text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
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))
```
No
| 104,822 | [
0.1888427734375,
-0.05499267578125,
-0.54296875,
0.638671875,
-0.50927734375,
-0.251708984375,
0.03594970703125,
0.2666015625,
-0.30224609375,
0.48828125,
0.273681640625,
-0.007556915283203125,
0.2325439453125,
-0.62060546875,
-0.5205078125,
-0.1947021484375,
-0.3515625,
-0.9345703... | 11 |
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 =" ")
```
No
| 104,823 | [
0.232177734375,
-0.0277557373046875,
-0.51416015625,
0.60693359375,
-0.458740234375,
-0.311767578125,
0.04754638671875,
0.29052734375,
-0.35546875,
0.56201171875,
0.268798828125,
0.01416778564453125,
0.2078857421875,
-0.58203125,
-0.5283203125,
-0.1905517578125,
-0.303466796875,
-0... | 11 |
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
"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]))
```
| 105,132 | [
0.54052734375,
-0.039337158203125,
-0.325927734375,
0.27783203125,
-0.68310546875,
-0.6025390625,
-0.2137451171875,
0.257568359375,
0.04718017578125,
0.7646484375,
0.38232421875,
-0.374755859375,
0.441162109375,
-0.810546875,
-0.053070068359375,
-0.215576171875,
-0.50537109375,
-0.... | 11 |
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
"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)
```
| 105,133 | [
0.5439453125,
-0.045318603515625,
-0.357666015625,
0.294189453125,
-0.62255859375,
-0.5888671875,
-0.2064208984375,
0.2362060546875,
0.048919677734375,
0.7861328125,
0.368896484375,
-0.402099609375,
0.431396484375,
-0.82080078125,
-0.08807373046875,
-0.2119140625,
-0.505859375,
-0.... | 11 |
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
"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)
```
| 105,134 | [
0.53759765625,
0.002536773681640625,
-0.329833984375,
0.3603515625,
-0.7333984375,
-0.6025390625,
-0.228515625,
0.263671875,
0.0245208740234375,
0.78369140625,
0.43212890625,
-0.375,
0.414306640625,
-0.81640625,
-0.09613037109375,
-0.1873779296875,
-0.5009765625,
-0.884765625,
-0... | 11 |
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
"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)
```
| 105,135 | [
0.59814453125,
-0.02862548828125,
-0.3662109375,
0.25927734375,
-0.6708984375,
-0.54833984375,
-0.1907958984375,
0.2822265625,
0.07086181640625,
0.787109375,
0.401611328125,
-0.349609375,
0.341064453125,
-0.857421875,
-0.07208251953125,
-0.2281494140625,
-0.50244140625,
-0.90722656... | 11 |
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
"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)
```
| 105,136 | [
0.5693359375,
-0.0562744140625,
-0.32958984375,
0.3017578125,
-0.689453125,
-0.6123046875,
-0.249755859375,
0.2705078125,
0.040496826171875,
0.80224609375,
0.4111328125,
-0.392578125,
0.419677734375,
-0.80517578125,
-0.03594970703125,
-0.2191162109375,
-0.5322265625,
-0.876953125,
... | 11 |
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
"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)
```
| 105,137 | [
0.51220703125,
0.003421783447265625,
-0.357666015625,
0.302001953125,
-0.6630859375,
-0.583984375,
-0.2744140625,
0.295654296875,
0.0308685302734375,
0.6865234375,
0.350830078125,
-0.355224609375,
0.400146484375,
-0.7958984375,
-0.0626220703125,
-0.25341796875,
-0.491455078125,
-0.... | 11 |
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
"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)
```
| 105,138 | [
0.5810546875,
-0.0316162109375,
-0.348876953125,
0.294677734375,
-0.671875,
-0.56884765625,
-0.2734375,
0.26171875,
0.009857177734375,
0.7958984375,
0.37255859375,
-0.3876953125,
0.431884765625,
-0.7607421875,
-0.046051025390625,
-0.205322265625,
-0.48681640625,
-0.86328125,
-0.4... | 11 |
