] = v
seg.update(sz-a[k], sz, 1)
if t == 2:
seg.update(sz-b[k], sz, 1)
b[k] = v
seg.update(sz-b[k], sz, -1)
pos = seg.find(0, check)
print(pos if pos < 0 else sz - pos)
def main():
CFR569_C()
if __name__ == '__main__':
main()
```
No
| 91,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
from sys import stdin, stdout
mod = 10**9+7
n = int(input())
dolls = []
for i in range(n):
o, i = map(int, stdin.readline().split())
dolls.append((o, i))
dolls.sort()
dolls = [(i, o) for (o, i) in dolls]
#print(dolls)
def bin_search(i):
lo = -1
hi = n-1
while lo+1 < hi:
mid = (lo+hi-1)//2 + 1
top = dolls[mid][1]
if top <= i:
lo = mid
else:
hi = mid
return hi
counts = [1]*(n+1)
dp = [0]*(n+1)
for k in range(n):
i, o = dolls[k]
m_prev = dp[k]+o-(dolls[k-1][1] if k > 0 else 0)
kk = bin_search(i)
m_with = dp[kk] + i-(dolls[kk-1][1] if kk > 0 else 0)
if m_prev < m_with:
dp[k+1] = m_prev
counts[k+1] = counts[k]
elif m_prev > m_with:
dp[k+1] = m_with
counts[k+1] = counts[kk]
else:
dp[k+1] = m_prev
counts[k+1] = (counts[k]+counts[kk]) % mod
#print(dp)
#print(counts)
best = 10**9+10
best_count = 0
maximal = max([i for i, o in dolls])
for i in range(bin_search(maximal), n):
ii = bin_search(dolls[i][0])
cur = dp[ii] + dolls[i][0] - (dolls[ii-1][1] if ii > 0 else 0)
#print(cur, "via", ii)
if cur < best:
best = cur
best_count = counts[ii]
elif cur == best:
best_count += counts[ii]
best_count %= mod
print(best_count)
```
| 91,004 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import io, sys
input = lambda f=io.StringIO(sys.stdin.buffer.read().decode()).readline: f().rstrip()
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp = [None] * n
dp_min = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
dp[i] = in_, 1
else:
empty, count = dp_min[j]
dp[i] = empty - out + in_, count
dp_min[i] = list(dp[i])
if i < n - 1:
if dp_min[i + 1][0] < dp_min[i][0]:
dp_min[i] = dp_min[i + 1]
elif dp_min[i + 1][0] == dp_min[i][0]:
dp_min[i][1] += dp_min[i + 1][1]
dp_min[i][1] %= MOD
print(dp_min[0][1])
```
| 91,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import io, sys
input = lambda f=io.StringIO(sys.stdin.buffer.read().decode()).readline: f().rstrip()
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp_suf = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
empty, count = in_, 1
else:
empty, count = dp_suf[j]
empty -= out - in_
if i < n - 1:
if empty > dp_suf[i + 1][0]:
empty, count = dp_suf[i + 1]
elif empty == dp_suf[i + 1][0]:
count += dp_suf[i + 1][1]
dp_suf[i] = empty, count % MOD
print(dp_suf[0][1])
```
| 91,006 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = []
B = []
val = set()
val.add(0)
a_max = -1
for _ in range(N):
a, b = map(int, input().split())
a, b = b, a
A.append(a)
B.append(b)
val.add(a)
val.add(b)
a_max = max(a_max, a)
val = sorted(list(val))
val2idx = {v: i for i, v in enumerate(val)}
NN = len(val)
adj = [[] for _ in range(NN)]
for i in range(NN-1):
adj[i].append((i+1, val[i+1] - val[i]))
for a_, b_ in zip(A, B):
a = val2idx[a_]
b = val2idx[b_]
adj[a].append((b, 0))
dist = [10**10] * NN
dist[0] = 0
for i in range(NN-1):
for b, cost in adj[i]:
dist[b] = min(dist[b], dist[i] + cost)
min_space = 10**10
B_set = set(B)
for b in B_set:
if b <= a_max:
continue
ib = val2idx[b]
min_space = min(min_space, dist[ib])
dp = [0] * NN
dp[0] = 1
for i in range(NN - 1):
for b, cost in adj[i]:
if dist[i] + cost == dist[b]:
dp[b] = (dp[b] + dp[i])%mod
ans = 0
for b in B_set:
if b <= a_max:
continue
ib = val2idx[b]
if dist[ib] == min_space:
ans = (ans + dp[ib])%mod
print(ans)
if __name__ == '__main__':
main()
```
| 91,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
from bisect import bisect_right as br
P = 10**9+7
N = int(input())
X = []
maxinn = 0
for _ in range(N):
a, b = map(int, input().split())
maxinn = max(maxinn, b)
X.append((a, b))
X = sorted(X)
OUT = [0]
VOL = [0]
CNT = [1]
for out, inn in X:
i = br(OUT, inn) - 1
vol = VOL[i] + out - inn
if OUT[-1] != out:
OUT.append(out)
VOL.append(VOL[-1] if len(CNT)>1 else 0)
CNT.append(CNT[-1] if len(CNT)>1 else 0)
if VOL[-1] < vol:
VOL[-1] = vol
CNT[-1] = CNT[i]
elif VOL[-1] == vol:
CNT[-1] += CNT[i]
CNT[-1] %= P
mi = min([OUT[i]-VOL[i] for i in range(len(CNT)) if OUT[i] > maxinn])
print(sum([CNT[i] for i in range(len(CNT)) if OUT[i] > maxinn and OUT[i]-VOL[i] == mi])%P)
```
| 91,008 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Tags: binary search, combinatorics, data structures, dp, shortest paths, sortings
Correct Solution:
```
from bisect import bisect_left as lb
MOD = 10 ** 9 + 7
n = int(input())
a = [tuple(map(int, input().split())) for _ in range(n)]
a = sorted((in_, out) for out, in_ in a)
dp_suf = [None] * n
for i in range(n - 1, -1, -1):
in_, out = a[i]
j = lb(a, (out, 0))
if j == n:
empty, count = in_, 1
else:
empty, count = dp_suf[j]
empty -= out - in_
if i < n - 1:
if empty > dp_suf[i + 1][0]:
empty, count = dp_suf[i + 1]
elif empty == dp_suf[i + 1][0]:
count += dp_suf[i + 1][1]
dp_suf[i] = empty, count % MOD
print(dp_suf[0][1])
```
| 91,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
from bisect import bisect_right, bisect_left
n = int(input())
gout = []
alls = []
mout = 0
for i in range(n):
a, b = map(int, input().split())
gout.append((a, b, i))
alls.append((a, b, i))
mout = max(mout, b)
gout.sort()
presets = [[]]
sus = []
for i in range(n):
sus.append(gout[i])
presets.append(sus.copy())
graph = {}
for a, b, i in alls:
z = bisect_right(gout, (b, 10 ** 10, 10 ** 10))
graph[(a, b, i)] = presets[z]
ans = 0
for y in graph:
if y[0] > mout:
h = graph[y]
b = y[1]
z = bisect_right(h, (b, 10 ** 10, 10 ** 10))
jz = bisect_left(h, (b, -10 ** 10, -10 ** 10))
ans += z - jz
ans %= (10 ** 9 + 7)
print(ans % (10 ** 9 + 7))
```
No
| 91,010 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
MOD = 10**9 + 7
n = int(input())
mat = []
vset = set()
for i in range(n):
out, inn = [int(item) for item in input().split()]
mat.append((inn, out))
vset.add(inn)
vset.add(out)
mat.sort()
vset = list(vset)
vset.sort()
vdic = dict()
for i, item in enumerate(vset):
vdic[item] = i
# print(mat)
# print(vdic)
edge = [[] for _ in range(len(vdic))]
for inn, out in mat:
edge[vdic[inn]].append(vdic[out])
dp = [0] * len(vdic)
survive = [0] * len(vdic)
dp[0] = 1
ans = 0
max_to = 0
for i, l in enumerate(edge):
for j in l:
dp[j] += dp[i]
dp[j] %= MOD
if dp[i] != 0:
max_to = max(max_to, j)
if i < len(edge)-1 and len(l) == 0 and max_to == i:
dp[i+1] = dp[i]
survive[i+1] = 1
print(dp)
for i, l in enumerate(edge[::-1]):
if len(l) == 0:
ans += dp[len(vdic) - 1 - i]
ans %= MOD
if survive[len(vdic) - 1 - i] == 1:
break
else:
if dp[len(vdic) - 1 - i] == 0:
continue
else:
break
print(ans)
```
No
| 91,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
MOD = 10**9 + 7
n = int(input())
mat = []
vset = set()
for i in range(n):
out, inn = [int(item) for item in input().split()]
mat.append((inn, out))
vset.add(inn)
vset.add(out)
mat.sort()
vset = list(vset)
vset.sort()
vdic = dict()
for i, item in enumerate(vset):
vdic[item] = i
# print(mat)
# print(vdic)
edge = [[] for _ in range(len(vdic))]
for inn, out in mat:
edge[vdic[inn]].append(vdic[out])
# print(edge)
dp = [0] * len(vdic)
dp[0] = 1
ans = 0
for i, l in enumerate(edge):
for j in l:
dp[j] += dp[i]
dp[j] %= MOD
for i, l in enumerate(edge[::-1]):
if len(l) == 0:
ans += dp[len(vdic) - 1 - i]
ans %= MOD
else:
break
print(ans)
```
No
| 91,012 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are famous Russian nesting dolls named matryoshkas sold in one of the souvenir stores nearby, and you'd like to buy several of them. The store has n different matryoshkas. Any matryoshka is a figure of volume out_i with an empty space inside of volume in_i (of course, out_i > in_i).
You don't have much free space inside your bag, but, fortunately, you know that matryoshkas can be nested one inside another. Formally, let's call a set of matryoshkas nested if we can rearrange dolls in such a way, that the first doll can be nested inside the second one, the second doll — inside the third one and so on. Matryoshka i can be nested inside matryoshka j if out_i ≤ in_j. So only the last doll will take space inside your bag.
Let's call extra space of a nested set of dolls as a total volume of empty space inside this structure. Obviously, it's equal to in_{i_1} + (in_{i_2} - out_{i_1}) + (in_{i_3} - out_{i_2}) + ... + (in_{i_k} - out_{i_{k-1}}), where i_1, i_2, ..., i_k are the indices of the chosen dolls in the order they are nested in each other.
