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stringlengths 2
39.6k
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int64 0
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int64 219
108k
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float64 11
11
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int64 438
217k
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|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest n problems were suggested and every problem had a cost β a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of points of a contestant was equal to the product of the costs of all the problems he/she had completed. If a person didn't solve anything, then he/she didn't even appear in final standings and wasn't considered as participant. Vasya understood that to get the maximal number of points it is not always useful to solve all the problems. Unfortunately, he understood it only after the contest was finished. Now he asks you to help him: find out what problems he had to solve to earn the maximal number of points.
Input
The first line contains an integer n (1 β€ n β€ 100) β the number of the suggested problems. The next line contains n space-separated integers ci ( - 100 β€ ci β€ 100) β the cost of the i-th task. The tasks' costs may coinΡide.
Output
Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them.
Examples
Input
5
1 2 -3 3 3
Output
3 1 2 3
Input
13
100 100 100 100 100 100 100 100 100 100 100 100 100
Output
100 100 100 100 100 100 100 100 100 100 100 100 100
Input
4
-2 -2 -2 -2
Output
-2 -2 -2 -2
Submitted Solution:
```
a= int(input())
t = list(map(int,input().split()))
ne =[]
po=[]
for s in t:
if s<0:
ne.append(s)
elif s>0:
po.append(s)
ne.sort()
if len(po)==0 and len(ne)==0:
if 0 in t:
print(0)
else:
if len(ne)%2==0:
print(*(ne+po))
else:
if len(ne)==1:
if len(po)>0:
print(*po)
else:
print(*ne)
else:
print(*(ne[:-1]+po))
```
|
instruction
| 0
| 104,822
| 11
| 209,644
|
No
|
output
| 1
| 104,822
| 11
| 209,645
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya has gotten interested in programming contests in TCMCF+++ rules. On the contest n problems were suggested and every problem had a cost β a certain integral number of points (perhaps, negative or even equal to zero). According to TCMCF+++ rules, only accepted problems can earn points and the overall number of points of a contestant was equal to the product of the costs of all the problems he/she had completed. If a person didn't solve anything, then he/she didn't even appear in final standings and wasn't considered as participant. Vasya understood that to get the maximal number of points it is not always useful to solve all the problems. Unfortunately, he understood it only after the contest was finished. Now he asks you to help him: find out what problems he had to solve to earn the maximal number of points.
Input
The first line contains an integer n (1 β€ n β€ 100) β the number of the suggested problems. The next line contains n space-separated integers ci ( - 100 β€ ci β€ 100) β the cost of the i-th task. The tasks' costs may coinΡide.
Output
Print space-separated the costs of the problems that needed to be solved to get the maximal possible number of points. Do not forget, please, that it was necessary to solve at least one problem. If there are several solutions to that problem, print any of them.
Examples
Input
5
1 2 -3 3 3
Output
3 1 2 3
Input
13
100 100 100 100 100 100 100 100 100 100 100 100 100
Output
100 100 100 100 100 100 100 100 100 100 100 100 100
Input
4
-2 -2 -2 -2
Output
-2 -2 -2 -2
Submitted Solution:
```
import sys
n=int(input())
a = list(map(int,input().strip().split()))[:n]
pos=0
neg=0
zero=0
for i in a :
if i>0:
pos=pos+1
if i<0:
