message stringlengths 2 20.2k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 757 108k | cluster float64 4 4 | __index_level_0__ int64 1.51k 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
You built an apartment. The apartment has a water tank with a capacity of L in order to store water for the residents. The tank works as a buffer between the water company and the residents.
It is required to keep the tank "not empty" at least during use of water. A pump is used to provide water into the tank. From the viewpoint of avoiding water shortage, a more powerful pump is better, of course. But such powerful pumps are expensive. That’s the life.
You have a daily schedule table of water usage. It does not differ over days. The table is composed of some schedules. Each schedule is indicated by the starting time of usage, the ending time and the used volume per unit of time during the given time span.
All right, you can find the minimum required speed of providing water for days from the schedule table. You are to write a program to compute it.
You can assume the following conditions.
* A day consists of 86,400 units of time.
* No schedule starts before the time 0 (the beginning of the day).
* No schedule ends after the time 86,400 (the end of the day).
* No two schedules overlap.
* Water is not consumed without schedules.
* The tank is full of water when the tank starts its work.
Input
The input is a sequence of datasets. Each dataset corresponds to a schedule table in the following format:
N L
s1 t1 u1
...
sN tN uN
The first line of a dataset contains two integers N and L (1 ≤ N ≤ 86400, 1 ≤ L ≤ 106), which represents the number of schedule in the table and the capacity of the tank, respectively.
The following N lines describe the N schedules. The (i + 1)-th line of the dataset corresponds to the i-th schedule, which consists of three integers si, ti and ui . The first two integers si and ti indicate the starting time and the ending time of the schedule. The last integer ui (1 ≤ ui ≤ 106 ) indicates the consumed volume per unit of time during the schedule. It is guaranteed that 0 ≤ s1 < t1 ≤ s2 < t2 ≤ ... ≤ sn < tn ≤ 86400.
The input is terminated by a line with two zeros. This line should not be processed.
Output
For each case, print the minimum required amount of water per unit of time provided by the pump in a line. The amount may be printed with an arbitrary number of digits after the decimal point, but should not contain an absolute error greater than 10-6.
Example
Input
1 100
0 86400 1
1 100
43200 86400 1
0 0
Output
1.000000
0.997685 | instruction | 0 | 79,077 | 4 | 158,154 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
def solve():
N, L = map(int, readline().split())
if N == 0:
return False
ma = 0
P = [list(map(int, readline().split())) for i in range(N)]
ma = max(u for s, t, u in P)
K = 86400
EPS = 1e-8
def check(x, M = 2):
rest = L
R = [0]*M
for i in range(M):
prv = 0
for s, t, u in P:
rest = min(rest + (s - prv) * x, L)
rest = min(rest + (t - s) * (x - u), L)
prv = t
if rest < 0:
return 0
rest = min(rest + (K - prv) * x, L)
R[i] = rest
return R[-2] - EPS < R[-1]
left = 0; right = ma
while left + EPS < right:
mid = (left + right) / 2
if check(mid):
right = mid
else:
left = mid
write("%.016f\n" % right)
return True
while solve():
...
``` | output | 1 | 79,077 | 4 | 158,155 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Mr. Takatsuki, who is planning to participate in the Aizu training camp, has a poor house and does not have much money. Therefore, he is trying to save money by using the Seishun 18 Ticket. With only one 18 ticket, you can ride a local train all day long, and you can enter and exit the ticket gates freely (detailed usage rules are omitted).
The attraction of the 18 Ticket is that you can use your spare time to see local souvenirs at the station where you change trains. She wants to visit various stations during her trip because it is a great opportunity. However, since I don't want to miss the next train, I decided to look around the station only when the time between the time I arrived at the transfer station and the time I left the transfer station was T minutes or more.
You will be given a transfer plan using Mr. Takatsuki's 18 ticket, so output the name and time of the station you can look around. Please note that the station that departs first and the station that arrives last are not candidates for exploration.
Constraints
* 1 <= N <= 10
* 1 <= T <= 180
* st_timei, ar_timei
* Represented by "HH: MM", HH is 00 or more and 23 or less, and MM is 00 or more and 59 or less. HH is hours and MM is minutes.
* 00:00 <= st_time1 <ar_time1 <st_time2 <ar_time2 <... <st_timeN <ar_timeN <= 23:59
* st_namei, ar_namei
* A character string represented by uppercase and lowercase letters.
* 1 <= string length <= 50
* The names of the i-th arrival station ar_namei and the i + 1th departure station st_namei + 1 match.
* The names of st_name1, ar_nameN, and the transfer station are different character strings.
Input
Each data set is input in the following format.
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
N is an integer representing the number of times the train is boarded, and T is an integer representing the permissible time (minutes) to see the transfer station. Subsequently, train departure and arrival pairs are given over N lines. The input of each line means that the train that Mr. Takatsuki rides departs from the station of st_namei at the time of st_timei, and that the train that Mr. Takatsuki rides arrives at the station of ar_namei at the time of ar_timei.
Output
Output in the following format for each data set.
M
stay_name1 stay_time1
stay_name2 stay_time2
...
stay_nameM stay_timeM
M (0 <= M <= N -1) is an integer that represents the number of stations you can look around. Then, over M lines, a list of stations that can be visited is output in ascending order of time. Each line means that you can walk around the station of stay_namei for stay_timei minutes.
Examples
Input
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
Output
2
Kanazawa 55
Niitsu 24
Input
8 24
05:30 Kyoto 06:37 Maibara
06:50 Maibara 07:36 Tsuruga
07:42 Tsuruga 10:03 Kanazawa
10:58 Kanazawa 12:07 Toyama
12:15 Toyama 14:12 Naoetsu
14:29 Naoetsu 15:57 Nagaoka
16:11 Nagaoka 17:14 Niitsu
17:38 Niitsu 20:06 AizuWakamatsu
Output
2
Kanazawa 55
Niitsu 24
Input
1 180
10:44 Koriyama 11:52 AizuWakamatsu
Output
0 | instruction | 0 | 79,082 | 4 | 158,164 |
"Correct Solution:
```
def f(t,_t):
h,m=[int(_t[i:i+2])-int(t[i:i+2])for i in[0,3]]
return 60*h+m
n,t=map(int,input().split())
prv=input().split()
a=[]
for _ in [0]*(n-1):
now=input().split()
d=f(prv[2],now[0])
if d>=t:
a.append(now[1]+' '+str(d))
prv=now
print(len(a))
for x in a:print(x)
``` | output | 1 | 79,082 | 4 | 158,165 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Mr. Takatsuki, who is planning to participate in the Aizu training camp, has a poor house and does not have much money. Therefore, he is trying to save money by using the Seishun 18 Ticket. With only one 18 ticket, you can ride a local train all day long, and you can enter and exit the ticket gates freely (detailed usage rules are omitted).
The attraction of the 18 Ticket is that you can use your spare time to see local souvenirs at the station where you change trains. She wants to visit various stations during her trip because it is a great opportunity. However, since I don't want to miss the next train, I decided to look around the station only when the time between the time I arrived at the transfer station and the time I left the transfer station was T minutes or more.