Finally, let's call a nested subset of the given sequence as big enough if there isn't any doll from the sequence that can be added to the nested subset without breaking its nested property.
You want to buy many matryoshkas, so you should choose a big enough nested subset to buy it. But you will be disappointed if too much space in your bag will be wasted, so you want to choose a big enough subset so that its extra space is minimum possible among all big enough subsets. Now you wonder, how many different nested subsets meet these conditions (they are big enough, and there is no big enough subset such that its extra space is less than the extra space of the chosen subset). Two subsets are considered different if there exists at least one index i such that one of the subsets contains the i-th doll, and another subset doesn't.
Since the answer can be large, print it modulo 10^9 + 7.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of matryoshkas.
The next n lines contain a description of each doll: two integers out_i and in_i (1 ≤ in_i < out_i ≤ 10^9) — the outer and inners volumes of the i-th matryoshka.
Output
Print one integer — the number of big enough nested subsets such that extra space of each of these subsets is minimum possible. Since the answer can be large, print it modulo 10^9 + 7.
Example
Input
7
4 1
4 2
4 2
2 1
5 4
6 4
3 2
Output
6
Note
There are 6 big enough nested subsets with minimum possible extra space in the example:
* \{1, 5\}: we can't add any other matryoshka and keep it nested; it's extra space is 1;
* \{1, 6\};
* \{2, 4, 5\};
* \{2, 4, 6\};
* \{3, 4, 5\};
* \{3, 4, 6\}.
There are no more "good" subsets because, for example, subset \{6, 7\} is not big enough (we can add the 4-th matryoshka to it) or subset \{4, 6, 7\} has extra space equal to 2.
Submitted Solution:
```
from bisect import bisect_right, bisect_left
n = int(input())
gout = []
alls = []
mout = 0
A = []
B = []
for i in range(n):
a, b = map(int, input().split())
gout.append((a, b, i))
alls.append((a, b, i))
mout = max(mout, b)
A.append(a)
B.append(b)
A.sort()
B.sort()
kr = 10 ** 9
u1 = 0
u2 = 0
while u1 < len(A) and u2 < len(A):
if A[u1] <= B[u2]:
kr = min(kr, B[u2] - A[u1])
u1 += 1
else:
u2 += 1
if kr == 10 ** 9:
print(0)
else:
gout.sort()
presets = [[]]
sus = []
for i in range(n):
sus.append(gout[i])
presets.append(sus.copy())
graph = {}
for a, b, i in alls:
z = bisect_right(gout, (b, 10 ** 10, 10 ** 10))
graph[(a, b, i)] = presets[z]
ans = 0
for y in graph:
if y[0] > mout:
h = graph[y]
b = y[1]
z = bisect_right(h, (b - kr, 10 ** 10, 10 ** 10))
jz = bisect_left(h, (b, -10 ** 10, -10 ** 10))
ans += z - jz
ans %= (10 ** 9 + 7)
print(ans % (10 ** 9 + 7))
```
No
| 91,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two large companies "Cecsi" and "Poca Pola" are fighting against each other for a long time. In order to overcome their competitor, "Poca Pola" started a super secret project, for which it has total n vacancies in all of their offices. After many tests and interviews n candidates were selected and the only thing left was their employment.
Because all candidates have the same skills, it doesn't matter where each of them will work. That is why the company decided to distribute candidates between workplaces so that the total distance between home and workplace over all candidates is minimal.
It is well known that Earth is round, so it can be described as a circle, and all m cities on Earth can be described as points on this circle. All cities are enumerated from 1 to m so that for each i (1 ≤ i ≤ m - 1) cities with indexes i and i + 1 are neighbors and cities with indexes 1 and m are neighbors as well. People can move only along the circle. The distance between any two cities equals to minimal number of transitions between neighboring cities you have to perform to get from one city to another. In particular, the distance between the city and itself equals 0.
The "Poca Pola" vacancies are located at offices in cities a_1, a_2, …, a_n. The candidates live in cities b_1, b_2, …, b_n. It is possible that some vacancies are located in the same cities and some candidates live in the same cities.
The "Poca Pola" managers are too busy with super secret project, so you were asked to help "Poca Pola" to distribute candidates between workplaces, so that the sum of the distance between home and workplace over all candidates is minimum possible.
Input
The first line contains two integers m and n (1 ≤ m ≤ 10^9, 1 ≤ n ≤ 200 000) — the number of cities on Earth and the number of vacancies.
The second line contains n integers a_1, a_2, a_3, …, a_n (1 ≤ a_i ≤ m) — the cities where vacancies are located.
The third line contains n integers b_1, b_2, b_3, …, b_n (1 ≤ b_i ≤ m) — the cities where the candidates live.
Output
The first line should contain the minimum total distance between home and workplace over all candidates.
The second line should contain n different integers from 1 to n. The i-th of them should be the index of candidate that should work at i-th workplace.
Examples
Input
10 3
1 5 5
10 4 6
Output
3
1 2 3
Input
10 3
1 4 8
8 3 6
Output
4
2 3 1
Note
In the first example, the distance between each candidate and his workplace equals to 1 (from 1 to 10, from 4 to 5 and from 6 to 5).
In the second example:
* The second candidate works at first workplace, the distance between cities 3 and 1 equals to 2.
* The third candidate works at second workplace, the distance between cities 6 and 4 equals to 2.
* The first candidate works at third workplace, the distance between cities 8 and 8 equals to 0.
Submitted Solution:
```
p , q = map(int , input().split())
a = list(map(int , input().split()))
b = list(map(int , input().split()))
c = []
count = 0
for i in range (q) :
c.append(0)
for i in range (q) :
min = p
for j in range(q) :
r = abs(a[i] - b[j])
z = abs(r - p)
if (r > z) :
s = z
else :
s = r
if (s == 0 and j+1 not in c) :
c[i] = j+1
min = 0
break
elif (s<min and j+1 not in c) :
min = s
c[i] = j+1
else :
pass
count = count + min
print(count)
print(*c)
```
No
| 91,014 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two large companies "Cecsi" and "Poca Pola" are fighting against each other for a long time. In order to overcome their competitor, "Poca Pola" started a super secret project, for which it has total n vacancies in all of their offices. After many tests and interviews n candidates were selected and the only thing left was their employment.
Because all candidates have the same skills, it doesn't matter where each of them will work. That is why the company decided to distribute candidates between workplaces so that the total distance between home and workplace over all candidates is minimal.
It is well known that Earth is round, so it can be described as a circle, and all m cities on Earth can be described as points on this circle. All cities are enumerated from 1 to m so that for each i (1 ≤ i ≤ m - 1) cities with indexes i and i + 1 are neighbors and cities with indexes 1 and m are neighbors as well. People can move only along the circle. The distance between any two cities equals to minimal number of transitions between neighboring cities you have to perform to get from one city to another. In particular, the distance between the city and itself equals 0.
The "Poca Pola" vacancies are located at offices in cities a_1, a_2, …, a_n. The candidates live in cities b_1, b_2, …, b_n. It is possible that some vacancies are located in the same cities and some candidates live in the same cities.
The "Poca Pola" managers are too busy with super secret project, so you were asked to help "Poca Pola" to distribute candidates between workplaces, so that the sum of the distance between home and workplace over all candidates is minimum possible.
Input
The first line contains two integers m and n (1 ≤ m ≤ 10^9, 1 ≤ n ≤ 200 000) — the number of cities on Earth and the number of vacancies.
The second line contains n integers a_1, a_2, a_3, …, a_n (1 ≤ a_i ≤ m) — the cities where vacancies are located.
The third line contains n integers b_1, b_2, b_3, …, b_n (1 ≤ b_i ≤ m) — the cities where the candidates live.
Output
The first line should contain the minimum total distance between home and workplace over all candidates.
The second line should contain n different integers from 1 to n. The i-th of them should be the index of candidate that should work at i-th workplace.
Examples
Input
10 3
1 5 5
10 4 6
Output
3
1 2 3
Input
10 3
1 4 8
8 3 6
Output
4
2 3 1
Note
In the first example, the distance between each candidate and his workplace equals to 1 (from 1 to 10, from 4 to 5 and from 6 to 5).
In the second example:
* The second candidate works at first workplace, the distance between cities 3 and 1 equals to 2.
* The third candidate works at second workplace, the distance between cities 6 and 4 equals to 2.
* The first candidate works at third workplace, the distance between cities 8 and 8 equals to 0.
Submitted Solution:
```
m,n = map(int,input().split())
a = sorted(zip(map(int,input().split()),range(1,n+1)))
b = sorted(zip(map(int,input().split()),range(1,n+1)))
t1 = sum([abs(x-y)for (x,i),(y,j) in zip(a,b)])
t2 = sum([min(abs(x+m-y),abs(x-y))for (x,i),(y,j) in zip(a,b[::-1])])
if t1<=t2:
print(t1)
print(' '.join(map(str, list(zip(*sorted([(i, j) for (x, i), (y, j) in zip(a, b)])))[1])))
else:
print(t2)
print(' '.join( map(str,list(zip(* sorted( [(i, j) for (x, i), (y, j) in zip(a, b[::-1])] )))[1])))
```
No
| 91,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two large companies "Cecsi" and "Poca Pola" are fighting against each other for a long time. In order to overcome their competitor, "Poca Pola" started a super secret project, for which it has total n vacancies in all of their offices. After many tests and interviews n candidates were selected and the only thing left was their employment.
Because all candidates have the same skills, it doesn't matter where each of them will work. That is why the company decided to distribute candidates between workplaces so that the total distance between home and workplace over all candidates is minimal.
It is well known that Earth is round, so it can be described as a circle, and all m cities on Earth can be described as points on this circle. All cities are enumerated from 1 to m so that for each i (1 ≤ i ≤ m - 1) cities with indexes i and i + 1 are neighbors and cities with indexes 1 and m are neighbors as well. People can move only along the circle. The distance between any two cities equals to minimal number of transitions between neighboring cities you have to perform to get from one city to another. In particular, the distance between the city and itself equals 0.
The "Poca Pola" vacancies are located at offices in cities a_1, a_2, …, a_n. The candidates live in cities b_1, b_2, …, b_n. It is possible that some vacancies are located in the same cities and some candidates live in the same cities.