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
"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])
```
| 105,139 | [
0.52734375,
-0.04296875,
-0.3154296875,
0.299072265625,
-0.669921875,
-0.603515625,
-0.2242431640625,
0.2471923828125,
0.06890869140625,
0.78857421875,
0.37939453125,
-0.371337890625,
0.429931640625,
-0.7939453125,
-0.049530029296875,
-0.2073974609375,
-0.483154296875,
-0.879882812... | 11 |
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)
```
Yes
| 105,140 | [
0.578125,
-0.148193359375,
-0.38232421875,
0.336181640625,
-0.63623046875,
-0.57177734375,
-0.27587890625,
0.2099609375,
0.07879638671875,
0.73193359375,
0.30078125,
-0.41845703125,
0.339111328125,
-0.810546875,
-0.0679931640625,
-0.2381591796875,
-0.467529296875,
-0.8486328125,
... | 11 |
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])
```
Yes
| 105,141 | [
0.59814453125,
-0.124267578125,
-0.403564453125,
0.368896484375,
-0.6181640625,
-0.58349609375,
-0.25439453125,
0.25537109375,
0.0853271484375,
0.74658203125,
0.337890625,
-0.35791015625,
0.318115234375,
-0.82275390625,
-0.058746337890625,
-0.2181396484375,
-0.4384765625,
-0.853515... | 11 |
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])
```
Yes
| 105,142 | [
0.57958984375,
-0.1373291015625,
-0.43310546875,
0.3671875,
-0.63818359375,
-0.5771484375,
-0.292236328125,
0.2327880859375,
0.0980224609375,
0.76318359375,
0.320556640625,
-0.412109375,
0.347412109375,
-0.80126953125,
-0.08367919921875,
-0.219482421875,
-0.424072265625,
-0.8320312... | 11 |
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)
```
Yes
| 105,143 | [
0.56298828125,
-0.138916015625,
-0.39892578125,
0.35595703125,
-0.619140625,
-0.57666015625,
-0.277587890625,
0.217041015625,
0.0946044921875,
0.75341796875,
0.29248046875,
-0.414306640625,
0.3466796875,
-0.82275390625,
-0.0654296875,
-0.23828125,
-0.433837890625,
-0.83056640625,
... | 11 |
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)
```
No
| 105,144 | [
0.57861328125,
-0.12371826171875,
-0.38720703125,
0.350830078125,
-0.623046875,
-0.5556640625,
-0.2467041015625,
0.238037109375,
0.0692138671875,
0.7392578125,
0.328125,
-0.40380859375,
0.345458984375,
-0.80419921875,
-0.1036376953125,
-0.2386474609375,
-0.452880859375,
-0.80859375... | 11 |
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)
```
No
| 105,145 | [
0.58837890625,
-0.138671875,
-0.39599609375,
0.340087890625,
-0.6865234375,
-0.55810546875,
-0.255859375,
0.27587890625,
0.06341552734375,
0.79931640625,
0.31884765625,
-0.331787109375,
0.32666015625,
-0.751953125,
-0.07318115234375,
-0.2154541015625,
-0.40625,
-0.77880859375,
-0... | 11 |
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)
```
No
| 105,146 | [
0.57373046875,
-0.127685546875,
-0.377685546875,
0.335693359375,
-0.66064453125,
-0.55810546875,
-0.251953125,
0.25048828125,
0.07293701171875,
0.80859375,
0.34716796875,
-0.376220703125,
0.34130859375,
-0.775390625,
-0.0865478515625,
-0.2078857421875,
-0.435546875,
-0.76953125,
... | 11 |
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)
```
No
| 105,147 | [
0.59326171875,
-0.133056640625,
-0.404296875,
0.37255859375,
-0.67919921875,
-0.53955078125,
-0.2481689453125,
0.28271484375,
0.0894775390625,
0.7998046875,
0.298583984375,
-0.3486328125,
0.31298828125,
-0.7607421875,
-0.06439208984375,
-0.228759765625,
-0.408935546875,
-0.79931640... | 11 |
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))
```
No
| 105,474 | [
0.365966796875,
-0.029296875,
0.15771484375,
-0.0184173583984375,
-0.5029296875,
-0.13818359375,
-0.1715087890625,
0.38671875,
0.1732177734375,
0.8310546875,
0.80419921875,
0.05126953125,
0.32958984375,
-0.78759765625,
-0.64013671875,
0.4345703125,
0.0634765625,
-0.80859375,
-0.0... | 11 |
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.