You will be given a transfer plan using Mr. Takatsuki's 18 ticket, so output the name and time of the station you can look around. Please note that the station that departs first and the station that arrives last are not candidates for exploration.
Constraints
* 1 <= N <= 10
* 1 <= T <= 180
* st_timei, ar_timei
* Represented by "HH: MM", HH is 00 or more and 23 or less, and MM is 00 or more and 59 or less. HH is hours and MM is minutes.
* 00:00 <= st_time1 <ar_time1 <st_time2 <ar_time2 <... <st_timeN <ar_timeN <= 23:59
* st_namei, ar_namei
* A character string represented by uppercase and lowercase letters.
* 1 <= string length <= 50
* The names of the i-th arrival station ar_namei and the i + 1th departure station st_namei + 1 match.
* The names of st_name1, ar_nameN, and the transfer station are different character strings.
Input
Each data set is input in the following format.
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
N is an integer representing the number of times the train is boarded, and T is an integer representing the permissible time (minutes) to see the transfer station. Subsequently, train departure and arrival pairs are given over N lines. The input of each line means that the train that Mr. Takatsuki rides departs from the station of st_namei at the time of st_timei, and that the train that Mr. Takatsuki rides arrives at the station of ar_namei at the time of ar_timei.
Output
Output in the following format for each data set.
M
stay_name1 stay_time1
stay_name2 stay_time2
...
stay_nameM stay_timeM
M (0 <= M <= N -1) is an integer that represents the number of stations you can look around. Then, over M lines, a list of stations that can be visited is output in ascending order of time. Each line means that you can walk around the station of stay_namei for stay_timei minutes.
Examples
Input
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
Output
2
Kanazawa 55
Niitsu 24
Input
8 24
05:30 Kyoto 06:37 Maibara
06:50 Maibara 07:36 Tsuruga
07:42 Tsuruga 10:03 Kanazawa
10:58 Kanazawa 12:07 Toyama
12:15 Toyama 14:12 Naoetsu
14:29 Naoetsu 15:57 Nagaoka
16:11 Nagaoka 17:14 Niitsu
17:38 Niitsu 20:06 AizuWakamatsu
Output
2
Kanazawa 55
Niitsu 24
Input
1 180
10:44 Koriyama 11:52 AizuWakamatsu
Output
0 | instruction | 0 | 79,083 | 4 | 158,166 |
"Correct Solution:
```
n, t = map(int, input().split())
def f(i, j):
return 60*j[0]+j[1]-60*i[0]-i[1]
res = []
tmp = input().split()
name, bef = tmp[3], list(map(int, tmp[2].split(":")))
for i in range(n-1):
tmp = input().split()
aft = list(map(int, tmp[0].split(":")))
tim = f(bef, aft)
if tim >= t : res.append([name, str(tim)])
name, bef = tmp[3], list(map(int, tmp[2].split(":")))
print(len(res))
if len(res) != 0 : print("\n".join([" ".join(x) for x in res]))
``` | output | 1 | 79,083 | 4 | 158,167 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Mr. Takatsuki, who is planning to participate in the Aizu training camp, has a poor house and does not have much money. Therefore, he is trying to save money by using the Seishun 18 Ticket. With only one 18 ticket, you can ride a local train all day long, and you can enter and exit the ticket gates freely (detailed usage rules are omitted).
The attraction of the 18 Ticket is that you can use your spare time to see local souvenirs at the station where you change trains. She wants to visit various stations during her trip because it is a great opportunity. However, since I don't want to miss the next train, I decided to look around the station only when the time between the time I arrived at the transfer station and the time I left the transfer station was T minutes or more.
You will be given a transfer plan using Mr. Takatsuki's 18 ticket, so output the name and time of the station you can look around. Please note that the station that departs first and the station that arrives last are not candidates for exploration.
Constraints
* 1 <= N <= 10
* 1 <= T <= 180
* st_timei, ar_timei
* Represented by "HH: MM", HH is 00 or more and 23 or less, and MM is 00 or more and 59 or less. HH is hours and MM is minutes.
* 00:00 <= st_time1 <ar_time1 <st_time2 <ar_time2 <... <st_timeN <ar_timeN <= 23:59
* st_namei, ar_namei
* A character string represented by uppercase and lowercase letters.
* 1 <= string length <= 50
* The names of the i-th arrival station ar_namei and the i + 1th departure station st_namei + 1 match.
* The names of st_name1, ar_nameN, and the transfer station are different character strings.
Input
Each data set is input in the following format.
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
N is an integer representing the number of times the train is boarded, and T is an integer representing the permissible time (minutes) to see the transfer station. Subsequently, train departure and arrival pairs are given over N lines. The input of each line means that the train that Mr. Takatsuki rides departs from the station of st_namei at the time of st_timei, and that the train that Mr. Takatsuki rides arrives at the station of ar_namei at the time of ar_timei.
Output
Output in the following format for each data set.
M
stay_name1 stay_time1
stay_name2 stay_time2
...
stay_nameM stay_timeM
M (0 <= M <= N -1) is an integer that represents the number of stations you can look around. Then, over M lines, a list of stations that can be visited is output in ascending order of time. Each line means that you can walk around the station of stay_namei for stay_timei minutes.
Examples
Input
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
Output
2
Kanazawa 55
Niitsu 24
Input
8 24
05:30 Kyoto 06:37 Maibara
06:50 Maibara 07:36 Tsuruga
07:42 Tsuruga 10:03 Kanazawa
10:58 Kanazawa 12:07 Toyama
12:15 Toyama 14:12 Naoetsu
14:29 Naoetsu 15:57 Nagaoka
16:11 Nagaoka 17:14 Niitsu
17:38 Niitsu 20:06 AizuWakamatsu
Output
2
Kanazawa 55
Niitsu 24
Input
1 180
10:44 Koriyama 11:52 AizuWakamatsu
Output
0 | instruction | 0 | 79,084 | 4 | 158,168 |
"Correct Solution:
```
N,T=map(int,input().split())
a,b,s,name1=input().split()
def func(s):
a=s[:2]
b=s[3:]
return int(a)*60+int(b)
s=func(s)
l=[]
for i in range(N-1):
t,name1,s2,name2=input().split()
t=func(t)
s2=func(s2)
if t-s>=T:
l.append([name1,t-s])
s=s2
print(len(l))
for i,j in l:
print(i,j)
``` | output | 1 | 79,084 | 4 | 158,169 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Mr. Takatsuki, who is planning to participate in the Aizu training camp, has a poor house and does not have much money. Therefore, he is trying to save money by using the Seishun 18 Ticket. With only one 18 ticket, you can ride a local train all day long, and you can enter and exit the ticket gates freely (detailed usage rules are omitted).
The attraction of the 18 Ticket is that you can use your spare time to see local souvenirs at the station where you change trains. She wants to visit various stations during her trip because it is a great opportunity. However, since I don't want to miss the next train, I decided to look around the station only when the time between the time I arrived at the transfer station and the time I left the transfer station was T minutes or more.
You will be given a transfer plan using Mr. Takatsuki's 18 ticket, so output the name and time of the station you can look around. Please note that the station that departs first and the station that arrives last are not candidates for exploration.