neg=neg+1
if i==0:
zero=zero+1
if neg==n:
for i in a:
print(i," ",end =" ")
sys.exit()
if zero==n:
for i in a:
print(i," ",end =" ")
sys.exit()
if pos==n:
for i in a:
print(i," ",end =" ")
sys.exit()
for i in a:
if i>=0:
print(i," ",end =" ")
```
|
instruction
| 0
| 104,823
| 11
| 209,646
|
No
|
output
| 1
| 104,823
| 11
| 209,647
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,132
| 11
| 210,264
|
"Correct Solution:
```
n=int(input())
A=list(map(int,input().split()))
C=[0]*9
for i in A:
x=i//400
if x<8:
C[x]=1
else:
C[8]+=1
b=sum(C[:8])
print(str(max(b,1))+' '+str(b+C[8]))
```
|
output
| 1
| 105,132
| 11
| 210,265
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,133
| 11
| 210,266
|
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
cnt=0
l=[]
for i in a:
p=i//400
if p<=7:
l.append(p)
else:
cnt+=1
d=len(set(l))
if d!=0:
print(d,d+cnt)
else:
print(1,cnt)
```
|
output
| 1
| 105,133
| 11
| 210,267
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,134
| 11
| 210,268
|
"Correct Solution:
```
n=int(input())
A=list(map(int,input().split()))
A=[a for a in A if a<3200]
free=n-len(A)
A=set(map(lambda x: x//400,A))
print(max(1,len(A)),len(A)+free)
```
|
output
| 1
| 105,134
| 11
| 210,269
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,135
| 11
| 210,270
|
"Correct Solution:
```
import math
n=input()
a= [min(math.floor(int(i)//400),8) for i in input().split()]
a7= len(set([i for i in a if i<=7]))
a8= len([i for i in a if i==8])
print(max(a7,1),a7+a8)
```
|
output
| 1
| 105,135
| 11
| 210,271
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,136
| 11
| 210,272
|
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
o=0
x=[]
for i in range(n):
if a[i]>=3200:
o+=1
else:
x.append(a[i]//400)
p=len(set(x))
q=max(1,p)
print(q,o+p)
```
|
output
| 1
| 105,136
| 11
| 210,273
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,137
| 11
| 210,274
|
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
c=[0]*8
sums=0
sai=0
for i in a:
if i<3200:
c[i//400]=1
else:
sai+=1
sums=sum(c)
print(max(1,sums),sums+sai)
```
|
output
| 1
| 105,137
| 11
| 210,275
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,138
| 11
| 210,276
|
"Correct Solution:
```
n, *a = map(int, open(0).read().split())
l = [i//400 for i in a if i<3200]
k = len(set(l))
print(k or 1, n - len(l) + k)
```
|
output
| 1
| 105,138
| 11
| 210,277
|
Provide a correct Python 3 solution for this coding contest problem.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
|
instruction
| 0
| 105,139
| 11
| 210,278
|
"Correct Solution:
```
v = [0]*9
n = int(input())
a = list(map(int,input().split()))
for i in a:
idx = i//400
if idx >= 8: idx = 8
v[idx] += 1
r = 0
for i in range(8):
if v[i] >= 1: r+= 1
print(max(1,r), r+v[8])
```
|
output
| 1
| 105,139
| 11
| 210,279
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
n=int(input())
a=[int(i)//400 for i in input().split()]
b=[]
c=0
for i in a:
if i<8:
b.append(i)
else:
c+=1
b=set(b)
print(max(len(b),1),len(b)+c)
```
|
instruction
| 0
| 105,140
| 11
| 210,280
|
Yes
|
output
| 1
| 105,140
| 11
| 210,281
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
input();a=[0]*9
for x in input().split():
x=int(x)
if x>3599:x=3200
a[x//400]+=1
b=sum([1 if x else 0 for x in a[:8]])
print(max(1,b),b+a[8])
```
|
instruction
| 0
| 105,141
| 11
| 210,282
|
Yes
|
output
| 1
| 105,141
| 11
| 210,283
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[0]*9
for i in a:
b[min(i//400,8)]+=1
print(max(8-b[:8].count(0),1),8-b[:8].count(0)+b[8])
```
|
instruction
| 0
| 105,142
| 11
| 210,284
|
Yes
|
output
| 1
| 105,142
| 11
| 210,285
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
a=list(map(lambda x:x//400,a))
cnt=0
z=[]
for i in a:
if i>=8:
cnt+=1
else:
z.append(i)