The "Poca Pola" managers are too busy with super secret project, so you were asked to help "Poca Pola" to distribute candidates between workplaces, so that the sum of the distance between home and workplace over all candidates is minimum possible.
Input
The first line contains two integers m and n (1 ≤ m ≤ 10^9, 1 ≤ n ≤ 200 000) — the number of cities on Earth and the number of vacancies.
The second line contains n integers a_1, a_2, a_3, …, a_n (1 ≤ a_i ≤ m) — the cities where vacancies are located.
The third line contains n integers b_1, b_2, b_3, …, b_n (1 ≤ b_i ≤ m) — the cities where the candidates live.
Output
The first line should contain the minimum total distance between home and workplace over all candidates.
The second line should contain n different integers from 1 to n. The i-th of them should be the index of candidate that should work at i-th workplace.
Examples
Input
10 3
1 5 5
10 4 6
Output
3
1 2 3
Input
10 3
1 4 8
8 3 6
Output
4
2 3 1
Note
In the first example, the distance between each candidate and his workplace equals to 1 (from 1 to 10, from 4 to 5 and from 6 to 5).
In the second example:
* The second candidate works at first workplace, the distance between cities 3 and 1 equals to 2.
* The third candidate works at second workplace, the distance between cities 6 and 4 equals to 2.
* The first candidate works at third workplace, the distance between cities 8 and 8 equals to 0.
Submitted Solution:
```
from collections import deque
n, m = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a = [[x, i+1] for i, x in enumerate(a)]
b = [[x, i+1] for i, x in enumerate(b)]
a.sort()
b.sort()
a = deque(a)
b = deque(b)
ans = []
cost = 0
l = 1
r = m
while True:
if len(a) == 1:
cost += min(abs(a[0][0] - b[0][0]), abs(n - a[0][0] - b[0][0]))
ans.append([a[0][1], str(b[0][1])])
break
c1 = min(abs(a[0][0] - b[0][0]), abs(n - a[0][0] - b[0][0]))
c2 = min(abs(a[-1][0] - b[0][0]), abs(n - a[-1][0] - b[0][0]))
c3 = min(abs(b[-1][0] - a[0][0]), abs(n - b[-1][0] - a[0][0]))
c4 = min(abs(b[-1][0] - a[-1][0]), abs(n - b[-1][0] - a[-1][0]))
c = min(c1, c2, c3, c4)
if c == c1:
a_ = a.popleft()
b_ = b.popleft()
cost += c1
ans.append([a_[1], str(b_[1])])
elif c == c2:
a_ = a.pop()
b_ = b.popleft()
cost += c2
ans.append([a_[1], str(b_[1])])
elif c == c3:
a_ = a.popleft()
b_ = b.pop()
cost += c3
ans.append([a_[1], str(b_[1])])
else:
a_ = a.pop()
b_ = b.pop()
cost += c4
ans.append([a_[1], str(b_[1])])
ans.sort()
ans = [x[1] for x in ans]
print(cost)
print(" ".join(ans))
```
No
| 91,016 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have two strings a and b of equal even length n consisting of characters 0 and 1.
We're in the endgame now. To finally make the universe perfectly balanced, you need to make strings a and b equal.
In one step, you can choose any prefix of a of even length and reverse it. Formally, if a = a_1 a_2 … a_n, you can choose a positive even integer p ≤ n and set a to a_p a_{p-1} … a_1 a_{p+1} a_{p+2} … a_n.
Find a way to make a equal to b using at most n + 1 reversals of the above kind, or determine that such a way doesn't exist. The number of reversals doesn't have to be minimized.
Input
The first line contains a single integer t (1 ≤ t ≤ 2000), denoting the number of test cases.
Each test case consists of two lines. The first line contains a string a of length n, and the second line contains a string b of the same length (2 ≤ n ≤ 4000; n mod 2 = 0). Both strings consist of characters 0 and 1.
The sum of n over all t test cases doesn't exceed 4000.
Output
For each test case, if it's impossible to make a equal to b in at most n + 1 reversals, output a single integer -1.
Otherwise, output an integer k (0 ≤ k ≤ n + 1), denoting the number of reversals in your sequence of steps, followed by k even integers p_1, p_2, …, p_k (2 ≤ p_i ≤ n; p_i mod 2 = 0), denoting the lengths of prefixes of a to be reversed, in chronological order.
Note that k doesn't have to be minimized. If there are many solutions, output any of them.
Example
Input
4
0100011011
1101011000
10101010
10101010
0011
1001
100011
110010
Output
3
6 4 10
0
-1
7
2 6 2 6 2 2 6
Note
In the first test case, string a changes as follows:
* after the first reversal: 1000101011;
* after the second reversal: 0001101011;
* after the third reversal: 1101011000.
Tags: constructive algorithms
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def solve_case():
a = [int(x) for x in input()]
b = [int(x) for x in input()]
n = len(a)
a_rev = []
b_rev = []
def rev_a(x):
if not x:
return
nonlocal a
a = a[:x][::-1] + a[x:]
a_rev.append(x)
def rev_b(x):
if not x:
return
nonlocal b
b = b[:x][::-1] + b[x:]
b_rev.append(x)
def answer():
#cleaning up
nonlocal a_rev
nonlocal b_rev
b_rev.reverse()
a_rev += b_rev
while True:
final = []
i = 0
repl = False
while i < len(a_rev):
if i < len(a_rev) - 1 and a_rev[i] == a_rev[i + 1]:
repl = True
i += 2
else:
final.append(a_rev[i])
i += 1
a_rev = final
if not repl:
break
print(len(a_rev))
print(*a_rev)
a_occ = [[0, 0], [0, 0]]
b_occ = [[0, 0], [0, 0]]
for i in range(0, n, 2):
a_occ[a[i]][a[i + 1]] += 1
for i in range(0, n, 2):
b_occ[b[i]][b[i + 1]] += 1
if a_occ[0][0] != b_occ[0][0] or a_occ[1][1] != b_occ[1][1]:
print(-1)
return
balanced = a_occ[0][1] == b_occ[1][0]
if not balanced:
zero, one = 0, 0
for i in range(0, n, 2):
if a[i] + a[i + 1] == 1:
zero, one = zero + a[i], one + a[i + 1]
if zero + (a_occ[0][1] - one) == b_occ[1][0]:
balanced = True
rev_a(i + 2)
break
if not balanced:
zero, one = 0, 0
for i in range(0, n, 2):
if b[i] + b[i + 1] == 1:
zero, one = zero + b[i], one + b[i + 1]
if zero + (b_occ[0][1] - one) == a_occ[1][0]:
balanced = True
rev_b(i + 2)
break
for i in range(0, n, 2):
for j in range(i, n, 2):
if a[j] == b[n - i - 1] and a[j + 1] == b[n - i - 2]:
rev_a(j)
rev_a(j + 2)
break
answer()
def main():
for _ in range(int(input())):
solve_case()
main()
```
| 91,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
t = int(input())
for _ in range(t):
a,b,n,s = list(map(int, input().split()))
coins = min(s//n,a)
s = s - coins*n
if(s<=b):
print("YES")
else:
print("NO")
```
| 91,018 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
""" should avoid multiplaction"""
import math
q=int(input())
for _ in range(q):
a , b, n, s=map(float,input().split())
t = s//n
t2 = s%n
if t <= a :
if t2 <= b:
print("YES")
else:
print("NO")
elif (abs(t-a) * n + t2) <= b:
print("YES")
else:
print("NO")
```
| 91,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
for _ in range(int(input())):
a , b , n , s = map(int,input().split())
au=min(a,s//n)
if s-(au)*n <=b:
print("YES")
else:
print("NO")
```
| 91,020 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
# cook your dish here
for _ in range(int(input())):
a,b,n,s = map(int ,input().split())
ans = s//n
su = 0
flag = 0
if(ans<=a):
su = su + ans*n
if(s-su<=b):
flag = 1
elif(ans> a):
su = su + a*n
if(s-su<=b):
flag = 1
print("YES" if flag==1 else "NO")
```
| 91,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
T = int(input())
for O in range(T):
a, b, n, s = map(int, input().split(' '))
c = s % n
if (c <= b and a*n+b >= s):
print("YES")
else:
print("NO")
```
| 91,022 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
t=int(input())
while t>0:
a,b,n,s=map(int,input().split())
if s//n <=a and s-((s//n)*n) <=b:
print("YES")
elif s//n > a and s-(a*n)<=b:
print('YES')
else:
print('NO')
t-=1
```
| 91,023 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
t=int(input())
while(t):
t-=1
q=list(map(int,input().split()))
a=q[0]
b=q[1]
n=q[2]
s=q[3]
if(b+(a*n)<s):
print("NO")
else:
if(s//n <= a):
if(s%n==0):
print("YES")
else:
s=s%n
if(b>=s):
print("YES")
else:
print("NO")
elif(s//n > a):
s=s%n
if(b>=s):
print("YES")
else:
print("NO")
```
| 91,024 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Tags: math
Correct Solution:
```
from collections import defaultdict as dd
from collections import deque
import bisect
import heapq
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
def solve():
a, b, n, s = rl()
if a * n + b < s:
print ("NO")
return
d = n * (s // n)
if s - d <= b:
print ("YES")
else:
print ("NO")
mode = 'T'
if mode == 'T':
t = ri()
for i in range(t):
solve()
else:
solve()
```
| 91,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
##for debug comment out
import sys,atexit
from io import BytesIO
inp = BytesIO(sys.stdin.buffer.read())
input = lambda:inp.readline().decode('ascii')
buf = BytesIO()
#sys.stdout.write = lambda s: buf.write(s.encode('ascii'))
#print = lambda s: buf.write(s.encode('ascii'))
atexit.register(lambda:sys.__stdout__.buffer.write(buf.getvalue()))
for i in range(int(input())):
a,b,n,S=map(int,input().split())
a=min(S//n,a)
S-=n*a
#print(S)
if b>=S:
print("YES")
else:
print("NO")
```
Yes
| 91,026 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
T = int(input())
for _ in range(T):
a, b, n, S = map(int, input().split())
if min(S//n, a) * n + b >= S:
print('YES')
else:
print('NO')
```
Yes
| 91,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
q = int(input())
for _ in range(q):
a, b, n, s = list(map(int, input().split()))
if a * n == s or b == s or (a*n < s and b >= (s-(a*n))):
print('YES')
elif b >= (s - (n * min((s // n), a))):
print('YES')
else:
print('NO')
```
Yes
| 91,028 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
#RAVENS
#TEAM_2
#ESSI-DAYI_MOHSEN-LORENZO
#STARTED IN 5:15 :)
check=False
for i in range(int(input())):
check=False
a,b,n,s=map(int,input().split())
if (a * n >= s):
if(s%n<=b):
print("YES")
else:
print("NO")
elif (((a * n) + b) >= s):
tmp=s%(a*n)
if (tmp <= b):
print("YES")
else:
print("NO")
else:
print("NO")
'''if(a*n>=s):
print("s : ",s)
print(((s//n)*a))
s=s-((s//n)*a)
if(b>=s):
check=True
else:
check=False
elif(((a*n)+b)>=s):
print("s : ",s)
s-=a*n
print("s : ",s)
if (b >= s):
check = True
else:
check = False
print("YES" if check else "NO")'''
```
Yes
| 91,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
q = int(input())
for tests in range(q):
inputs = input()
inputs = inputs.split(" ")
a = int(inputs[0])
b = int(inputs[1])
n = int(inputs[2])
S = int(inputs[3])
if(b > S):
print("yes")
continue
a = a + b//n
b = b % n
if(b != 0):
if(S > a*n + b):
print("NO")
continue
quotient = S // n
remainder = S - n * quotient
if(quotient > a):
print("NO")
continue
elif(remainder > b):
print("NO")
continue
print("YES")
```
No
| 91,030 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
for _ in range(int(input())):
a,b,n,s=map(int,input().split())
value=int(s/n)
if value<=a:
rem=value%n
if rem<=b:
print("YES")
else:
print("NO")
else:
s=s-(n*a)
if s>b:
print("NO")
else:
print("YES")
```
No
| 91,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
from __future__ import print_function
q=int(input())
while q > 0:
q-=1
lst=input()
#taking intgers as input
#a,b,n,s
#print("yes")
arr=list(map(int,lst.split()))
s=arr[3]
if arr[0]*arr[2] > s:
if s%arr[0] <= arr[1]:
print("YES")
else:
print("NO")
else :
if arr[0]*arr[2]+arr[1] < s:
print("NO")
else:
print("YES")
```
No
| 91,032 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0 ≤ x ≤ a) coins of value n and y (0 ≤ y ≤ b) coins of value 1, then the total value of taken coins will be S.