Constraints
* 1 <= N <= 10
* 1 <= T <= 180
* st_timei, ar_timei
* Represented by "HH: MM", HH is 00 or more and 23 or less, and MM is 00 or more and 59 or less. HH is hours and MM is minutes.
* 00:00 <= st_time1 <ar_time1 <st_time2 <ar_time2 <... <st_timeN <ar_timeN <= 23:59
* st_namei, ar_namei
* A character string represented by uppercase and lowercase letters.
* 1 <= string length <= 50
* The names of the i-th arrival station ar_namei and the i + 1th departure station st_namei + 1 match.
* The names of st_name1, ar_nameN, and the transfer station are different character strings.
Input
Each data set is input in the following format.
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
N is an integer representing the number of times the train is boarded, and T is an integer representing the permissible time (minutes) to see the transfer station. Subsequently, train departure and arrival pairs are given over N lines. The input of each line means that the train that Mr. Takatsuki rides departs from the station of st_namei at the time of st_timei, and that the train that Mr. Takatsuki rides arrives at the station of ar_namei at the time of ar_timei.
Output
Output in the following format for each data set.
M
stay_name1 stay_time1
stay_name2 stay_time2
...
stay_nameM stay_timeM
M (0 <= M <= N -1) is an integer that represents the number of stations you can look around. Then, over M lines, a list of stations that can be visited is output in ascending order of time. Each line means that you can walk around the station of stay_namei for stay_timei minutes.
Examples
Input
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
Output
2
Kanazawa 55
Niitsu 24
Input
8 24
05:30 Kyoto 06:37 Maibara
06:50 Maibara 07:36 Tsuruga
07:42 Tsuruga 10:03 Kanazawa
10:58 Kanazawa 12:07 Toyama
12:15 Toyama 14:12 Naoetsu
14:29 Naoetsu 15:57 Nagaoka
16:11 Nagaoka 17:14 Niitsu
17:38 Niitsu 20:06 AizuWakamatsu
Output
2
Kanazawa 55
Niitsu 24
Input
1 180
10:44 Koriyama 11:52 AizuWakamatsu
Output
0 | instruction | 0 | 79,085 | 4 | 158,170 |
"Correct Solution:
```
def f(t,_t):
h,m=[int(_t[i:i+2])-int(t[i:i+2])for i in[0,3]]
return 60*h+m
n,t=map(int,input().split())
prv=input().split()
a=[]
for _ in [0]*(n-1):
now=input().split()
d=f(prv[2],now[0])
if d>=t:
a+=[[now[1],d]]
prv=now
print(len(a))
for x in a:print(*x)
``` | output | 1 | 79,085 | 4 | 158,171 |
Provide a correct Python 3 solution for this coding contest problem.
Problem Statement
Mr. Takatsuki, who is planning to participate in the Aizu training camp, has a poor house and does not have much money. Therefore, he is trying to save money by using the Seishun 18 Ticket. With only one 18 ticket, you can ride a local train all day long, and you can enter and exit the ticket gates freely (detailed usage rules are omitted).
The attraction of the 18 Ticket is that you can use your spare time to see local souvenirs at the station where you change trains. She wants to visit various stations during her trip because it is a great opportunity. However, since I don't want to miss the next train, I decided to look around the station only when the time between the time I arrived at the transfer station and the time I left the transfer station was T minutes or more.
You will be given a transfer plan using Mr. Takatsuki's 18 ticket, so output the name and time of the station you can look around. Please note that the station that departs first and the station that arrives last are not candidates for exploration.
Constraints
* 1 <= N <= 10
* 1 <= T <= 180
* st_timei, ar_timei
* Represented by "HH: MM", HH is 00 or more and 23 or less, and MM is 00 or more and 59 or less. HH is hours and MM is minutes.
* 00:00 <= st_time1 <ar_time1 <st_time2 <ar_time2 <... <st_timeN <ar_timeN <= 23:59
* st_namei, ar_namei
* A character string represented by uppercase and lowercase letters.
* 1 <= string length <= 50
* The names of the i-th arrival station ar_namei and the i + 1th departure station st_namei + 1 match.
* The names of st_name1, ar_nameN, and the transfer station are different character strings.
Input
Each data set is input in the following format.
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
N is an integer representing the number of times the train is boarded, and T is an integer representing the permissible time (minutes) to see the transfer station. Subsequently, train departure and arrival pairs are given over N lines. The input of each line means that the train that Mr. Takatsuki rides departs from the station of st_namei at the time of st_timei, and that the train that Mr. Takatsuki rides arrives at the station of ar_namei at the time of ar_timei.
Output
Output in the following format for each data set.
M
stay_name1 stay_time1
stay_name2 stay_time2
...
stay_nameM stay_timeM
M (0 <= M <= N -1) is an integer that represents the number of stations you can look around. Then, over M lines, a list of stations that can be visited is output in ascending order of time. Each line means that you can walk around the station of stay_namei for stay_timei minutes.
Examples
Input
N T
st_time1 st_name1 ar_time1 ar_name1
st_time2 st_name2 ar_time2 ar_name2
...
st_timeN st_nameN ar_timeN ar_nameN
Output
2
Kanazawa 55
Niitsu 24
Input
8 24
05:30 Kyoto 06:37 Maibara
06:50 Maibara 07:36 Tsuruga
07:42 Tsuruga 10:03 Kanazawa
10:58 Kanazawa 12:07 Toyama
12:15 Toyama 14:12 Naoetsu
14:29 Naoetsu 15:57 Nagaoka
16:11 Nagaoka 17:14 Niitsu
17:38 Niitsu 20:06 AizuWakamatsu
Output
2
Kanazawa 55
Niitsu 24
Input
1 180
10:44 Koriyama 11:52 AizuWakamatsu
Output
0 | instruction | 0 | 79,086 | 4 | 158,172 |
"Correct Solution:
```
def main():
n,t = map(int, input().split())
star = [list(map(str,input().split())) for i in range(n)]
ans = []
for i in range(n-1):
a = list(map(int,star[i][2].split(":")))
b = list(map(int,star[i+1][0].split(":")))
ai = a[0]*60+a[1]
bi = b[0]*60+b[1]
x = bi-ai
if x >= t:
ans.append((star[i][3],x))
print(len(ans))
for a,b in ans:
print(a,b)
main()
``` | output | 1 | 79,086 | 4 | 158,173 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,797 | 4 | 159,594 |
"Correct Solution:
```
N, T = map(int, input().split())
t = list(map(int, input().split()))
ans = T*N
for i in range(N-1):
ans -= max(T-(t[i+1]-t[i]), 0)
print(ans)
``` | output | 1 | 79,797 | 4 | 159,595 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,798 | 4 | 159,596 |
"Correct Solution:
```
n, t = map(int, input().split())
t_list = list(map(int, input().split()))
ans = t
for i in range(1, n):
ans += min(t, abs(t_list[i] - t_list[i-1]))
print(ans)
``` | output | 1 | 79,798 | 4 | 159,597 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,799 | 4 | 159,598 |
"Correct Solution:
```
n,T=map(int,input().split())
t=list(map(int,input().split()))
t.append(t[-1]+T+1)
ans=0
for i in range(n):
ans += min(T,t[i+1]-t[i])
print(ans)
``` | output | 1 | 79,799 | 4 | 159,599 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,800 | 4 | 159,600 |
"Correct Solution:
```
n,t = map(int,input().split())
tt = list(map(int,input().split()))
tt = [0]+tt
a = 0
for i in range(n):
if(tt[i+1]-tt[i]>=t):
a +=t
else:
a += tt[i+1]-tt[i]
print(a+t)
``` | output | 1 | 79,800 | 4 | 159,601 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,801 | 4 | 159,602 |
"Correct Solution:
```
n, s= map(int, input().split())
t = list(map(int, input().split()))
ans=0
for i in range(1,n):
ans += min(t[i]-t[i-1] , s)
print(ans+s)
``` | output | 1 | 79,801 | 4 | 159,603 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,802 | 4 | 159,604 |
"Correct Solution:
```
N, T = map(int, input().split())
ts = list(map(int, input().split()))
ans = 0
for i in range(N - 1):
ans += min(T, ts[i + 1] - ts[i])
# 最後にN秒流れて終わり
ans += T
print(ans)
``` | output | 1 | 79,802 | 4 | 159,605 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,803 | 4 | 159,606 |
"Correct Solution:
```
N,T = map(int,input().split())
t =list(map(int,input().split()))
X = T
for i in range(N-1):
X += min(T,t[i+1] -t[i])
print(X)
``` | output | 1 | 79,803 | 4 | 159,607 |
Provide a correct Python 3 solution for this coding contest problem.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67 | instruction | 0 | 79,804 | 4 | 159,608 |
"Correct Solution:
```
N,T = map(int, input().split())
A = list(map(int, input().split()))
cnt = 0
for i in range(N-1):
# print(i, A[i+1]-A[i], 10)
cnt += min(T, A[i+1]-A[i])
print(cnt+T)
``` | output | 1 | 79,804 | 4 | 159,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In a public bath, there is a shower which emits water for T seconds when the switch is pushed.