print(max(1,len(set(z))),len(set(z))+cnt)
```
|
instruction
| 0
| 105,143
| 11
| 210,286
|
Yes
|
output
| 1
| 105,143
| 11
| 210,287
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
N = int(input())
a = list(map(int,input().split()))
C = [0 for _ in range(9)]
for i in range(N):
if a[i] // 400 < 8:
C[a[i]//400] = 1
else:
C[8] += 1
ans_min = sum(C[:8])
ans_max = min(8,sum(C))
print(ans_min,ans_max)
```
|
instruction
| 0
| 105,144
| 11
| 210,288
|
No
|
output
| 1
| 105,144
| 11
| 210,289
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
color = [0] * 8
any = 0
for i in a:
if 1 <= i <= 399:
color[0] = 1
elif 400 <= i <= 799:
color[1] = 1
elif 800 <= i <= 1199:
color[2] = 1
elif 1200 <= i <= 1599:
color[3] = 1
elif 1600 <= i <= 1999:
color[4] = 1
elif 2000 <= i <= 2399:
color[5] = 1
elif 2400 <= i <= 2799:
color[6] = 1
elif 2800 <= i <= 3199:
color[7] = 1
else:
any += 1
m = max(sum(color), 1)
M = min(sum(color) + any, 8)
print(m, M)
```
|
instruction
| 0
| 105,145
| 11
| 210,290
|
No
|
output
| 1
| 105,145
| 11
| 210,291
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
ans = [0]*9
leng = 0
for i in range(N):
if 1 <= A[i] <= 399:
ans[0] = 1
elif 400 <= A[i] <= 799:
ans[1] = 1
elif 800 <= A[i] <= 1199:
ans[2] = 1
elif 1200 <= A[i] <= 1599:
ans[3] = 1
elif 1600 <= A[i] <= 1999:
ans[4] = 1
elif 2000 <= A[i] <= 2399:
ans[5] = 1
elif 2400 <= A[i] <= 2799:
ans[6] = 1
elif 2800 <= A[i] <= 3199:
ans[7] = 1
else:
leng += 1
print(sum(ans), sum(ans)+leng)
```
|
instruction
| 0
| 105,146
| 11
| 210,292
|
No
|
output
| 1
| 105,146
| 11
| 210,293
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In AtCoder, a person who has participated in a contest receives a color, which corresponds to the person's rating as follows:
* Rating 1-399 : gray
* Rating 400-799 : brown
* Rating 800-1199 : green
* Rating 1200-1599 : cyan
* Rating 1600-1999 : blue
* Rating 2000-2399 : yellow
* Rating 2400-2799 : orange
* Rating 2800-3199 : red
Other than the above, a person whose rating is 3200 or higher can freely pick his/her color, which can be one of the eight colors above or not.
Currently, there are N users who have participated in a contest in AtCoder, and the i-th user has a rating of a_i.
Find the minimum and maximum possible numbers of different colors of the users.
Constraints
* 1 β€ N β€ 100
* 1 β€ a_i β€ 4800
* a_i is an integer.
Input
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum possible number of different colors of the users, and the maximum possible number of different colors, with a space in between.
Examples
Input
4
2100 2500 2700 2700
Output
2 2
Input
5
1100 1900 2800 3200 3200
Output
3 5
Input
20
800 810 820 830 840 850 860 870 880 890 900 910 920 930 940 950 960 970 980 990
Output
1 1
Submitted Solution:
```
n = int(input())
a = [0]*9
i = input().split()
for x in range(n):
y = int(i[x])
if 1<= y <=399:
a[0] += 1
elif 400<= y <=799:
a[1] += 1
elif 800 <= y <= 1199:
a[2] += 1
elif 1200 <= y <= 1599:
a[3] += 1
elif 1600 <= y <= 1999:
a[4] += 1
elif 2000 <= y <= 2399:
a[5] += 1
elif 2400 <= y<= 2799:
a[6] += 1
elif 2800 <= y <= 3199:
a[7] += 1
else :
a[8] += 1
count = 0
for x in range(8):
if a[x] > 0:
count += 1
max = count + a[8]
if max > 8:
max = 8
if count == 0 and a[8] > 0:
count = 1
print(count,max)
```
|
instruction
| 0
| 105,147
| 11
| 210,294
|
No
|
output
| 1
| 105,147
| 11
| 210,295
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Bubble Cup hypothesis stood unsolved for 130 years. Who ever proves the hypothesis will be regarded as one of the greatest mathematicians of our time! A famous mathematician Jerry Mao managed to reduce the hypothesis to this problem:
Given a number m, how many polynomials P with coefficients in set {\{0,1,2,3,4,5,6,7\}} have: P(2)=m?