You have to answer q independent test cases.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of test cases. Then q test cases follow.
The only line of the test case contains four integers a, b, n and S (1 ≤ a, b, n, S ≤ 10^9) — the number of coins of value n, the number of coins of value 1, the value n and the required total value.
Output
For the i-th test case print the answer on it — YES (without quotes) if there exist such x and y that if you take x coins of value n and y coins of value 1, then the total value of taken coins will be S, and NO otherwise.
You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).
Example
Input
4
1 2 3 4
1 2 3 6
5 2 6 27
3 3 5 18
Output
YES
NO
NO
YES
Submitted Solution:
```
t=int(input())
for _ in range(t):
a,b,n,s=map(int,input().split())
f='NO'
for i in range(1,b+1):
if (s-i)%n==0 and (s-i)//n<=a and (s-i)//n>=0:
f='YES'
break;
print(f)
```
No
| 91,033 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
n, k = map(int, input().split())
l = []
for i in range(n):
l.append(list(map(int, input().split())))
s = 0
for i in range(1, n):
s += ((l[i][0]-l[i-1][0])**2 + (l[i][1]-l[i-1][1])**2)**0.5
print(s*k/50)
```
| 91,034 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
'''input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
'''
n, k = map(int, input().split())
a, b = map(int, input().split())
d = 0
for _ in range(n-1):
x, y = map(int, input().split())
d += ((x-a)**2+(y-b)**2)**0.5
a, b = x, y
print(k*d/50)
```
| 91,035 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
from math import *
n,k = map(int,input().split())
pts = []
for _ in range(n):
x,y = map(int,input().split())
pts.append((x,y))
s = 0
for i in range(n-1):
x,y = pts[i]
u,v = pts[i+1]
s += hypot(x-u,y-v)
print(s*k/50)
# C:\Users\Usuario\HOME2\Programacion\ACM
```
| 91,036 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
import math
def dist(x0, y0, x1, y1):
return math.sqrt(math.pow((x1-x0), 2) + math.pow((y1-y0), 2))
tmp = input().split()
n, k = int(tmp[0]), float(tmp[1])
list_l = []
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum = 0
for i in range(1, n):
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum += dist(list_l[i-1][0], list_l[i-1][1], list_l[i][0], list_l[i][1])
dist_sum /= 50
dist_sum *= k
print(dist_sum)
```
| 91,037 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
class point:
x= 0
y = 0
def __init__(self,a,b):
self.x = a
self.y = b
def dist(a,b):
return (((b.x-a.x)*(b.x-a.x))+((b.y-a.y)*(b.y-a.y)))**.5
arr = []
n,k = list(map(int,input().split()))
for i in range(0,n,1):
a,b = list(map(int,input().split()))
p = point(a,b)
arr.append(p)
dista = 0.0
for i in range(0, n-1,1):
dista += (dist(arr[i],arr[(i+1)]))
#print(i,(i+1)%tam)
print('{0:.8f}'.format((dista*k)/50))
```
| 91,038 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
import math
n,l=list(map(int,input().split()))
p=[]
for k in range(n):
p.append(list(map(int,input().split())))
s=0
for i in range(1,n):
s+=math.sqrt(abs(p[i-1][0]-p[i][0])**2+abs(p[i-1][1]-p[i][1])**2)
print(s*l/50)
```
| 91,039 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
from math import sqrt
def distance(x1,y1,x2,y2):
xdis=(x1-x2)**2
ydis=(y1-y2)**2
dis=sqrt(xdis+ydis)
return dis
n,k=map(int,input().split())
points=[]
tot = 0
for i in range(n):
x,y=map(int,input().split())
points.append((x,y))
for j in range(len(points)-1):
tot += distance(points[j][0],points[j][1],points[j+1][0],points[j+1][1])
est = (tot / 50)
waste = est *k
print('%.9f'%(waste))
##print('%.9f'%(5/2))
```
| 91,040 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Tags: geometry
Correct Solution:
```
n, k = map(int, input().split())
d = 0
a,b = map(int, input().split())
for _ in range(n-1):
x,y = map(int, input().split())
d += ((x-a)**2 + (y-b)**2)**0.5
a,b = x,y
print((d * k)/50)
```
| 91,041 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
n,k=map(int,input().split())
lis=[]
while n>0:
x,y=map(int,input().split())
lis.append(x)
lis.append(y)
n=n-1
dist=0.0
i=0
while ( (i+3) <= len(lis)):
dist=dist+((lis[i]-lis[i+2])**2+(lis[i+1]-lis[i+3])**2)**0.5
i=i+2
r=(dist*k/50)
print(r)
```
Yes
| 91,042 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
n,k=map(int,input().split())
x,y=map(int,input().split())
s=0
for i in range(n-1):
xx,yy=map(int,input().split())
s+=(((x-xx)**2+(y-yy)**2)**0.5)/50
x,y=xx,yy
print(s*k)
```
Yes
| 91,043 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
n_k = list(map(int, input().split()))
n = n_k[0]
k = n_k[1]
length = 0
last_point = list(map(int, input().split()))
for i in range(n-1):
this_point = list(map(int, input().split()))
length += ((last_point[0]-this_point[0])**2 + (last_point[1]-this_point[1])**2)**0.5
last_point = this_point
time = length/50
print(time*k)
```
Yes
| 91,044 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
n,k=map(int,input().split())
a=[]
for i in range(n):
x,y=map(int,input().split())
a.append([x,y])
s=0
for i in range(n-1):
s+=((a[i][0]-a[i+1][0])**2+(a[i][1]-a[i+1][1])**2)**0.5
print((s/50)*k)
```
Yes
| 91,045 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
n, k = map(int, input().split())
coords = []
for _ in range(n):
x, y = map(int, input().split())
coords.append((x, y))
px, py = coords[0]
space = 0
for i in range(1, n):
cx, cy = coords[i]
dist = ((cx - px) * (cx - px) + (cy - py) * (cy - py)) ** 0.5
space += dist
px, py = cx, cy
ans = space / 50
print(ans)
```
No
| 91,046 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
import math
pts, no = input().split()
pts = int(pts)
no = int(no)
x, y = input().split()
x = int(x)
y = int(y)
dist = 0
for i in range(pts - 1):
x1, y1 = input().split()
x1 = int(x1)
y1 = int(y1)
dist += math.sqrt((x-x1)*(x-x1) + (y-y1)*(y-y1))
x = x1
y = y1
print(dist/(50*no))
```
No
| 91,047 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
import math
n, k = map(int, input().split())
a = 0
x, y = map(int, input().split())
for i in range(n-1):
p, q = map(int, input().split())
a += math.sqrt((x - p)* (x - p) + (y - q) * (y - q))
print(a * k / 50)
```
No
| 91,048 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mr. Scrooge, a very busy man, decided to count the time he wastes on all sorts of useless stuff to evaluate the lost profit. He has already counted the time he wastes sleeping and eating. And now Mr. Scrooge wants to count the time he has wasted signing papers.
Mr. Scrooge's signature can be represented as a polyline A1A2... An. Scrooge signs like that: first it places a pen at the point A1, then draws a segment from point A1 to point A2, then he draws a segment from point A2 to point A3 and so on to point An, where he stops signing and takes the pen off the paper. At that the resulting line can intersect with itself and partially repeat itself but Scrooge pays no attention to it and never changes his signing style. As Scrooge makes the signature, he never takes the pen off the paper and his writing speed is constant — 50 millimeters per second.
Scrooge signed exactly k papers throughout his life and all those signatures look the same.
Find the total time Scrooge wasted signing the papers.
Input
The first line contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ 1000). Each of the following n lines contains the coordinates of the polyline's endpoints. The i-th one contains coordinates of the point Ai — integers xi and yi, separated by a space.
All points Ai are different. The absolute value of all coordinates does not exceed 20. The coordinates are measured in millimeters.
Output
Print one real number — the total time Scrooges wastes on signing the papers in seconds. The absolute or relative error should not exceed 10 - 6.