If the switch is pushed when the shower is already emitting water, from that moment it will be emitting water for T seconds. Note that it does not mean that the shower emits water for T additional seconds.
N people will push the switch while passing by the shower. The i-th person will push the switch t_i seconds after the first person pushes it.
How long will the shower emit water in total?
Constraints
* 1 ≤ N ≤ 200,000
* 1 ≤ T ≤ 10^9
* 0 = t_1 < t_2 < t_3 < , ..., < t_{N-1} < t_N ≤ 10^9
* T and each t_i are integers.
Input
Input is given from Standard Input in the following format:
N T
t_1 t_2 ... t_N
Output
Assume that the shower will emit water for a total of X seconds. Print X.
Examples
Input
2 4
0 3
Output
7
Input
2 4
0 5
Output
8
Input
4 1000000000
0 1000 1000000 1000000000
Output
2000000000
Input
1 1
0
Output
1
Input
9 10
0 3 5 7 100 110 200 300 311
Output
67
Submitted Solution:
```
n, time = map(int, input().split())
T = list(map(int, input().split()))
cnt = time
tmp = T[0]
for t in T[1:]:
if t <= time + time:
cnt += t - tmp
else:
cnt += time
tmp = t
print(cnt)
``` | instruction | 0 | 79,812 | 4 | 159,624 |
No | output | 1 | 79,812 | 4 | 159,625 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5 | instruction | 0 | 79,813 | 4 | 159,626 |
"Correct Solution:
```
n = int(input())
x,num = [],0
for i in range(n):
a,b = (int(j) for j in input().split())
x.append((num+a)%86400)
num = (num+a+b)%86400
x,ans = sorted(x),0
from bisect import bisect
for i in range(n):
ans = max(ans,bisect(x,x[i]+10800)-i)
if x[i]<=10800: x.append(x[i]+86400)
print(ans)
``` | output | 1 | 79,813 | 4 | 159,627 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5 | instruction | 0 | 79,814 | 4 | 159,628 |
"Correct Solution:
```
N = int(input())
T = 86400
W = 10801
mem = [0 for i in range(T)]
time = 0
for i in range(N):
a,b = map(int,input().split())
time += a
mem[time % T] += 1
time += b
mem += mem
cums = [0]
for t in mem:
cums.append(cums[-1] + t)
ans = 0
for t in range(T):
ans = max(ans, cums[t+W] - cums[t])
print(ans)
``` | output | 1 | 79,814 | 4 | 159,629 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5 | instruction | 0 | 79,815 | 4 | 159,630 |
"Correct Solution:
```
mod = 3600 * 24
wakeup = [0] * (mod + 11000)
n = int(input())
now = 0
for i in range(n):
a, b = [int(item) for item in input().split()]
now += a
now %= mod
wakeup[now] += 1
wakeup[now + 10801] -= 1
now += b
for i in range(1, mod + 11000):
wakeup[i] += wakeup[i-1]
if i != i%mod:
wakeup[i%mod] += wakeup[i]
print(max(wakeup))
``` | output | 1 | 79,815 | 4 | 159,631 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5 | instruction | 0 | 79,816 | 4 | 159,632 |
"Correct Solution:
```
import sys
import bisect
input = sys.stdin.readline
def cumsum(inlist):
s = 0
outlist = []
for i in inlist:
s += i
outlist.append(s)
return outlist
oneday = 86400
time = [ 0 for i in range(oneday) ]
n = int(input())
t = 0
for i in range(n):
a, b = [ int(v) for v in input().split() ]
t = (t+a) % oneday
time[t] += 1
t = (t+b) % oneday
time = time * 2
timesum = cumsum(time)
ans_list = []
for i in range(oneday*2-20000):
ans_list.append(timesum[i+10801] - timesum[i])
print(max(ans_list))
``` | output | 1 | 79,816 | 4 | 159,633 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5
Submitted Solution:
```
n = int(input())
x,num = [],0
for i in range(n):
a,b = (int(j) for j in input().split())
x.append((num+a)%86400)
num = (num+a+b)%86400
x,ans = sorted(x),0
from bisect import bisect
for i in range(n): ans = max(ans,bisect(x,x[i]+10800)-i)
print(ans)
``` | instruction | 0 | 79,817 | 4 | 159,634 |
No | output | 1 | 79,817 | 4 | 159,635 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi recorded his daily life for the last few days as a integer sequence of length 2N, as follows:
* a_1, b_1, a_2, b_2, ... , a_N, b_N
This means that, starting from a certain time T, he was:
* sleeping for exactly a_1 seconds
* then awake for exactly b_1 seconds
* then sleeping for exactly a_2 seconds
* :
* then sleeping for exactly a_N seconds
* then awake for exactly b_N seconds
In this record, he waked up N times.
Takahashi is wondering how many times he waked up early during the recorded period.
Here, he is said to wake up early if he wakes up between 4:00 AM and 7:00 AM, inclusive.
If he wakes up more than once during this period, each of these awakenings is counted as waking up early.
Unfortunately, he forgot the time T.
Find the maximum possible number of times he waked up early during the recorded period.
For your information, a day consists of 86400 seconds, and the length of the period between 4:00 AM and 7:00 AM is 10800 seconds.