Help Jerry Mao solve the long standing problem!
Input
The first line contains a single integer t (1 β€ t β€ 5β
10^5) - number of test cases.
On next line there are t numbers, m_i (1 β€ m_i β€ 10^{18}) - meaning that in case i you should solve for number m_i.
Output
For each test case i, print the answer on separate lines: number of polynomials P as described in statement such that P(2)=m_i, modulo 10^9 + 7.
Example
Input
2
2 4
Output
2
4
Note
In first case, for m=2, polynomials that satisfy the constraint are x and 2.
In second case, for m=4, polynomials that satisfy the constraint are x^2, x + 2, 2x and 4.
Submitted Solution:
```
import sys
input=sys.stdin.readline
def f(n):
return ((n+4)//2)**2//4
input()
for x in list(map(int,input().split())):
print(f(x))
```
|
instruction
| 0
| 105,474
| 11
| 210,948
|
No
|
output
| 1
| 105,474
| 11
| 210,949
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,526
| 11
| 211,052
|
Tags: implementation, math
Correct Solution:
```
n,k=input().split(" ")
n=int(n)
k=int(k)
if 3*n>k:
print((3*n)-k)
else:
print(0)
```
|
output
| 1
| 105,526
| 11
| 211,053
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,527
| 11
| 211,054
|
Tags: implementation, math
Correct Solution:
```
z=input
mod = 10**9 + 7
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
def lcd(xnum1,xnum2):
return (xnum1*xnum2//gcd(xnum1,xnum2))
################################################################################
"""
n=int(z())
for _ in range(int(z())):
x=int(z())
l=list(map(int,z().split()))
n=int(z())
l=sorted(list(map(int,z().split())))[::-1]
a,b=map(int,z().split())
l=set(map(int,z().split()))
led=(6,2,5,5,4,5,6,3,7,6)
vowel={'a':0,'e':0,'i':0,'o':0,'u':0}
color-4=["G", "GB", "YGB", "YGBI", "OYGBI" ,"OYGBIV",'ROYGBIV' ]
"""
###########################---START-CODING---###############################################
for _ in range(1):
n,k=map(int,input().split())
if 3*n<=k:
print(0)
else:print(3*n-k)
```
|
output
| 1
| 105,527
| 11
| 211,055
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,528
| 11
| 211,056
|
Tags: implementation, math
Correct Solution:
```
n,k=map(int,input().split())
k=k-2*n
print(max(n-k,0))
```
|
output
| 1
| 105,528
| 11
| 211,057
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,529
| 11
| 211,058
|
Tags: implementation, math
Correct Solution:
```
n,k=[int(x) for x in input().split()]
if k//n>2:
print(0)
else:
print(n-(k%n))
```
|
output
| 1
| 105,529
| 11
| 211,059
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,530
| 11
| 211,060
|
Tags: implementation, math
Correct Solution:
```
def readln(inp=None): return tuple(map(int, (inp or input()).split()))
n, k = readln()
ans = 100000
for c5 in range(n + 1):
for c4 in range(n - c5 + 1):
for c3 in range(n - c5 - c4 + 1):
c2 = n - c3 - c4 - c5
if c2 >= 0 and 2 * c2 + 3 * c3 + 4 * c4 + 5 * c5 == k and c2 < ans:
ans = c2
print(ans)
```
|
output
| 1
| 105,530
| 11
| 211,061
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,531
| 11
| 211,062
|
Tags: implementation, math
Correct Solution:
```
n, k = list(map(int, input().split()))
a = k - 2*n
if a >= n:
print(0)
else:
print(n - a)
```
|
output
| 1
| 105,531
| 11
| 211,063
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,532
| 11
| 211,064
|
Tags: implementation, math
Correct Solution:
```
n,k = map(int,input().split())
if k/n >= 3:
print(0)
else:
c = 0
while k%n != 0:
k -= 1
c += 1
print(n-c)
```
|
output
| 1
| 105,532
| 11
| 211,065
|
Provide tags and a correct Python 3 solution for this coding contest problem.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
|
instruction
| 0
| 105,533
| 11
| 211,066
|
Tags: implementation, math
Correct Solution:
```
n, k = map(int, input().split())
print(n*3-k if k <= n * 3 else 0)
```
|
output
| 1
| 105,533
| 11
| 211,067
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
nk = input().split(' ')
n = int(nk[0])
k = int(nk[1])
for a in range(0,n+1):