Examples
Input
2 1
0 0
10 0
Output
0.200000000
Input
5 10
3 1
-5 6
-2 -1
3 2
10 0
Output
6.032163204
Input
6 10
5 0
4 0
6 0
3 0
7 0
2 0
Output
3.000000000
Submitted Solution:
```
import math
def dist(x0, y0, x1, y1):
return math.sqrt(math.pow((x1-x0), 2) + math.pow((y1-y0), 2))
tmp = input().split()
n, k = int(tmp[0]), float(tmp[1])
list_l = []
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum = 0
for i in range(1, int(n/2)):
tmp = input().split()
list_l.append((float(tmp[0]), float(tmp[1])))
dist_sum += dist(list_l[i-1][0], list_l[i-1][1], list_l[i][0], list_l[i][1])
dist_sum /= 50
dist_sum *= k
print(dist_sum)
```
No
| 91,049 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
import math, collections, sys
input = sys.stdin.readline
factors = set()
def Divisors(n):
i = 1
while i <= math.sqrt(n):
if (n % i == 0):
factors.add(i)
factors.add(n//i)
i+=1
n, m, k = map(int, input().split())
Divisors(k)
a = list(map(int, input().split()))
b = list(map(int, input().split()))
countA = collections.defaultdict(int)
countB = collections.defaultdict(int)
za, zb = [0], [0]
for i in a:
za.append(za[-1]+i)
for i in b:
zb.append(zb[-1]+i)
for i in factors:
for j in range(n-i+1):
if za[j+i] - za[j] == i:
countA[i]+=1
for j in range(m-i+1):
if zb[j+i] - zb[j] == i:
countB[i]+=1
ans = 0
x = 1
for i in factors:
ans+=countA[i]*countB[k//i]
print(ans)
```
| 91,050 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
from sys import stdin
from math import sqrt, factorial, ceil
from collections import Counter, defaultdict
from heapq import heapify, heapreplace, heappush, heappop
from bisect import bisect_left
# def fac(n):
# f = []
# for i in range(1, int(sqrt(n)) + 1):
# if n % i == 0:
# f.append(i)
# return f
n, m, k = map(int, stdin.readline().split())
a = list(map(int, stdin.readline().split()))
b = list(map(int, stdin.readline().split()))
# mat = [[0 for _ in range(m)] for j in range(n)]
# for i in range(n):
# for j in range(m):
# mat[i][j] = a[i] * b[j]
sa,sb=[0]*50000, [0]*50000
i, j = 0, 0
while i<n:
j=i
while j<n:
if a[j]==0:break
j+=1
if j>i:
temp=j-i
c=1
for x in range(temp,0,-1):
sa[x]+=c
c+=1
i=j+1
i, j = 0, 0
while i<m:
j=i
while j<m:
if b[j]==0:break
j+=1
if j>i:
temp=j-i
c=1
for x in range(temp,0,-1):
sb[x]+=c
c+=1
i=j+1
rect=0
for x in range(1,int(sqrt(k))+1):
if k%x==0:
if max(x,k//x)<=max(n,m):
rect+=sa[x]*sb[k//x]
if x!=k//x:
rect+=sa[k//x]*sb[x]
print(rect)
```
| 91,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
from bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
#sys.setrecursionlimit(100000)
def fac(x):
f=set()
for i in range(1,int(x**0.5)+1):
if x%i==0:
f.add(i)
f.add(x//i)
return f
n,m,k=mdata()
a=list(mdata())
b=list(mdata())
f=sorted(fac(k))
t=0
a1={}
b1={}
fa={}
fb={}
for i in a:
if i==1:
t+=1
elif t!=0:
if a1.get(t)==None:
a1[t]=1
else:
a1[t]+=1
t=0
if t!=0:
if a1.get(t) == None:
a1[t] = 1
else:
a1[t] += 1
for i in f:
fa[i]=0
for j in a1:
if i<=j:
fa[i]+=(j-i+1)*a1[j]
t=0
for i in b:
if i==1:
t+=1
elif t!=0:
if b1.get(t)==None:
b1[t]=1
else:
b1[t]+=1
t=0
if t!=0:
if b1.get(t) == None:
b1[t] = 1
else:
b1[t] += 1
for i in f:
fb[i] = 0
for j in b1:
if i<=j:
fb[i]+=(j-i+1)*b1[j]
ans=0
for i in fa:
if k//i in fb:
ans+=fa[i]*fb[k//i]
print(ans)
```
| 91,052 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
import sys
import math
import itertools
import functools
import collections
import operator
import fileinput
import copy
ORDA = 97
def ii(): return int(input())
def mi(): return map(int, input().split())
def li(): return list(map(int, input().split()))
def lcm(a, b): return abs(a * b) // math.gcd(a, b)
def revn(n): return str(n)[::-1]
def dd(): return collections.defaultdict(int)
def ddl(): return collections.defaultdict(list)
def sieve(n):
if n < 2: return list()
prime = [True for _ in range(n + 1)]
p = 3
while p * p <= n:
if prime[p]:
for i in range(p * 2, n + 1, p):
prime[i] = False
p += 2
r = [2]
for p in range(3, n + 1, 2):
if prime[p]:
r.append(p)
return r
def divs(n, start=1):
r = []
for i in range(start, int(math.sqrt(n) + 1)):
if (n % i == 0):
if (n / i == i):
r.append(i)
else:
r.extend([i, n // i])
return r
def divn(n, primes):
divs_number = 1
for i in primes:
if n == 1:
return divs_number
t = 1
while n % i == 0:
t += 1
n //= i
divs_number *= t
def prime(n):
if n == 2: return True
if n % 2 == 0 or n <= 1: return False
sqr = int(math.sqrt(n)) + 1
for d in range(3, sqr, 2):
if n % d == 0: return False
return True
def convn(number, base):
newnumber = 0
while number > 0:
newnumber += number % base
number //= base
return newnumber
def cdiv(n, k): return n // k + (n % k != 0)
def pars(arr, num):
t = dd()
c = 0
for i in range(num):
if arr[i]:
c += 1
else:
if c: t[c] += 1
c = 0
if c:
t[c] += 1
return t
n, m, k = mi()
a = li()
b = li()
aones = pars(a, n)
bones = pars(b, m)
ans = 0
abones = {}
for i in aones:
for j in bones:
abones[(i, j)] = aones[i] * bones[j]
for t in abones:
for d in divs(k):
ans += abones[t] * max(0, t[0] - d + 1) * max(0, t[1] - k // d + 1)
print(ans)
```
| 91,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
from collections import *
n, m, k = map(int, input().split())
a = list(map(int, input().split()))+[0]
b = list(map(int, input().split()))+[0]
cnt1 = [0]*(n+1)
now = 0
for i in range(n+1):
if a[i]==0:
for j in range(1, now+1):
cnt1[j] += now-j+1
now = 0
else:
now += 1
cnt2 = [0]*(m+1)
now = 0
for i in range(m+1):
if b[i]==0:
for j in range(1, now+1):
cnt2[j] += now-j+1
now = 0
else:
now += 1
ans = 0
for i in range(1, int(k**0.5)+1):
if k%i==0:
if i<=n and k//i<=m:
ans += cnt1[i]*cnt2[k//i]
if i!=k//i:
if k//i<=n and i<=m:
ans += cnt1[k//i]*cnt2[i]
print(ans)
```
| 91,054 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
a_lenths = {}
curr_len = 0
for i in range(len(a)):
if a[i] == 1:
curr_len += 1
else:
a_lenths[curr_len] = a_lenths.get(curr_len, 0) + 1
curr_len = 0
if curr_len != 0:
a_lenths[curr_len] = a_lenths.get(curr_len, 0) + 1
b_lenths = {}
curr_len = 0
for i in range(len(b)):
if b[i] == 1:
curr_len += 1
else:
b_lenths[curr_len] = b_lenths.get(curr_len, 0) + 1
curr_len = 0
if curr_len != 0:
b_lenths[curr_len] = b_lenths.get(curr_len, 0) + 1
# print("a_lenths", a_lenths)
# print("b_lenths", b_lenths)
k_divs = []
for i in range(1, int(k ** 0.5)+1):
if k % i == 0:
k_divs.append(i)
if k // i != i:
k_divs.append(k // i)
ans = 0
for seg_a in a_lenths:
for div1 in k_divs:
if div1 <= seg_a:
div2 = k // div1
num_seg_a = (seg_a - div1 + 1) * a_lenths[seg_a]
for seg_b in b_lenths:
if div2 <= seg_b:
num_seg_b = (seg_b - div2 + 1) * b_lenths[seg_b]
ans += num_seg_a * num_seg_b
# print("div1", div1)
# print("div2", div2)
# print("num_seg_a", num_seg_a)
# print("num_seg_b", num_seg_b)
# print("ans", ans)
print(ans)
```
| 91,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
import io, os
from sys import setrecursionlimit
from collections import defaultdict
from itertools import groupby
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def rint():
return map(int, input().split())
def oint():
return int(input())
def get_str():
return input().decode().strip()
n, m, k = rint()
fla= [len(a) for a in get_str().replace(' ', '').split('0') if len(a)]
flb= [len(a) for a in get_str().replace(' ', '').split('0') if len(a)]
ans = 0
i = 1
while i*i <= k:
if k%i == 0:
sr = i
sc = k//sr
ans += sum(r-sr+1 for r in fla if sr <= r) * sum(c-sc+1 for c in flb if sc <= c)
if sr == sc:
break
sr, sc = sc, sr
ans += sum(r-sr+1 for r in fla if sr <= r) * sum(c-sc+1 for c in flb if sc <= c)
i += 1
print(ans)
```
| 91,056 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Tags: binary search, greedy, implementation
Correct Solution:
```
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
aa=[]
bb=[]
c=0
f=0
for i in range(n):
if(a[i]==1):
if(f==0):
c=1
f=1
else:
c+=1
else:
if(f==1):
f=0
aa.append(c)