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)
```
| 105,526 | [
0.3447265625,
-0.039215087890625,
-0.049346923828125,
0.340087890625,
-0.60498046875,
-0.5,
0.1136474609375,
-0.1258544921875,
-0.34375,
0.79248046875,
0.83203125,
-0.1065673828125,
0.3408203125,
-0.76416015625,
-0.2291259765625,
0.303466796875,
-0.607421875,
-0.7978515625,
-0.45... | 11 |
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.
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)
```
| 105,527 | [
0.308349609375,
-0.08978271484375,
0.09466552734375,
0.330322265625,
-0.615234375,
-0.40869140625,
0.09283447265625,
-0.2451171875,
-0.349365234375,
0.82373046875,
0.72607421875,
-0.156005859375,
0.37841796875,
-0.73828125,
-0.2203369140625,
0.313720703125,
-0.5732421875,
-0.785644... | 11 |
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.
Tags: implementation, math
Correct Solution:
```
n,k=map(int,input().split())
k=k-2*n
print(max(n-k,0))
```
| 105,528 | [
0.347412109375,
-0.00833892822265625,
-0.0300140380859375,
0.38330078125,
-0.57080078125,
-0.5126953125,
0.054229736328125,
-0.1431884765625,
-0.34716796875,
0.79248046875,
0.79541015625,
-0.06085205078125,
0.394287109375,
-0.75390625,
-0.1866455078125,
0.342041015625,
-0.58935546875... | 11 |
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.
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))
```
| 105,529 | [
0.322265625,
-0.03826904296875,
-0.0288543701171875,
0.358154296875,
-0.56201171875,
-0.5087890625,
0.08868408203125,
-0.1336669921875,
-0.31201171875,
0.81689453125,
0.85400390625,
-0.10284423828125,
0.354248046875,
-0.75048828125,
-0.2174072265625,
0.32275390625,
-0.5908203125,
-... | 11 |
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.
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)
```
| 105,530 | [
0.385498046875,
-0.06585693359375,
0.0299072265625,
0.425537109375,
-0.58642578125,
-0.486328125,
0.1268310546875,
-0.2052001953125,
-0.315673828125,
0.81640625,
0.779296875,
-0.11279296875,
0.35986328125,
-0.7841796875,
-0.19384765625,
0.2103271484375,
-0.646484375,
-0.72021484375... | 11 |
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.
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)
```
| 105,531 | [
0.353515625,
-0.00786590576171875,
-0.039031982421875,
0.34521484375,
-0.58447265625,
-0.5009765625,
0.0992431640625,
-0.1202392578125,
-0.319091796875,
0.8251953125,
0.8291015625,
-0.09967041015625,
0.402099609375,
-0.75634765625,
-0.2156982421875,
0.32568359375,
-0.59375,
-0.7832... | 11 |
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.
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)
```
| 105,532 | [
0.33544921875,
-0.00612640380859375,
-0.0224151611328125,
0.353759765625,
-0.5556640625,
-0.493896484375,
0.10186767578125,
-0.1319580078125,
-0.372802734375,
0.767578125,
0.80810546875,
-0.105224609375,
0.36669921875,
-0.77685546875,
-0.2264404296875,
0.288818359375,
-0.611328125,
... | 11 |
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.