for b in range(n-a+1):
for c in range(n-b+1):
for d in range(n-c+1):
sum = ((a*2) + (b*3) + (c * 4) + (d * 5))
if sum == k:
# d = d / 5
if a + b + c + d == n:
print(a)
exit()
```
|
instruction
| 0
| 105,534
| 11
| 211,068
|
Yes
|
output
| 1
| 105,534
| 11
| 211,069
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
# import random
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a,b):
if a == 0:
return b
return gcd(b%a, a)
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b
# find function
def find(x, link):
while(x != link[x]):
x = link[x]
return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, size, link):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
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
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()))
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################# ---------------- TEMPLATE ENDS HERE ---------------- #################
n, k = int_array();
if k//n > 2:
print(0);
else:
print(n - k%n);
```
|
instruction
| 0
| 105,535
| 11
| 211,070
|
Yes
|
output
| 1
| 105,535
| 11
| 211,071
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
n, k=map(int, input().split())
if k<(3*n):
print(3*n-k)
else:
print(0)
```
|
instruction
| 0
| 105,536
| 11
| 211,072
|
Yes
|
output
| 1
| 105,536
| 11
| 211,073
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
n, k = input().split()
n = int(n)
k = int(k)
l = n*2
m = k - l
if m == 0:
print(n)
elif n >= m > 0:
print(n - m)
elif m > n:
print(0)
```
|
instruction
| 0
| 105,537
| 11
| 211,074
|
Yes
|
output
| 1
| 105,537
| 11
| 211,075
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
n, k = map(int, input().split())
m = 2*n
r = k-m
while r!=0:
n = n-1
r = r-1
print(n)
```
|
instruction
| 0
| 105,538
| 11
| 211,076
|
No
|
output
| 1
| 105,538
| 11
| 211,077
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
nk=input().split()
n=int(nk[0])
k=int(nk[1])
if(k==2*n):
print(n)
elif(k<2*n):
print(k//2)
else:
if(k>=3*n):
print(0)
else:
m=k-2*n
print(m)
```
|
instruction
| 0
| 105,539
| 11
| 211,078
|
No
|
output
| 1
| 105,539
| 11
| 211,079
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
n,k=map(int,input().split())
if k//n>2:
print(0)
elif k/n==2:
print(k//n)
elif k//n<2:
print(n)
else:
i=1
r=0
while(True):
r+=2
if (k-r)%3==0 and (k-r)/3==n-i:
print(i)
break
if r>=k:
print(i)
break
i+=1
```
|
instruction
| 0
| 105,540
| 11
| 211,080
|
No
|
output
| 1
| 105,540
| 11
| 211,081
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day the Codeforces round author sat exams. He had n exams and he needed to get an integer from 2 to 5 for each exam. He will have to re-sit each failed exam, i.e. the exam that gets mark 2.
The author would need to spend too much time and effort to make the sum of his marks strictly more than k. That could have spoilt the Codeforces round. On the other hand, if the sum of his marks is strictly less than k, the author's mum won't be pleased at all.
The Codeforces authors are very smart and they always get the mark they choose themselves. Also, the Codeforces authors just hate re-sitting exams.
Help the author and find the minimum number of exams he will have to re-sit if he passes the exams in the way that makes the sum of marks for all n exams equal exactly k.
Input
The single input line contains space-separated integers n and k (1 β€ n β€ 50, 1 β€ k β€ 250) β the number of exams and the required sum of marks.
It is guaranteed that there exists a way to pass n exams in the way that makes the sum of marks equal exactly k.
Output
Print the single number β the minimum number of exams that the author will get a 2 for, considering that the sum of marks for all exams must equal k.
Examples
Input
4 8
Output
4
Input
4 10
Output
2
Input
1 3
Output
0
Note
In the first sample the author has to get a 2 for all his exams.
In the second sample he should get a 3 for two exams and a 2 for two more.