c=0
else:
continue
if(f==1):
aa.append(c)
c=0
######
f=0
c=0
for i in range(m):
if(b[i]==1):
if(f==0):
c=1
f=1
else:
c+=1
else:
if(f==1):
f=0
bb.append(c)
c=0
else:
continue
if(f==1):
bb.append(c)
c=0
x=[]
for i in range(1,int(k**(0.5))+1):
if(k%i==0):
if(i!=k//i):
x.append(i)
x.append(k//i)
else:
x.append(i)
ans=0
for p in x:
q=k//p
y=0
z=0
for i in aa:
if(i>=p):
y+=i-p+1
for i in bb:
if(i>=q):
z+=i-q+1
ans+=y*z
print(int(ans))
```
| 91,057 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
from __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
#always use bit 1-indexed
n,m,k = mapi()
a = list(mapi())
b = list(mapi())
res=0
bit = [0]*(n+1)
nit = [0]*(m+1)
def update(bit,n,idx,val):
while idx<=n:
bit[idx]+=val
idx+=idx&(-idx)
def query(bit,idx):
res = 0
while idx>0:
res+=bit[idx]
idx-=idx&(-idx)
return res
cnt =0
for i in a:
if i==1:
cnt+=1
update(bit,n,1,1)
update(bit,n,cnt+1,-1)
else:
cnt = 0
cnt =0
for i in b:
if i==1:
cnt+=1
update(nit,m,1,1)
update(nit,m,cnt+1,-1)
else:
cnt = 0
for i in range(1,int(k**.5)+1):
if k%i!=0: continue
if i<=n and k//i<=m:
res+=query(bit,i)*query(nit,k//i)
if k//i!=i and k//i<=n and i<=m:
res+=query(bit,k//i)*query(nit,i)
print(res)
```
Yes
| 91,058 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
import math
# Method to print the divisors
def printDivisors(n,l) :
list = []
# List to store half of the divisors
for i in range(1, int(math.sqrt(n) + 1)) :
if (n % i == 0) :
# Check if divisors are equal
if (n / i == i) :
l.append(i)
else :
# Otherwise print both
l.append(i)
list.append(int(n / i))
# The list will be printed in reverse
for i in list[::-1] :
l.append(i)
n,m,k=map(int,input().split())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
seq1=[]
seq2=[]
l=0
for i in range(n):
if(l1[i]==1):
l+=1
if(i==n-1):
seq1.append(l)
else:
if(l>0):
seq1.append(l)
l=0
l=0
for i in range(m):
if(l2[i]==1):
l+=1
if(i==m-1):
seq2.append(l)
else:
if(l>0):
seq2.append(l)
l=0
list_of_div=[]
printDivisors(k,list_of_div)
#print(list_of_div)
#print(seq1)
#print(seq2)
summ=0
for i in range(len(list_of_div)):
a=0
b=0
for j in range(len(seq1)):
if(seq1[j]>=list_of_div[i]):
a+=seq1[j]-list_of_div[i]+1
for j in range(len(seq2)):
if(seq2[j]>=(k//list_of_div[i])):
b+=seq2[j]-(k//list_of_div[i])+1
summ+=(a*b)
print(summ)
```
Yes
| 91,059 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
from collections import defaultdict
from math import factorial
from sys import stdin
#input = stdin.readline
def read_tuple(): return map(int,input().split())
def read_int(): return int(input())
def read_list(): return list(map(int,input().split()))
def factors(n):
fact = []
for i in range(1, int(n**0.5)+1):
if n%i == 0:
if n//i == i:
fact.append(i)
else:
fact.append(i)
fact.append(n//i)
fact.sort()
return fact
def comb(n,r):
return factorial(n) // ( factorial(n-r)*factorial(r) )
n,m,k = read_tuple()
a = read_list()
b = read_list()
cnta, cntb = defaultdict(int), defaultdict(int)
cnta[0] = 0
cntb[0] = 0
cnt=0
for i in range(n):
if a[i]==1: cnt+=1
else:cnta[cnt]+=1; cnt=0
cnta[cnt]+=1
cnt = 0
for i in range(m):
if b[i]==1: cnt+=1
else: cntb[cnt]+=1; cnt=0
cntb[cnt]+=1
factor = factors(k)
ans = 0
for p1, cnt1 in cnta.items():
for p2, cnt2 in cntb.items():
if p1*p2 < k: continue
for f in factor:
f1, f2 = f, k//f
if p1 >= f1 and p2 >= f2:
ans += cnt1*cnt2*(p1-f1+1)*(p2-f2+1)
print(ans)
```
Yes
| 91,060 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
from math import sqrt
def gen_dzielniki(val):
ret = list()
for i in range(1, int(sqrt(val)) + 1):
if(not (val % i)):
ret.append(i)
for i in range(int(sqrt(val)), 0, -1):
if(not (val % i) and i * i != val):
ret.append(val//i)
return ret
def zamien(lista):
ret = []
last = -1
for i in range(len(lista)):
if lista[i] == 0 and i != 0 and lista[i-1] != 0:
ret.append(i - last - 1)
last = i
elif lista[i] == 1 and i == len(lista) - 1:
ret.append(i - last)
elif lista[i] == 0:
last = i
else:
pass
ret2 = {}
for u in ret:
if not ret2.get(u):
ret2[u] = 1
else:
ret2[u] += 1
return ret2
n, m, k = list(map(int, input().split()))
dzielniki = gen_dzielniki(k)
def policz(a, b):
# print(str(a) + ' ' + str(b) + " " + str(k))
ans = 0
for i in dzielniki:
A = a - i
B = b - k/i
# print(A, B, i)
if A < 0 or B < 0:
continue
ans += (A + 1) * (B + 1)
# print("wynik to " + str(ans))
return ans
mapa = {}
A = list(map(int, input().split()))
B = list(map(int, input().split()))
A = zamien(A)
B = zamien(B)
ans = 0
for u in A:
for v in B:
ans += A[u] * B[v] * policz(u, v)
print(int(ans))
```
Yes
| 91,061 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
d1={}
flag=False
ct=0
for i in range(len(a)):
if a[i]==1 and not flag:
ct=1
flag=True
elif a[i]==1:
ct+=1
else:
flag=False
if d1.get(ct)==None:
d1[ct]=1
else:
d1[ct]+=1
ct=0
if ct!=0 and flag:
if d1.get(ct)==None:
d1[ct]=1
else:
d1[ct]+=1
d2={}
ct=0
flag=False
for i in range(len(b)):
if b[i]==1 and not flag:
ct=1
flag=True
elif b[i]==1:
ct+=1
else:
flag=False
if d2.get(ct)==None:
d2[ct]=1
else:
d2[ct]+=1
ct=0
if ct!=0 and flag:
if d2.get(ct)==None:
d2[ct]=1
else:
d2[ct]+=1
ans=0
#print(d1)
for i in d1:
for j in d2:
if i*j>=k:
for p in range(1,i+1):
for q in range(1,j+1):
if p*q==k:
ans+=(i-p+1)*(j-q+1)
print(ans)
```
No
| 91,062 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
n,m,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
d1={}
flag=False
ct=0
for i in range(len(a)):
if a[i]==1 and not flag:
ct=1
flag=True
elif a[i]==1:
ct+=1
else:
flag=False
if d1.get(ct)==None:
d1[ct]=1
else:
d1[ct]+=1
ct=0
if ct!=0 and flag:
if d1.get(ct)==None:
d1[ct]=1
else:
d1[ct]+=1
d2={}
ct=0
flag=False
for i in range(len(b)):
if b[i]==1 and not flag:
ct=1
flag=True
elif b[i]==1:
ct+=1
else:
flag=False
if d2.get(ct)==None:
d2[ct]=1
else:
d2[ct]+=1
ct=0
if ct!=0 and flag:
if d2.get(ct)==None:
d2[ct]=1
else:
d2[ct]+=1
ans=0
#print(d1,d2)
for i in d1:
for j in d2:
if i*j>=k:
ans+=(i*j-k+1)*d1[i]*d2[j]
if i>=k:
ans+=(i-k+1)*d1[i]
if j>=k:
ans+=(j-k+1)*d2[j]
print(ans)
```
No
| 91,063 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
n, m, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
h = []
not_b = [-1]
ans = 0
w_i = 0
i = 0
while i < n:
while i < n and a[i] == 1:
w_i += 1
i += 1
if w_i:
h.append(w_i)
while i < n and a[i] == 0:
i += 1
w_i = 0
print(h)
w_j = 0
j = 0
while j < m:
while j < m and b[j] == 1:
w_j += 1
j += 1
if w_j:
print(w_j)
for x in h:
if w_j * x >= k:
for p in range(1, x + 1):
if k % p == 0 and k % p <= w_j:
w = k // p
print(w)
ans += (w_j - w + 1) * (x - p + 1)
while j < m and b[j] == 0:
j += 1
w_j = 0
print(ans)
```
No
| 91,064 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a of length n and array b of length m both consisting of only integers 0 and 1. Consider a matrix c of size n × m formed by following rule: c_{i, j} = a_i ⋅ b_j (i.e. a_i multiplied by b_j). It's easy to see that c consists of only zeroes and ones too.
How many subrectangles of size (area) k consisting only of ones are there in c?
A subrectangle is an intersection of a consecutive (subsequent) segment of rows and a consecutive (subsequent) segment of columns. I.e. consider four integers x_1, x_2, y_1, y_2 (1 ≤ x_1 ≤ x_2 ≤ n, 1 ≤ y_1 ≤ y_2 ≤ m) a subrectangle c[x_1 ... x_2][y_1 ... y_2] is an intersection of the rows x_1, x_1+1, x_1+2, ..., x_2 and the columns y_1, y_1+1, y_1+2, ..., y_2.
The size (area) of a subrectangle is the total number of cells in it.
Input
The first line contains three integers n, m and k (1 ≤ n, m ≤ 40 000, 1 ≤ k ≤ n ⋅ m), length of array a, length of array b and required size of subrectangles.
The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 1), elements of a.
The third line contains m integers b_1, b_2, …, b_m (0 ≤ b_i ≤ 1), elements of b.
Output
Output single integer — the number of subrectangles of c with size (area) k consisting only of ones.