Constraints
* 1 \leq N \leq 10^5
* 1 \leq a_i, b_i \leq 10^5
* a_i and b_i are integers.
Input
The input is given from Standard Input in the following format:
N
a_1 b_1
a_2 b_2
:
a_N b_N
Output
Print the maximum possible number of times he waked up early during the recorded period.
Examples
Input
3
28800 57600
28800 57600
57600 28800
Output
2
Input
10
28800 57600
4800 9600
6000 1200
600 600
300 600
5400 600
6000 5760
6760 2880
6000 12000
9000 600
Output
5
Submitted Solution:
```
import sys
import bisect
input = sys.stdin.readline
def cumsum(inlist):
s = 0
outlist = []
for i in inlist:
s += i
outlist.append(s)
return outlist
oneday = 86400
time = [ 0 for i in range(oneday) ]
n = int(input())
t = 0
for i in range(n):
a, b = [ int(v) for v in input().split() ]
t = (t+a) % oneday
time[t] += 1
t = (t+b) % oneday
time = time * 2
timesum = cumsum(time)
ans_list = []
for i in range(oneday*2-20000):
ans_list.append(timesum[i+10800] - timesum[i])
print(max(ans_list))
``` | instruction | 0 | 79,818 | 4 | 159,636 |
No | output | 1 | 79,818 | 4 | 159,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,815 | 4 | 161,630 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline
from heapq import heappush,heappop,heapify
n,m,d=map(int,input().split())
a=list(map(int,input().split()))
ans=[0 for i in range(n)]
arr=[]
he_one=[]
he_two=[]
for i in range(n):
heappush(he_one,[a[i],i])
heapify(he_one)
heapify(he_two)
day=1
while he_one:
value,indx=heappop(he_one)
if he_two and value -d > he_two[0][0]:
v,dd=heappop(he_two)
ans[indx] =dd
heappush(he_two,[value,dd])
else:
heappush(he_two,[value,day])
ans[indx] =day
day+=1
print(max(ans))
print(*ans)
``` | output | 1 | 80,815 | 4 | 161,631 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,816 | 4 | 161,632 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
from copy import deepcopy
import itertools
from bisect import bisect_left
from bisect import bisect_right
import math
from collections import deque
from collections import Counter
def read():
return int(input())
def readmap():
return map(int, input().split())
def readlist():
return list(map(int, input().split()))
n, m, d = readmap()
A = readlist()
Aind = dict([(A[i], i) for i in range(n)])
A.sort()
q = deque()
a = A[0]
ans = [0] * n
ans[Aind[a]] = 1
maxday = 1
q.append((a, 1))
for i in range(1, n):
if A[i] > q[0][0] + d:
ans[Aind[A[i]]] = q[0][1]
q.append((A[i], q[0][1]))
q.popleft()
else:
maxday += 1
ans[Aind[A[i]]] = maxday
q.append((A[i], maxday))
print(maxday)
print(" ".join(list(map(str, ans))))
``` | output | 1 | 80,816 | 4 | 161,633 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,817 | 4 | 161,634 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
n,m,D=map(int,input().split())
lst=[*map(int,input().split())]
d={}
for i,x in enumerate(lst):d[x]=i
lst.sort()
res,j,result=[0]*n,0,0
for i,x in enumerate(lst):
if x-lst[j]>D:
res[d[x]]=res[d[lst[j]]]
j+=1
else:
result+=1
res[d[x]]=result
print(result)
print(*res)
``` | output | 1 | 80,817 | 4 | 161,635 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,818 | 4 | 161,636 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
n,m,D=map(int,input().split())
lst=[*map(int,input().split())]
d={x:i for i,x in enumerate(lst)}
lst.sort()
res,j,result=[0]*n,0,0
for i,x in enumerate(lst):
if x-lst[j]>D:
res[d[x]]=res[d[lst[j]]]
j+=1
else:
result+=1
res[d[x]]=result
print(result)
print(*res)
``` | output | 1 | 80,818 | 4 | 161,637 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,819 | 4 | 161,638 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
n, m, k = list(map(int, input().split()))
l = list(map(int, input().split()))
for i in range(n):
l[i] = [l[i], i]
l.sort()
uk1 = 1
uk2 = n
def pos(n):
cnt = [-10000000001] * len(l)
global k
for i in range(len(l)):
if l[i][0] - cnt[i % n] <= k:
return False
else:
cnt[i % n] = l[i][0]
return True
while uk2 - uk1 > 1:
if pos((uk2 + uk1) // 2):
uk2 = (uk1 + uk2) // 2
else:
uk1 = (uk1 + uk2) // 2
ans = [0] * n
if not pos(uk1):
for i in range(n):
ans[l[i][1]] = i % uk2 + 1
print(uk2)
for i in range(n):
print(ans[i], end =' ')
else:
for i in range(n):
ans[l[i][1]] = i % uk1 + 1
print(uk1)
for i in range(n):
print(ans[i], end =' ')
``` | output | 1 | 80,819 | 4 | 161,639 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,820 | 4 | 161,640 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
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")
#-------------------game starts now-----------------------------------------------------
n , m , d = map(int , input().split())
a = list(map(int , input().split()))
a = sorted([(a[i] , i) for i in range(n)])
right = 0
left = 0
day = 0
ret = [0] * n
while right < n :
if a[right][0] - a[left][0] <= d :
day += 1
ret[a[right][1]] = day
else :
ret[a[right][1]] = ret[a[left][1]]
left += 1
right += 1
print(day)
print(*ret)
``` | output | 1 | 80,820 | 4 | 161,641 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,821 | 4 | 161,642 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
from collections import defaultdict, Counter
from math import sqrt, log10, log2, log, gcd, floor, factorial
from bisect import bisect_left, bisect_right
from itertools import combinations, combinations_with_replacement
import sys, io, os
input = sys.stdin.readline
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# print=sys.stdout.write
# sys.setrecursionlimit(10000)
mod = 10 ** 9 + 7;inf = float('inf')
def get_list(): return [int(i) for i in input().split()]
yn = lambda a: print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
t=1
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
for i in range(t):
n,m,d=[int(i) for i in input().split()]
l=[int(i) for i in input().split()]
dict=defaultdict(int)
s=SortedList(l)
currentday=0
while len(s):
currentday+=1
last=s.pop(0)
dict[last]=currentday
while 1:
indexa=s.bisect_left(last+d+1)
if indexa<len(s):
last=s.pop(indexa)
dict[last]=currentday
else:
break;
print(currentday)
print(*[dict[i] for i in l])
``` | output | 1 | 80,821 | 4 | 161,643 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day. | instruction | 0 | 80,822 | 4 | 161,644 |
Tags: binary search, data structures, greedy, two pointers
Correct Solution:
```
from sys import stdin
from collections import defaultdict