Tags: implementation, math
Correct Solution:
```
n, k = map(int, input().split())
print(n*3-k if k <= n * 3 else 0)
```
| 105,533 | [
0.3447265625,
-0.0164794921875,
-0.04376220703125,
0.34033203125,
-0.5517578125,
-0.50244140625,
0.06884765625,
-0.1265869140625,
-0.343505859375,
0.81298828125,
0.81591796875,
-0.0740966796875,
0.403564453125,
-0.7724609375,
-0.2091064453125,
0.30908203125,
-0.5927734375,
-0.77343... | 11 |
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()
```
Yes
| 105,534 | [
0.380126953125,
-0.05029296875,
-0.03631591796875,
0.27880859375,
-0.5791015625,
-0.33154296875,
0.0625,
-0.070556640625,
-0.31689453125,
0.74853515625,
0.75048828125,
-0.09222412109375,
0.248046875,
-0.8291015625,
-0.2159423828125,
0.2861328125,
-0.53759765625,
-0.7294921875,
-0... | 11 |
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);
```
Yes
| 105,535 | [
0.427490234375,
-0.039825439453125,
0.048736572265625,
0.322998046875,
-0.414794921875,
-0.213134765625,
0.013916015625,
-0.1668701171875,
-0.238037109375,
0.74609375,
0.9267578125,
-0.266357421875,
0.3720703125,
-0.96484375,
-0.2431640625,
0.3876953125,
-0.50341796875,
-0.63671875... | 11 |
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)
```
Yes
| 105,536 | [
0.420166015625,
-0.02496337890625,
-0.07867431640625,
0.293701171875,
-0.55615234375,
-0.31640625,
0.02056884765625,
-0.061248779296875,
-0.33837890625,
0.75439453125,
0.74462890625,
-0.071533203125,
0.30029296875,
-0.83349609375,
-0.197265625,
0.30029296875,
-0.50146484375,
-0.732... | 11 |
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)
```
Yes
| 105,537 | [
0.442138671875,
-0.055450439453125,
-0.09710693359375,
0.289794921875,
-0.5712890625,
-0.30419921875,
0.047698974609375,
-0.046600341796875,
-0.328369140625,
0.736328125,
0.75537109375,
-0.075927734375,
0.259033203125,
-0.849609375,
-0.2423095703125,
0.29833984375,
-0.5283203125,
-... | 11 |
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)
```
No
| 105,538 | [
0.427734375,
-0.032989501953125,
-0.07708740234375,
0.29931640625,
-0.55517578125,
-0.320068359375,
0.0152435302734375,
-0.0692138671875,
-0.335205078125,
0.7685546875,
0.7734375,
-0.050323486328125,
0.284423828125,
-0.82177734375,
-0.2095947265625,
0.313720703125,
-0.49169921875,
... | 11 |
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)
```
No
| 105,539 | [
0.375,
-0.05023193359375,
-0.0633544921875,
0.2763671875,
-0.59814453125,
-0.340576171875,
0.0894775390625,
-0.028594970703125,
-0.317138671875,
0.73974609375,
0.74609375,
-0.08868408203125,
0.297607421875,
-0.82470703125,
-0.209228515625,
0.286376953125,
-0.560546875,
-0.75390625,... | 11 |
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
```
No
| 105,540 | [
0.387939453125,
-0.04461669921875,
-0.0657958984375,
0.307861328125,
-0.5810546875,
-0.319091796875,
0.0531005859375,
-0.01219940185546875,
-0.337158203125,
0.75390625,
0.70751953125,
-0.11041259765625,
0.25634765625,
-0.791015625,
-0.201171875,
0.306884765625,
-0.53662109375,
-0.6... | 11 |
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)
```
No
| 105,541 | [
0.40869140625,
-0.055938720703125,
-0.077392578125,
0.301513671875,
-0.548828125,
-0.321044921875,
0.0863037109375,
-0.0352783203125,
-0.310546875,
0.7646484375,
0.7724609375,
-0.079345703125,
0.305908203125,
-0.8447265625,
-0.1878662109375,
0.297607421875,
-0.53955078125,
-0.73535... | 11 |
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.
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)
```
| 105,672 | [
0.290283203125,
-0.33251953125,
0.311279296875,
-0.0268707275390625,
-0.6259765625,
-0.5078125,
-0.051910400390625,
-0.08843994140625,
0.10235595703125,
0.94677734375,
0.54052734375,
0.0321044921875,
0.4091796875,
-0.68359375,
-0.19384765625,
0.09588623046875,
-0.67529296875,
-0.86... | 11 |
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.
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)
```
| 105,673 | [
0.309814453125,
-0.36865234375,
0.2666015625,
-0.02838134765625,
-0.60546875,
-0.55419921875,
-0.08502197265625,
-0.093994140625,
0.1573486328125,
0.93798828125,
0.49560546875,
0.020294189453125,
0.43359375,
-0.6044921875,
-0.1912841796875,
0.11968994140625,
-0.6494140625,
-0.80712... | 11 |
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.