Examples
Input
3 3 2
1 0 1
1 1 1
Output
4
Input
3 5 4
1 1 1
1 1 1 1 1
Output
14
Note
In first example matrix c is:
<image>
There are 4 subrectangles of size 2 consisting of only ones in it:
<image>
In second example matrix c is:
<image>
Submitted Solution:
```
R = lambda:map(int,input().split())
n, m, k = R()
a = list(R())
b = list(R())
a_counts = [0]*(n+1)
b_counts = [0]*(m+1)
count = 0
for i in range(n):
if a[i] == 1:
count += 1
if a[i]==0 or i==n-1:
for j in range(1, count+1):
a_counts[j] = a_counts[j] + count - j + 1
count = 0
count = 0
for i in range(m):
if b[i] == 1:
count += 1
if b[i]==0 or i==m-1:
for j in range(1, count+1):
b_counts[j] = b_counts[j] + count - j + 1
count = 0
ans = 0
for i in range(1, (int)(k**(1/2)) + 1) :
if (k % i == 0) :
if (k / i == i) and i<=min(m,n):
ans += a_counts[i]* b_counts[i]
else :
m2 = k//i
if i<n and k//i < m:
ans += a_counts[i]* b_counts[k // i]
if i<m and k//i < n:
ans += b_counts[i]* a_counts[k // i]
print(ans)
```
No
| 91,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
ans=[]
for _ in range(int(input())):
x,y=map(int,input().split())
a,b=map(int,input().split())
c1=(max(x,y)-min(x,y))*a+min(x,y)*b
c2=(x+y)*a
ans.append(min(c1,c2))
print("\n".join(map(str,ans)))
```
| 91,066 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
t = int(input())
for i in range(t):
x,y = map(int, input().split())
a,b = map(int, input().split())
print( min(abs(x-y)*a + min(x,y)*b, x*a+y*a) )
```
| 91,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
x,y=map(int,input().split())
a,b=map(int,input().split())
t=[]
if x==y==0:
print(0)
else:
z=abs(x-y)
z=z*a
l=min(x,y)
l=l*b
t.append(l+z)
t.append(a*(x+y))
print(min(t))
```
| 91,068 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
t=int(input())
for _ in range(t):
x,y=map(int,input().split())
a,b=map(int,input().split())
if x<y:
print(x*min(b,a+a) + (y-x)*a)
else:
print(y*min(b,a+a) + (x-y)*a)
```
| 91,069 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
mod=998244353
t=int(input())
for _ in range(t):
x,y=map(int,input().split())
a,b=map(int,input().split())
x1=(abs(x-y))*a + b*min(abs(x),abs(y))
x2=(x+y)*a
print(min(x1,x2))
```
| 91,070 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
t = int(input())
for i in range(t):
x, y = map(int, input().split())
a, b = map(int, input().split())
ans = 0
diff = abs(x-y)
if b > 2*a:
ans += x*a+y*a
else:
ans += diff*a + min(x, y)*b
print(ans)
```
| 91,071 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
if __name__ == "__main__":
for _ in range(int(input())):
x, y = map(int, input().split())
a, b = map(int, input().split())
b = min(b, a + a)
if x < y:
x, y= y, x
print(y * b + (x - y) * a)
```
| 91,072 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Tags: greedy, math
Correct Solution:
```
T = int(input())
for t in range(T):
x, y = list(map(int, input().split()))
a, b = list(map(int, input().split()))
if(2*a <= b):
print((abs(x)+abs(y))*a)
continue
else:
ans = min(x, y)*b
x, y =x-min(x,y), y- min(x,y)
ans += max(x,y)*a
print(ans)
```
| 91,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
t=int(input())
for _ in range(t):
x,y=map(int,input().split())
a,b=map(int,input().split())
diff=abs(x-y)
m=min(x,y)
print(diff*a+min(m*a*2,m*b))
```
Yes
| 91,074 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
a=int(input())
for i in range(0,a):
x,y=map(int,input().split())
a,b=map(int,input().split())
minn=min(x,y)
if(b>2*a):
print(x*a+y*a)
else:
print(a*(abs(y-x))+b*minn)
```
Yes
| 91,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
t=int(input())
for i in range(t):
num1=list(map(int,input().split()))
num2=list(map(int,input().split()))
x=num1[0]
y=num1[1]
a=num2[0]
b=num2[1]
z=(x+y)*a
c=min(x,y)
x-=c
y-=c
m=(c*b)+(x*a)+(y*a)
print(min(m,z))
```
Yes
| 91,076 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
for _ in range(int(input())):
x,y=list(map(int,input().split()))
a,b=list(map(int,input().split()))
ans1=(x+y)*a
if(x<y):
temp=x
t=y-x
else:
temp=y
t=x-y
ans2=temp*b+t*a
if(ans1<ans2):
print(ans1)
else:
print(ans2)
```
Yes
| 91,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
t=int(input())
for i in range(t):
x,y=map(int,input().split())
a,b=map(int,input().split())
print(min(x,y)*b+(max(x,y)-min(x,y))*a)
```
No
| 91,078 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
t = int(input())
for _ in range(t):
x,y = map(int,input().split(' '))
a,b = map(int,input().split(' '))
cost = 0
cost += min(x,y) * b
cost += abs(x-y) * a
print(cost)
```
No
| 91,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
for _ in range(int(input())):
x,y = map(int,input().split())
a,b = map(int,input().split())
k=x*a+x*b
if(x>y):
l=x-y
print(min((b*l+a),k))
elif(x<y):
l=y-x
print(min((a*l+b),k))
else:
print(min(y*b,2*x*a))
```
No
| 91,080 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two integers x and y. You can perform two types of operations:
1. Pay a dollars and increase or decrease any of these integers by 1. For example, if x = 0 and y = 7 there are four possible outcomes after this operation:
* x = 0, y = 6;
* x = 0, y = 8;
* x = -1, y = 7;
* x = 1, y = 7.
2. Pay b dollars and increase or decrease both integers by 1. For example, if x = 0 and y = 7 there are two possible outcomes after this operation:
* x = -1, y = 6;
* x = 1, y = 8.
Your goal is to make both given integers equal zero simultaneously, i.e. x = y = 0. There are no other requirements. In particular, it is possible to move from x=1, y=0 to x=y=0.
Calculate the minimum amount of dollars you have to spend on it.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of testcases.
The first line of each test case contains two integers x and y (0 ≤ x, y ≤ 10^9).
The second line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^9).
Output
For each test case print one integer — the minimum amount of dollars you have to spend.
Example
Input
2
1 3
391 555
0 0
9 4
Output
1337
0
Note
In the first test case you can perform the following sequence of operations: first, second, first. This way you spend 391 + 555 + 391 = 1337 dollars.
In the second test case both integers are equal to zero initially, so you dont' have to spend money.
Submitted Solution:
```
for i in range(int(input())):
a=list(map(int,input().strip().split()))[:2]
b=list(map(int,input().strip().split()))[:2]
if a[0]>a[1]:
count=(a[0]-a[1])*b[0]+(a[1])*b[1]
print(count)
elif a[0]<a[1]:
count=(a[1]-a[0])*b[0]+a[0]*b[1]
print(count)
else:
count=a[0]*b[1]
print(count)
```
No
| 91,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
from collections import defaultdict
for _ in range(int(input())):
n, x = map(int, input().split())
a = list(map(int, input().split()))
b = [0]*n
b[0] = a[0]%x
if a[0] % x == 0:
f = True
else:
f = False
for i in range(1, n):
b[i] = (b[i-1] + a[i] % x) % x
if a[i] %x != 0:
f = False
# print(b)
if b[-1] != 0:
print(n)
elif f:
print(-1)
else:
f1 = 0
f2 = 0
for i in range(n):
if b[i] != 0:
f1 = i
break
for i in range(n-1,-1,-1):
if b[i] != 0:
f2 = i
break
print(max(n - f1 - 1, f2 + 1))
```
| 91,082 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
t = int(input())
for i in range(t):
n = int(input())
arr = [int(x) for x in input().split()]
vis = {k:False for k in range(n)}
S = set()
res = 0
for i in range(n):
if(arr[i] not in S):
S.add(arr[i])
res += 1
print(res)
```
| 91,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
l = set(map(int, input().split()))
print(len(l))
```
| 91,084 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
for _ in range(int(input())):
n,x=map(int,input().split())
a=list(map(int,input().split()))
l=[]
s=0
for i in range(n):
l.append(a[i]%x)
s=(s+(a[i]%x))%x
if s!=0:
print(n)
elif l.count(0)==n:
print(-1)
else:
p=0
for i in range(n):
if l[i]!=0:
p=i+1
break
q=0
j=0
for i in range(n-1,-1,-1):
j+=1
if l[i]!=0:
q=j
break
print(max(n-p,n-q))
```
| 91,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
n=int(input())
for i in range(n):
a,b=map(int,input().split())
s=list(map(int,input().split()))
sum=0
t=-1
for i in range(a):
sum=sum+s[i]
s[i]=sum
if not sum%b==0:
print(a)
else:
for i in range(a):
if (sum-s[i])%b:
t=max(t,a-i-1)
break
for i in range(a-1,-1,-1):
if (s[i]%b):
t=max(t,i+1)
break
print(t)
```
| 91,086 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
from sys import stdin, stdout
"""
n = stdin.readline()
arr = [int(x) for x in stdin.readline().split()]
stdout.write(str(summation))
"""
t = int(stdin.readline())
for test in range(t) :
n = int(stdin.readline())
arr = [int(x) for x in stdin.readline().split()]
dicti = {}
for i in arr :
if i not in dicti :
dicti[i] = True
stdout.write(str(len(dicti))+"\n")
```
| 91,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
q = int(input())
for i in range(1,q+1):
n = int(input())
a = [int(c) for c in input().split()]
print(len(set(a)))
```
| 91,088 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Tags: brute force, data structures, number theory, two pointers
Correct Solution:
```
t = int(input())
while t>0:
n = int(input())
temp = list(map(int,input().split()))
ans = 0
a = []
#print(type(a))
for i in range(0,n,1):
a.insert(i,temp[i])
a.sort()
for i in range(1,n,1):
if a[i-1] < a[i]:
ans+=1
print(ans+1)
t -= 1
```
| 91,089 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
from sys import stdin
"""
import functools
mod = 998244353
def ncr(n,r):
r=min(r,n-r)
a=functools.reduce(lambda x,y: (x*y), range(n,n-r,-1),1)
b=functools.reduce(lambda x,y: (x*y), range(1,r+1),1)
return a//b
"""
def func(arr, l, r, x):
while(l<r):
mid = (l+r)//2
if arr[mid]+x > 0:
r = mid
else:
l = mid+1
return l
for _ in range(int(stdin.readline())):
x = int(stdin.readline())
#n, h, l, r = int(stdin.readline())
arr = list(map(int,stdin.readline().split()))
print(len(set(arr)))
```
Yes
| 91,090 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
for _ in range(int(input())):
n,x = map(int,input().split())
a = list(map(int,input().split()))
s = sum(a)
if s%x!=0:
print(n)
continue
elif x==1:
print(-1)
continue
i1,i2=-1,-1
for i in range(n):
if a[i]%x!=0:
i1=i
break
for i in range(n-1,-1,-1):
if a[i]%x!=0:
i2=i
break
if i1==-1:
print(-1)
else:
print(n-min(i1+1,n-i2))
```
Yes
| 91,091 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
def ii(): return int(input())
def mi(): return map(int, input().split())
def ai(): return list(map(int, input().split()))
for _ in range(ii()):
n = ii()
lst = list(set(ai()))
print(len(lst))
```
Yes
| 91,092 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
import sys
t=int(sys.stdin.readline())
for i in range (t):
a=int(sys.stdin.readline())
aa = (map(int, sys.stdin.readline().strip().split()))
b = set()
for num in aa:
b.add(num)
print(len(b))
```
Yes
| 91,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
n = int(input())
l = []
l = list(map(int,input().split()))
k = set(l)
m = len(k)
print(m)
```
No
| 91,094 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
for _ in range(int(input())):
n,d=map(int,input().split())
arr=list(map(int,input().split()))
ans=-1
for i in range(n):
if arr[i]%d:ans=max(ans,i+1)
for j in range(n-1,-1,-1):
if arr[j]%d:ans=max(ans,n-j-1)
print(ans)
```
No
| 91,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
import sys
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_array(): return list(map(int, sys.stdin.readline().strip().split()))
def input(): return sys.stdin.readline().strip()
# def maxlength(arr,x,i,j,n,s):
# if(i>=j):
# return
for _ in range(int(input())):
n,x = get_ints()
arr = get_array()
s=0
for i in arr:
s+=i
# ans = maxlenght(arr,x,0,n,n,s)
i=0
j=n-1
f=0
check=0
length = n
if(s%x!=0):
print(n)
else:
while(i<=j):
if(i==j):
if(arr[i]%x!=0):
n=1
else:
n=-1
break
if((s-arr[i])%x!=0):
s=s-arr[i]
n=n-1
break
elif((s-arr[j])%x!=0):
s=s-arr[j]
n=n-1
break
else:
if(arr[i]==0):
check=check+1
if(arr[j]==0):
check=check+1
i=i+1
j=j-1
n=n-2
s=s-arr[i]-arr[j]
if(length == n):
print(-1)
elif(n==0):
print(-1)
else:
print(n+check)
```
No
| 91,096 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ehab loves number theory, but for some reason he hates the number x. Given an array a, find the length of its longest subarray such that the sum of its elements isn't divisible by x, or determine that such subarray doesn't exist.