In the third sample he should get a 3 for one exam.
Submitted Solution:
```
n,k= map(int, input().split())
if n==1:
print(0)
exit()
if n*2==k:
print(n)
exit()
if 3*n>k:
print(3*n-k)
elif k>=3*n:
print(0)
```
|
instruction
| 0
| 105,541
| 11
| 211,082
|
No
|
output
| 1
| 105,541
| 11
| 211,083
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,672
| 11
| 211,344
|
Tags: bitmasks, brute force
Correct Solution:
```
n,l,r,x=map(int,input().split())
num=list(map(int,input().split()))
ans=0
for i in range(2**n):
st=bin(i)[2:]
st='0'*(n-len(st))+st
if st.count('1')>=2:
pt=[]
for i in range(len(st)):
if st[i]=='1':
pt.append(num[i])
if sum(pt)<=r and sum(pt)>=l and max(pt)-min(pt)>=x:
ans+=1
print(ans)
```
|
output
| 1
| 105,672
| 11
| 211,345
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,673
| 11
| 211,346
|
Tags: bitmasks, brute force
Correct Solution:
```
import itertools
n,l,x1,r = map(int,input().split())
li = [int(i) for i in input().split()]
a =[]
for i in range(1,n+1):
x = itertools.combinations(li,i)
a+=x
count = 0
for j in a:
if(sum(list(j))<=x1 and sum(list(j))>=l):
if(max(j)-min(j)>=r):
count+=1
print(count)
```
|
output
| 1
| 105,673
| 11
| 211,347
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,674
| 11
| 211,348
|
Tags: bitmasks, brute force
Correct Solution:
```
a = []
def fn(x,n,l,r,k) :
sum = 0
v = []
for j in range(n) :
if (1<<j)&x :
v.append(int(a[j]))
sum+=int(a[j])
if len(v)<2 :
return 0
else : return bool(v[len(v)-1]-v[0]>=k and sum >= l and sum <= r)
n ,l,r,k = input().split()
b = input().split()
for i in b :
a.append(int(i))
a.sort()
ans = 0
for i in range(1<<int(n)) :
ans = ans + int(fn(i,int(n),int(l),int(r),int(k)))
print(ans)
```
|
output
| 1
| 105,674
| 11
| 211,349
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,675
| 11
| 211,350
|
Tags: bitmasks, brute force
Correct Solution:
```
import itertools
q=0
n, l, r, x=map(int,input().split())
c=list(map(int,input().split()))
for j in range(2,n+1):
for i in itertools.combinations(c,j):
if l<=sum(i)<=r and max(i)-min(i)>=x:
q+=1
print(q)
```
|
output
| 1
| 105,675
| 11
| 211,351
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,676
| 11
| 211,352
|
Tags: bitmasks, brute force
Correct Solution:
```
# Description of the problem can be found at http://codeforces.com/problemset/problem/550/B
def b_f(l_v, c_v, c_i, l, h, l_p, h_p, x):
if c_i == len(l_v):
return 0
else:
n_l = l_v[c_i] if not l_p else min(l_v[c_i], l_p)
n_h = l_v[c_i] if not l_p else max(l_v[c_i], h_p)
return (1 if c_v + l_v[c_i] >= l and
c_v + l_v[c_i] <= h and
n_h - n_l >= x else 0) + (
b_f(l_v, c_v, c_i + 1, l, h, l_p, h_p, x) +
b_f(l_v, c_v + l_v[c_i], c_i + 1, l, h, n_l, n_h, x))
n, l, r, x = map(int, input().split())
l_v = list(map(int, input().split()))
print(b_f(l_v, 0, 0, l, r, None, None, x))
```
|
output
| 1
| 105,676
| 11
| 211,353
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,677
| 11
| 211,354
|
Tags: bitmasks, brute force
Correct Solution:
```
IL = lambda: list(map(int, input().split()))
IS = lambda: input().split()
I = lambda: int(input())
S = lambda: input()
n, l, r, x = IL()
cArr = sorted(IL())
ans = 0
for i in range(1, 2**n):
csub = [cArr[j] for j in range(n) if i & 2**j]