import heapq
input=stdin.readline
n,m,d=map(int,input().split())
a=list(map(int,input().split()))
idx=defaultdict(int)
for i in range(n):
idx[a[i]]=i
group=[]
a.sort()
num=1
ans=[0]*n
for aa in a:
if len(group)==0:
heapq.heappush(group,(aa,num))
ans[idx[aa]]=num
continue
last_num,group_num=heapq.heappop(group)
if aa-last_num>d:
ans[idx[aa]]=group_num
heapq.heappush(group,(aa,group_num))
else:
num+=1
ans[idx[aa]]=num
heapq.heappush(group,(last_num,group_num))
heapq.heappush(group,(aa,num))
print(num)
print(*ans)
``` | output | 1 | 80,822 | 4 | 161,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
R = lambda: map(int, input().split())
n, m, d = R()
a = sorted((x, i) for i, x in enumerate(R()))
res = [-1] * len(a)
res[0] = 0
cnt = 0
l = 0
for r in range(1, n):
if a[r][0] - d <= a[l][0]:
cnt += 1
res[r] = cnt
else:
res[r] = res[l]
l += 1
print(max(res) + 1)
for t in sorted(zip(a, res), key=lambda x: x[0][1]):
print(t[1] + 1, end=' ')
``` | instruction | 0 | 80,823 | 4 | 161,646 |
Yes | output | 1 | 80,823 | 4 | 161,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
from collections import defaultdict, Counter
from math import sqrt, log10, log2, log, gcd, floor, factorial
from bisect import bisect_left, bisect_right
from itertools import combinations, combinations_with_replacement
import sys, io, os
input = sys.stdin.readline
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# print=sys.stdout.write
# sys.setrecursionlimit(10000)
mod = 10 ** 9 + 7;inf = float('inf')
def get_list(): return [int(i) for i in input().split()]
yn = lambda a: print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
t=1
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
for i in range(t):
n,m,d=get_list()
l=get_list()
dict=defaultdict(int)
s=SortedList(l)
currentday=0
while len(s):
currentday+=1
last=s.pop(0)
dict[last]=currentday
while 1:
indexa=s.bisect_left(last+d+1)
if indexa<len(s):
last=s.pop(indexa)
dict[last]=currentday
else:
break;
print(currentday)
print(*[dict[i] for i in l])
``` | instruction | 0 | 80,824 | 4 | 161,648 |
Yes | output | 1 | 80,824 | 4 | 161,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
from collections import defaultdict, Counter
from math import sqrt, log10, log2, log, gcd, floor, factorial
from bisect import bisect_left, bisect_right
from itertools import combinations, combinations_with_replacement
import sys, io, os
input = sys.stdin.readline
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
# print=sys.stdout.write
# sys.setrecursionlimit(10000)
mod = 10 ** 9 + 7;inf = float('inf')
def get_list(): return [int(i) for i in input().split()]
yn = lambda a: print("YES" if a else "NO")
ceil = lambda a, b: (a + b - 1) // b
t=1
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index + 1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
for i in range(t):
n,m,d=[int(i) for i in input().split()]
l=[int(i) for i in input().split()]
dict=defaultdict(int)
s=SortedList(l)
currentday=0
while len(s):
currentday+=1
last=s.pop(0)
dict[last]=currentday
while 1:
indexa=s.bisect_left(last+d+1)
if indexa<len(s):
last=s.pop(indexa)
dict[last]=currentday
else:
break;
print(currentday)
for i in l:
print(dict[i],end=" ")
``` | instruction | 0 | 80,825 | 4 | 161,650 |
Yes | output | 1 | 80,825 | 4 | 161,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
n,m,d=map(int,input().split())
b=sorted([[int(s),i] for i,s in enumerate(input().split())])
ans=[0]*n
day=0
i=j=0
while i<n:
if b[i][0]-d<=b[j][0]:day+=1;ans[b[i][1]]=day
else:ans[b[i][1]]=ans[b[j][1]];j+=1
i+=1
print(day)
print(*ans)
``` | instruction | 0 | 80,826 | 4 | 161,652 |
Yes | output | 1 | 80,826 | 4 | 161,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
'''
import sys
import math
import bisect
n=int(input())
arr=list(map(int,input().split()))
n,m=map(int,input().split())
t=sys.stdin.buffer.readline()
sys.stdin=open("input.txt")
sys.stdout=open("output.txt", 'w')
sys.stdout.write("Yes" + '\n')
s="abcdefghijklmnopqrstuvwxyz"
mod=1000000007
mod=998244353
vow=['a','e','i','o','u']
t=[[0 for i in range(n)]for j in range(n)]
pow(4,7)
def gcd(x,y):
while y:
x,y=y%x,x
return y'''
n,l,g=map(int,input().split())
ar=list(map(int,input().split()))
d={}
arr=sorted(ar)
day=1
visited=[False]*(l+1)
for i in range(n):
if visited[arr[i]]==False:
visited[arr[i]] = True
var = arr[i]+g+1
d[arr[i]] = day
while var<=l:
if var in arr:
if visited[var] == False:
d[var] = day
visited[var]=True
#print(d)
var+=g+1
day += 1
print(max(d.values()))
for i in ar:
print(d[i],end=" ")
``` | instruction | 0 | 80,827 | 4 | 161,654 |
No | output | 1 | 80,827 | 4 | 161,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
n,m,de=map(int,input().split())
de+=1
l=list(map(int,input().split()))
a=[]
cop=[]+l
l.sort()
d=dict()
u=[0]*n
for i in range(n):
t=l[i]-de
print(t)
if len(a)==0 or t<a[0]:
a.append(l[i])
d.update({l[i]:len(a)})
else:
r=a[0]
a.pop(0)
a.append(l[i])
d.update({l[i]:d[r]})
#print(a)
print(len(a))
for j in cop:
print(d[j],end=" ")
``` | instruction | 0 | 80,828 | 4 | 161,656 |
No | output | 1 | 80,828 | 4 | 161,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
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 value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
n,m,d=value()
a=array()
timing=sorted([(a[i],i) for i in range(n)])
#print(timing)
day={}
cur=1
coffee=sorted([i for i,j in timing])
#print(coffee)
for i in coffee:
ind=bisect_right(coffee,i-d-1)
if(ind==0):
day[i]=cur
cur+=1
else:
day[i]=day[coffee[ind-1]]
for i in a:
print(day[i],end=" ")
``` | instruction | 0 | 80,829 | 4 | 161,658 |
No | output | 1 | 80,829 | 4 | 161,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute).
However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.
For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent.
Input
The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.
The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.
Output
In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes.
In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.
Examples
Input
4 5 3
3 5 1 2
Output
3
3 1 1 2
Input
10 10 1
10 5 7 4 6 3 2 1 9 8
Output
2
2 1 1 2 2 1 2 1 1 2
Note
In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).
In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.