An array a is a subarray of an array b if a can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains an integer t (1 ≤ t ≤ 5) — the number of test cases you need to solve. The description of the test cases follows.
The first line of each test case contains 2 integers n and x (1 ≤ n ≤ 10^5, 1 ≤ x ≤ 10^4) — the number of elements in the array a and the number that Ehab hates.
The second line contains n space-separated integers a_1, a_2, …, a_{n} (0 ≤ a_i ≤ 10^4) — the elements of the array a.
Output
For each testcase, print the length of the longest subarray whose sum isn't divisible by x. If there's no such subarray, print -1.
Example
Input
3
3 3
1 2 3
3 4
1 2 3
2 2
0 6
Output
2
3
-1
Note
In the first test case, the subarray [2,3] has sum of elements 5, which isn't divisible by 3.
In the second test case, the sum of elements of the whole array is 6, which isn't divisible by 4.
In the third test case, all subarrays have an even sum, so the answer is -1.
Submitted Solution:
```
x=int(input())
for i in range(0,x):
s=int(input())
lst=[int(x) for x in input().split()]
n=min(lst)
s=0
for i in range(len(lst)):
if lst[i]>n:
s+=1
print(s+1)
```
No
| 91,097 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
for _ in range(int(input())):
N, K, L = map(int, input().split())
A = list(map(int, input().split()))
A.append(-K)
if max(A) > L:
print("No")
continue
ok = 1
k = -K-1
for a in A:
k += 1
if k < 0:
if a + K <= L:
k = -K-1
if a - k > L:
k = -(L - a)
else:
if a + k > L:
ok = 0
break
elif a + K <= L:
k = -K-1
if ok:
print("Yes")
else:
print("No")
if __name__ == '__main__':
main()
```
| 91,098 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is on constraints. In this version constraints are higher. You can make hacks only if all versions of the problem are solved.
Koa the Koala is at the beach!
The beach consists (from left to right) of a shore, n+1 meters of sea and an island at n+1 meters from the shore.
She measured the depth of the sea at 1, 2, ..., n meters from the shore and saved them in array d. d_i denotes the depth of the sea at i meters from the shore for 1 ≤ i ≤ n.
Like any beach this one has tide, the intensity of the tide is measured by parameter k and affects all depths from the beginning at time t=0 in the following way:
* For a total of k seconds, each second, tide increases all depths by 1.
* Then, for a total of k seconds, each second, tide decreases all depths by 1.
* This process repeats again and again (ie. depths increase for k seconds then decrease for k seconds and so on ...).
Formally, let's define 0-indexed array p = [0, 1, 2, …, k - 2, k - 1, k, k - 1, k - 2, …, 2, 1] of length 2k. At time t (0 ≤ t) depth at i meters from the shore equals d_i + p[t mod 2k] (t mod 2k denotes the remainder of the division of t by 2k). Note that the changes occur instantaneously after each second, see the notes for better understanding.
At time t=0 Koa is standing at the shore and wants to get to the island. Suppose that at some time t (0 ≤ t) she is at x (0 ≤ x ≤ n) meters from the shore:
* In one second Koa can swim 1 meter further from the shore (x changes to x+1) or not swim at all (x stays the same), in both cases t changes to t+1.
* As Koa is a bad swimmer, the depth of the sea at the point where she is can't exceed l at integer points of time (or she will drown). More formally, if Koa is at x (1 ≤ x ≤ n) meters from the shore at the moment t (for some integer t≥ 0), the depth of the sea at this point — d_x + p[t mod 2k] — can't exceed l. In other words, d_x + p[t mod 2k] ≤ l must hold always.
* Once Koa reaches the island at n+1 meters from the shore, she stops and can rest.
Note that while Koa swims tide doesn't have effect on her (ie. she can't drown while swimming). Note that Koa can choose to stay on the shore for as long as she needs and neither the shore or the island are affected by the tide (they are solid ground and she won't drown there).
Koa wants to know whether she can go from the shore to the island. Help her!
Input
The first line of the input contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.
The first line of each test case contains three integers n, k and l (1 ≤ n ≤ 3 ⋅ 10^5; 1 ≤ k ≤ 10^9; 1 ≤ l ≤ 10^9) — the number of meters of sea Koa measured and parameters k and l.
The second line of each test case contains n integers d_1, d_2, …, d_n (0 ≤ d_i ≤ 10^9) — the depths of each meter of sea Koa measured.
It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case:
Print Yes if Koa can get from the shore to the island, and No otherwise.
You may print each letter in any case (upper or lower).
Example
Input
7
2 1 1
1 0
5 2 3
1 2 3 2 2
4 3 4
0 2 4 3
2 3 5
3 0
7 2 3
3 0 2 1 3 0 1
7 1 4
4 4 3 0 2 4 2
5 2 3
1 2 3 2 2
Output
Yes
No
Yes
Yes
Yes
No
No
Note
In the following s denotes the shore, i denotes the island, x denotes distance from Koa to the shore, the underline denotes the position of Koa, and values in the array below denote current depths, affected by tide, at 1, 2, ..., n meters from the shore.
In test case 1 we have n = 2, k = 1, l = 1, p = [ 0, 1 ].
Koa wants to go from shore (at x = 0) to the island (at x = 3). Let's describe a possible solution:
* Initially at t = 0 the beach looks like this: [\underline{s}, 1, 0, i].
* At t = 0 if Koa would decide to swim to x = 1, beach would look like: [s, \underline{2}, 1, i] at t = 1, since 2 > 1 she would drown. So Koa waits 1 second instead and beach looks like [\underline{s}, 2, 1, i] at t = 1.
* At t = 1 Koa swims to x = 1, beach looks like [s, \underline{1}, 0, i] at t = 2. Koa doesn't drown because 1 ≤ 1.
* At t = 2 Koa swims to x = 2, beach looks like [s, 2, \underline{1}, i] at t = 3. Koa doesn't drown because 1 ≤ 1.
* At t = 3 Koa swims to x = 3, beach looks like [s, 1, 0, \underline{i}] at t = 4.
* At t = 4 Koa is at x = 3 and she made it!
We can show that in test case 2 Koa can't get to the island.
Tags: constructive algorithms, dp, greedy, implementation
Correct Solution:
```
def main():
t = int(input())
for i in range(t):
n, k, l = map(int, input().split())
d = tuple(map(int, input().split()))
output = "Yes" if _can_get_to_finish(d, k, l) else "No"
print(output)
def _can_get_to_finish(depths, max_tide_height, max_height_can_swim):
tide_heights_allowed = [max_height_can_swim - depth for depth in depths]
if any(allowed_height < 0 for allowed_height in tide_heights_allowed):
return False
sea_size = len(tide_heights_allowed)
def is_checkpoint(pos):
return tide_heights_allowed[pos] >= max_tide_height
get_tide_height = lambda time: _get_tide_height_by_time(time, max_tide_height)
next_pos = 0
curr_checkpoint = -1
while next_pos < sea_size:
# Time when we will be at the next position.
# Because at the checkpoint we can spend as much time as we need,
# initially it is time of the first moment when can visit next position.
# It is negative because position becomes accessible before new period starts.
time_of_next_pos = -tide_heights_allowed[next_pos]
while next_pos < sea_size and not is_checkpoint(next_pos):
max_allowed_height = tide_heights_allowed[next_pos]
if get_tide_height(time_of_next_pos) > max_allowed_height:
how_much_must_wait = get_tide_height(time_of_next_pos) - max_allowed_height
if time_of_next_pos + how_much_must_wait > 0:
return False
time_of_next_pos += how_much_must_wait
next_pos += 1
time_of_next_pos += 1
while next_pos < sea_size and is_checkpoint(next_pos):
curr_checkpoint = next_pos
next_pos += 1
return True
def _get_tide_height_by_time(time, max_tide_height):
time %= 2*max_tide_height
if time < max_tide_height:
result = time
else:
time_since_max = time - max_tide_height
result = max_tide_height - time_since_max
return result
if __name__ == '__main__':
main()
```
| 91,099 |
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