if l <= sum(csub) <= r and csub[-1] - csub[0] >= x:
ans += 1
print(ans)
```
|
output
| 1
| 105,677
| 11
| 211,355
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,678
| 11
| 211,356
|
Tags: bitmasks, brute force
Correct Solution:
```
n, l, r, x = map(int, input().split())
c = list(map(int, input().split()))
ans = 0
for i in range(2 ** n):
mask = bin(i)[2:].zfill(n)
temp = [c[j] for j in range(n) if int(mask[j])]
ans += (l <= sum(temp) <= r) and ((max(temp) - min(temp)) >= x)
print(ans)
```
|
output
| 1
| 105,678
| 11
| 211,357
|
Provide tags and a correct Python 3 solution for this coding contest problem.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
|
instruction
| 0
| 105,679
| 11
| 211,358
|
Tags: bitmasks, brute force
Correct Solution:
```
n, l, r, x = map(int, input().split())
arr = list(map(int, input().split()))
answer = 0
for i in range(2 ** n):
arr1 = []
for j in range(n):
if (i // 2 ** j) % 2 == 1:
arr1.append(arr[j])
if len(arr1) >= 2 and l <= sum(arr1) <= r and max(arr1) - min(arr1) >= x:
answer += 1
print(answer)
```
|
output
| 1
| 105,679
| 11
| 211,359
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
"""
Brandt Smith, Lemuel Gorion and Peter Haddad
codeforces.com
Problem 12455
"""
import sys
def set(mask, pos):
return mask | (1 << pos)
def isOn(mask, pos):
return mask & ( 1 << pos) > 0
n, l, r, x = map(int, input().split(' '))
dif = list(map(int, input().split(' ')))
count, mask = 0, 0
while mask <= 2**n:
summ, bit = [], 0
while bit < n:
if isOn(mask, bit):
summ.append(dif[bit])
bit += 1
if sum(summ) <= r and sum(summ) >= l and max(summ) - min(summ) >= x:
count += 1
mask += 1
print(count)
```
|
instruction
| 0
| 105,680
| 11
| 211,360
|
Yes
|
output
| 1
| 105,680
| 11
| 211,361
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
def s():
[n,l,r,x] = list(map(int,input().split(' ')))
l -= 1
r += 1
a = list(map(int,input().split(' ')))
s = 0
for i in range(1, 1<<n):
ind = 0
k = []
while i:
if i & 1:
k.append(a[ind])
ind += 1
i >>= 1
if max(k)-min(k) >= x and l < sum(k) < r:
s += 1
print(s)
s()
```
|
instruction
| 0
| 105,681
| 11
| 211,362
|
Yes
|
output
| 1
| 105,681
| 11
| 211,363
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
# Program to print all combination
# of size r in an array of size n
# The main function that prints
# all combinations of size r in
# arr[] of size n. This function
# mainly uses combinationUtil()
ans=0
n, l, w, x = map(int,input().split())
def printCombination(arr, n, r):
# A temporary array to
# store all combination
# one by one
data = [0]*r;
# Print all combination
# using temprary array 'data[]'
combinationUtil(arr, data, 0, n - 1, 0,r );
# arr[] ---> Input Array
# data[] ---> Temporary array to
# store current combination
# start & end ---> Staring and Ending
# indexes in arr[]
# index ---> Current index in data[]
# r ---> Size of a combination
# to be printed
def combinationUtil(arr, data, start,
end, index, r):
# Current combination is ready
# to be printed, print it
if (index == r):
# print('data=',data)
# print('l w ',l,w)
if l<=sum(data)<=w and (max(data)-min(data))>=x:
# for j in range(r):
# print(data[j], end = " ");
# print();
global ans
ans+=1
return;