Submitted Solution:
```
n, m, d = map(int, input().split())
a = list(map(int, input().split()))
ans = [0 for i in range(n)]
q = sorted(a)
day = 1
Ans = ""
while len(q):
last = q[0]
ans[a.index(q[0])] = day
del q[0]
i = 0
while i < len(q):
print(q, i, last)
if q[i] - last - 1 >= d:
ans[a.index(q[i])] = day
last = q[i]
del q[i]
else:
i += 1
day += 1
Ans += str(day - 1) + "\n"
for i in range(n):
Ans += str(ans[i]) + " "
print(Ans)
``` | instruction | 0 | 80,830 | 4 | 161,660 |
No | output | 1 | 80,830 | 4 | 161,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,081 | 4 | 164,162 |
Tags: implementation
Correct Solution:
```
from itertools import product
from math import ceil, gcd, sqrt
import string
from decimal import Decimal
def binary_table(string_with_all_characters, length_to_make):
return [''.join(x) for x in product(string_with_all_characters, repeat=length_to_make)]
def all_possible_substrings(string):
return [int(string[i: j]) for i in range(len(string)) for j in range(i + 1, len(string) + 1)]
def number_of_substrings(length):
return int(length * (length + 1) / 2)
num_of_z_lines, num_of_x_lines, l, r = map(int, input().split())
z_lines = []
x_lines = []
for i in range(num_of_z_lines):
x, y = map(int, input().split())
z_lines.extend([x for x in range(x, y + 1)])
for i in range(num_of_x_lines):
x, y = map(int, input().split())
x_lines.extend([x for x in range(x, y + 1)])
num = 0
for i in range(l, r + 1):
for j in x_lines:
if j + i in z_lines:
num += 1
break
print(num)
``` | output | 1 | 82,081 | 4 | 164,163 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,082 | 4 | 164,164 |
Tags: implementation
Correct Solution:
```
x1=[]
x2=[]
p, q, l, r = map(int, input().split(' '))
for i in range(p):
a, b = map(int, input().split(' '))
x1.append([a, b])
for i in range(q):
a, b = map(int, input().split(' '))
x2.append([a, b])
ok2 = []
for a in x1:
for b in x2:
lo = a[0]-b[1]
hi = a[1]-b[0]
if hi >= r:
hi = r
if lo <= l:
lo = l
ok2.append([lo, hi])
ok = [i for i in ok2 if i[1]>=i[0]]
ok.sort()
total = 0
if len(ok)==0:
print(0)
else:
intv = ok[0]
for i in range(1, len(ok)):
curr = ok[i]
if curr[0] > intv[1]:
if intv[1] == intv[0]:
total += 1
else:
total += intv[1]-intv[0]+1
intv = curr
else:
minim = [min(curr[0], intv[0]), max(curr[1], intv[1])]
intv = minim
if intv[1] == intv[0]:
total += 1
else:
total += intv[1]-intv[0]+1
print(total)
``` | output | 1 | 82,082 | 4 | 164,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,083 | 4 | 164,166 |
Tags: implementation
Correct Solution:
```
input1 = list(map(lambda x: int(x), input().split()))
p = input1[0]
q = input1[1]
l = input1[2]
r = input1[3]
zTimes = []
xTimes = []
for i in range(p):
z = tuple(map(lambda x: int(x), input().split()))
zTimes.append(z)
for i in range(q):
x = tuple(map(lambda x: int(x), input().split()))
xTimes.append(x)
suitTimes = []
for z in zTimes:
for x in xTimes:
# if x[1] < z[0]:
# pass
# elif x[0] < z[0] and x[1] >= z[0] and x[1] <= z[1]:
# pass
# elif x[0] >= z[0] and x[1] <= z[1]:
# pass
# elif x[0] >= z[0] and x[0] <= z[1] and x[1] > z[1]:
# pass
# elif x[0] > z[0]:
# pass
left = z[0] - x[1]
right = z[1] - x[0]
suitTimes.append((left, right))
resultSet = set([])
for s in suitTimes:
for i in range(s[0], s[1]+1):
if i >= l and i <= r:
resultSet.add(i)
print(len(resultSet))
``` | output | 1 | 82,083 | 4 | 164,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,084 | 4 | 164,168 |
Tags: implementation
Correct Solution:
```
# Fast IO Region
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# Get out of main function
def main():
pass
# decimal to binary
def binary(n):
return (bin(n).replace("0b", ""))
# binary to decimal
def decimal(s):
return (int(s, 2))
# power of a number base 2
def pow2(n):
p = 0
while n > 1:
n //= 2
p += 1
return (p)
# if number is prime in √n time
def isPrime(n):
if (n == 1):
return (False)
else:
root = int(n ** 0.5)
root += 1
for i in range(2, root):
if (n % i == 0):
return (False)
return (True)
# list to string ,no spaces
def lts(l):
s = ''.join(map(str, l))
return s
# String to list
def stl(s):
# for each character in string to list with no spaces -->
l = list(s)
# for space in string -->
# l=list(s.split(" "))
return l
# Returns list of numbers with a particular sum
def sq(a, target, arr=[]):
s = sum(arr)
if (s == target):
return arr
if (s >= target):
return
for i in range(len(a)):
n = a[i]
remaining = a[i + 1:]
ans = sq(remaining, target, arr + [n])
if (ans):
return ans
# Sieve for prime numbers in a range
def SieveOfEratosthenes(n):
cnt = 0
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
for p in range(2, n + 1):
if prime[p]:
cnt += 1
# print(p)
return (cnt)
# for positive integerse only
def nCr(n, r):
f = math.factorial
return f(n) // f(r) // f(n - r)
# 1000000007
mod = int(1e9) + 7
def ssinp(): return input()
# s=input()
def iinp(): return int(input())
# n=int(input())
def nninp(): return map(int, input().split())
# a,b,c=map(int,input().split())
def llinp(): return list(map(int, input().split()))
# a=list(map(int,input().split()))
def p(xyz): print(xyz)
def p2(a, b): print(a, b)
import math
# import random
# sys.setrecursionlimit(300000)
# from fractions import Fraction
from collections import OrderedDict
# from collections import deque
######################## mat=[[0 for i in range(n)] for j in range(m)] ########################
######################## list.sort(key=lambda x:x[1]) for sorting a list according to second element in sublist ########################
######################## Speed: STRING < LIST < SET,DICTIONARY ########################
######################## from collections import deque ########################
######################## ASCII of A-Z= 65-90 ########################
######################## ASCII of a-z= 97-122 ########################
######################## d1.setdefault(key, []).append(value) ########################
#for __ in range(iinp()):
x,y,r,l=nninp()
time1=[]
time2=[]
for i in range(x):
a,b=nninp()
for j in range(a,b+1):
time1.append(j)
for i in range(y):
c,d=nninp()
for j in range(c,d+1):
time2.append(j)
ans=0
for t in range(r,l+1):
temp=list(map(lambda x:x+t,time2))
for c in temp:
if(c in time1):
ans+=1
break
p(ans)
``` | output | 1 | 82,084 | 4 | 164,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,085 | 4 | 164,170 |
Tags: implementation
Correct Solution:
```
p,q,l,r=map(int, input().strip().split())
hz=[0 for a in range(0, 10000)]
hx=[0 for a in range(0, 10000)]
for i in range(0, p):
a, b = map(int, input().strip().split())
for i in range(a, b+1):
hz[i]=1
for i in range(0, q):
a, b = map(int, input().strip().split())
for i in range(a, b+1):
hx[i]=1
res=0
for i in range(l, r+1):
hx=[0]*i+hx
for j in range(0, len(hz)):
if hz[j]==1 and hx[j]== 1:
res+=1
break
hx=hx[i:]
print(res)
``` | output | 1 | 82,085 | 4 | 164,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,086 | 4 | 164,172 |
Tags: implementation
Correct Solution:
```
p, q, l, r = map(int, input().split())
abs = []
for i in range(p):
abs.append(list(map(int, input().split())))
cds = []
for i in range(q):
cds.append(list(map(int, input().split())))
shifts = [False for i in range(1001)]
for ab in abs:
for i in range(ab[0], ab[1]+1):
for cd in cds:
for j in range(cd[0], min(cd[1], i)+1):
shift = i-j
if shift >= 0:
shifts[shift] = True
sol = 0
for i in range(l, r+1):
if shifts[i]:
sol += 1
print(sol)
``` | output | 1 | 82,086 | 4 | 164,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,087 | 4 | 164,174 |
Tags: implementation
Correct Solution:
```
# link: https://codeforces.com/contest/469/problem/B
if __name__ == "__main__":
p,q,l,r = map(int,input().split())
Z = []
X = []
while p:
Z.append(list(map(int,input().split())))
p -= 1
while q:
X.append(list(map(int,input().split())))
q -= 1
Z.sort()
X.sort()
# l and r are known, so do not exceed r
overlap = 0
while l<=r:
for value in Z:
# values[0] and value[1]
flag = False
for values in X:
start = values[0] + l
end = values[1] + l
if start>value[1]:
break
if not (value[0] > end):
overlap += 1
flag = True
break
if flag:
break
l += 1
print(overlap)
"""
5 2 27 452
148 154
421 427
462 470
777 786
969 978
245 247
313 322"""
``` | output | 1 | 82,087 | 4 | 164,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20 | instruction | 0 | 82,088 | 4 | 164,176 |
Tags: implementation
Correct Solution:
```
p , q , l , r =map(int,input().split())
x=[ [int(j) for j in input().split()] for i in range(p)]
z=[ [int(j) for j in input().split()] for i in range(q)]
moment=[0]*1001;ans=0
for i in x:
for j in z:
if i[1]-j[0] <= r or i[0]-j[1] <= r :
for e in range(max(l , i[0]-j[1]) , min(i[1]-j[0] , r)+1):
if moment[e]==0:
ans+=1
moment[e]=1
print(ans)
``` | output | 1 | 82,088 | 4 | 164,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20
Submitted Solution:
```
p,q,l,r = map(int,input().split())
hz = [0 for a in range(10000)]
hx = [0 for b in range(10000)]
for i in range(p):
a,b = map(int,input().strip().split())
for j in range(a,b+1):
hz[j] = 1
for _ in range(q):
a,b = map(int,input().strip().split())
for j in range(a,b+1):
hx[j] = 1
res = 0
for i in range(l,r+1):
hx = [0]*i+hx
for j in range(0,len(hz)):
if hz[j] == 1 and hx[j] == 1:
res += 1
break
hx = hx[i:]
print(res)
``` | instruction | 0 | 82,089 | 4 | 164,178 |
Yes | output | 1 | 82,089 | 4 | 164,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20
Submitted Solution:
```
def solution():
p,q,l,r = [int(x) for x in input().split(' ')]
x = []
y = []
ans = 0
for i in range(p):
x.append([int(_x) for _x in input().split(' ')])
for i in range(q):
y.append([int(_x) for _x in input().split(' ')])
for t in range(l,r+1):
current = 0
found = False
for a,b in x:
if found:
break
for c,d in y:
c += t
d += t
if not(a>d or b<c):
current += max(min(b,d)-max(a,c)+1,0)
if current>0:
ans +=1
found = True
break
return ans
print(solution())
``` | instruction | 0 | 82,090 | 4 | 164,180 |
Yes | output | 1 | 82,090 | 4 | 164,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20
Submitted Solution:
```
p , q , l , r = map(int,input().split())
l1 = set()
l2 = set()
count = 0
for i in range(p):
a, b = map(int,input().split())
[l1.add(n) for n in range(a , b+1)]
for i in range(q):
c , d = map(int,input().split())
[l2.add(n) for n in range(c,d+1)]
for i in range(l , r+1):
for x in l2 :
if i + x in l1 :
count +=1
break
print(count)
``` | instruction | 0 | 82,091 | 4 | 164,182 |
Yes | output | 1 | 82,091 | 4 | 164,183 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20
Submitted Solution:
```
p, q, l, r = map(int, input().split())
ptimes = []
qtimes = []
genp = []
genq = []
for inp in range(p):
ptimes.append([int(z) for z in input().split()])
for i in range(ptimes[inp][0], ptimes[inp][1]+1):
genp.append(i)
for inp in range(q):
qtimes.append([int(z) for z in input().split()])
for i in range(qtimes[inp][0], qtimes[inp][1]+1):
genq.append(i)
cnt = 0
#print(genp, genq)
for up in range(l, r+1):
genq1 = genq[::]
for i in range(len(genq)):
genq1[i] += up
lst = genp + genq1
if len(set(lst)) < len(lst):
cnt += 1
#print(lst)
print(cnt)
``` | instruction | 0 | 82,092 | 4 | 164,184 |
Yes | output | 1 | 82,092 | 4 | 164,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X and Little Z are good friends. They always chat online. But both of them have schedules.
Little Z has fixed schedule. He always online at any moment of time between a1 and b1, between a2 and b2, ..., between ap and bp (all borders inclusive). But the schedule of Little X is quite strange, it depends on the time when he gets up. If he gets up at time 0, he will be online at any moment of time between c1 and d1, between c2 and d2, ..., between cq and dq (all borders inclusive). But if he gets up at time t, these segments will be shifted by t. They become [ci + t, di + t] (for all i).
If at a moment of time, both Little X and Little Z are online simultaneosly, they can chat online happily. You know that Little X can get up at an integer moment of time between l and r (both borders inclusive). Also you know that Little X wants to get up at the moment of time, that is suitable for chatting with Little Z (they must have at least one common moment of time in schedules). How many integer moments of time from the segment [l, r] suit for that?
Input
The first line contains four space-separated integers p, q, l, r (1 ≤ p, q ≤ 50; 0 ≤ l ≤ r ≤ 1000).
Each of the next p lines contains two space-separated integers ai, bi (0 ≤ ai < bi ≤ 1000). Each of the next q lines contains two space-separated integers cj, dj (0 ≤ cj < dj ≤ 1000).
It's guaranteed that bi < ai + 1 and dj < cj + 1 for all valid i and j.
Output
Output a single integer — the number of moments of time from the segment [l, r] which suit for online conversation.
Examples
Input
1 1 0 4
2 3
0 1
Output
3
Input
2 3 0 20
15 17
23 26
1 4
7 11
15 17
Output
20
Submitted Solution:
```
def t(i):
c = 0
for j in fq:
for k in fp:
if (j[0]+i >= k[0] and j[0]+i <= k[1]) or (j[1]+i >= k[0] and j[1]+i <= k[1]):
c = 1
return 1
return 0
p, q, l, r = map(int, input().split())
fp = []
fq = []
count = 0
for i in range(p):
fp.append(list(map(int, input().split())))
for i in range(q):
fq.append(list(map(int, input().split())))
for i in range(l, r+1):
count += t(i)
print(count)
``` | instruction | 0 | 82,093 | 4 | 164,186 |
No | output | 1 | 82,093 | 4 | 164,187 |
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