# replace index with all possible elements.
#The condition "end-i+1 >= r-index"
# makes sure that including one element at index will make a combination
# with remaining elements at remaining positions
i = start;
while(i <= end and end - i + 1 >= r - index):
data[index] = arr[i];
combinationUtil(arr, data, i + 1,
end, index + 1, r);
i += 1;
# Driver Code
arr=list(map(int,input().split()))
#arr = [1,2,3,4,5];
for r in range(2,n+1):
#r = 3;
n = len(arr);
printCombination(arr, n, r);
# This code is contributed by mits
print(ans)
```
|
instruction
| 0
| 105,682
| 11
| 211,364
|
Yes
|
output
| 1
| 105,682
| 11
| 211,365
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
def check(j):
if sum(j)>=l and sum(j)<=r and (max(j)-min(j))>=x:
return 1
return 0
from itertools import combinations
n,l,r,x=list(map(int,input().split()))
c=list(map(int,input().rstrip().split()))
count=0
for i in range(2,n+1):
a=list(combinations(c,i))
for j in a:
if check(j):
count+=1
print(count)
```
|
instruction
| 0
| 105,683
| 11
| 211,366
|
Yes
|
output
| 1
| 105,683
| 11
| 211,367
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
n , l , r , x = map(int,input().split(' '))
list_num = list(map(int,input().split(' ')))
ans = 0
for m in range((1 << n)): # 2**n
mn = -9223372036854775807 # like integer min
mx = 9223372036854775807
count = 0
sum = 0
for i in range(n):
if m&(1<<i) != 0:
count +=1
sum += list_num[i]
mn = min(mn , list_num[i])
mx = min(mx , list_num[i])
if mx - mn >= x and sum >= l and sum <=r and count>= 2:
ans +=1
print(ans)
```
|
instruction
| 0
| 105,684
| 11
| 211,368
|
No
|
output
| 1
| 105,684
| 11
| 211,369
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
from itertools import combinations
n,l,x,r=map(int,input().split())
problems=[int(x) for x in input().split()]
result=0
for i in range(2,n+1):
for comb in combinations(problems, i):
summ = sum(comb)
mini = min(comb)
maxx = max(comb)
if summ>=l or summ<=r and maxx-mini>=x:
result += 1
print(result)
```
|
instruction
| 0
| 105,685
| 11
| 211,370
|
No
|
output
| 1
| 105,685
| 11
| 211,371
|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have n problems. You have estimated the difficulty of the i-th one as integer ci. Now you want to prepare a problemset for a contest, using some of the problems you've made.
A problemset for the contest must consist of at least two problems. You think that the total difficulty of the problems of the contest must be at least l and at most r. Also, you think that the difference between difficulties of the easiest and the hardest of the chosen problems must be at least x.
Find the number of ways to choose a problemset for the contest.
Input
The first line contains four integers n, l, r, x (1 β€ n β€ 15, 1 β€ l β€ r β€ 109, 1 β€ x β€ 106) β the number of problems you have, the minimum and maximum value of total difficulty of the problemset and the minimum difference in difficulty between the hardest problem in the pack and the easiest one, respectively.
The second line contains n integers c1, c2, ..., cn (1 β€ ci β€ 106) β the difficulty of each problem.
Output
Print the number of ways to choose a suitable problemset for the contest.
Examples
Input
3 5 6 1
1 2 3
Output
2
Input
4 40 50 10
10 20 30 25
Output
2
Input
5 25 35 10
10 10 20 10 20
Output
6
Note
In the first example two sets are suitable, one consisting of the second and third problem, another one consisting of all three problems.
In the second example, two sets of problems are suitable β the set of problems with difficulties 10 and 30 as well as the set of problems with difficulties 20 and 30.
In the third example any set consisting of one problem of difficulty 10 and one problem of difficulty 20 is suitable.
Submitted Solution:
```
inp = input()
nums = [int(i) for i in inp.split(" ")]
inp1 = input()
nums1 = [int(j) for j in inp1.split(" ")]
l = nums[1]
r = nums[2]
x = nums[3]
def checker(array, x):
sums = []
used = []
for i in range(len(array)):
for j in range(len(array)):
if i == j:
continue
if [i,j] in used:
continue
if (array[i] - array[j] == x) or (array[j] - array[i] == x):
sums.append(array[i]+array[j])
used.append([i, j])
return sums
sums = checker(nums1, x)
c = 0
for k in sums:
if k >= l and k <= r:
c += 1
print(c)
```
|
instruction
| 0
| 105,686
| 11
| 211,372
|
No
|
output
| 1
| 105,686
| 11
| 211,373
|
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