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)
```
| 105,674 | [
0.3212890625,
-0.311279296875,
0.32373046875,
0.029693603515625,
-0.591796875,
-0.50341796875,
-0.033905029296875,
-0.097900390625,
0.11846923828125,
0.91845703125,
0.52978515625,
0.01474761962890625,
0.395751953125,
-0.62744140625,
-0.173828125,
0.1651611328125,
-0.650390625,
-0.8... | 11 |
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.
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)
```
| 105,675 | [
0.336669921875,
-0.359619140625,
0.2666015625,
-0.044647216796875,
-0.5908203125,
-0.55859375,
-0.0819091796875,
-0.093017578125,
0.128662109375,
0.8994140625,
0.497314453125,
0.0445556640625,
0.43505859375,
-0.62939453125,
-0.2254638671875,
0.11749267578125,
-0.69677734375,
-0.825... | 11 |
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.
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))
```
| 105,676 | [
0.34814453125,
-0.2978515625,
0.35595703125,
0.0210418701171875,
-0.6328125,
-0.5400390625,
-0.04888916015625,
-0.06439208984375,
0.10406494140625,
0.92626953125,
0.4921875,
0.039642333984375,
0.40625,
-0.6162109375,
-0.21044921875,
0.11480712890625,
-0.66650390625,
-0.80859375,
... | 11 |
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.
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)
```
| 105,677 | [
0.343505859375,
-0.265869140625,
0.37353515625,
0.025543212890625,
-0.53857421875,
-0.451171875,
-0.08624267578125,
-0.09039306640625,
0.1214599609375,
0.93701171875,
0.54443359375,
-0.024139404296875,
0.34619140625,
-0.734375,
-0.2279052734375,
0.04095458984375,
-0.5576171875,
-0.... | 11 |
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.
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)
```
| 105,678 | [
0.32666015625,
-0.349853515625,
0.33740234375,
-0.044342041015625,
-0.60400390625,
-0.5615234375,
-0.059326171875,
-0.0946044921875,
0.101806640625,
0.9677734375,
0.53271484375,
0.056396484375,
0.42724609375,
-0.66748046875,
-0.2442626953125,
0.1287841796875,
-0.63232421875,
-0.833... | 11 |
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.
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)
```
| 105,679 | [
0.328125,
-0.316650390625,
0.315185546875,
-0.016082763671875,
-0.5859375,
-0.52734375,
-0.05450439453125,
-0.0750732421875,
0.11688232421875,
0.95361328125,
0.544921875,
0.044036865234375,
0.41796875,
-0.67333984375,
-0.208984375,
0.107177734375,
-0.66796875,
-0.83740234375,
-0.... | 11 |
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)
```
Yes
| 105,680 | [
0.412109375,
-0.26416015625,
0.1505126953125,
0.07220458984375,
-0.75048828125,
-0.314208984375,
-0.1126708984375,
0.032470703125,
0.0255279541015625,
0.96533203125,
0.5458984375,
-0.052581787109375,
0.35009765625,
-0.61865234375,
-0.26416015625,
-0.02496337890625,
-0.73583984375,
... | 11 |
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()
```
Yes
| 105,681 | [
0.351806640625,
-0.24658203125,
0.1629638671875,
0.015899658203125,
-0.77734375,
-0.31787109375,
-0.0963134765625,
0.10986328125,
0.02752685546875,
0.9306640625,
0.53857421875,
0.0057220458984375,
0.37646484375,
-0.6923828125,
-0.2232666015625,
-0.095703125,
-0.75390625,
-0.9208984... | 11 |
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)
```
Yes
| 105,682 | [
0.312255859375,
-0.29052734375,
0.2164306640625,
0.1405029296875,
-0.65087890625,
-0.428955078125,
-0.161376953125,
0.07183837890625,
0.0799560546875,
0.88818359375,
0.60302734375,
-0.0212249755859375,
0.39208984375,
-0.69580078125,
-0.351806640625,
-0.10369873046875,
-0.6220703125,
... | 11 |
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)
```
Yes
| 105,683 | [
0.35107421875,
-0.279296875,
0.1820068359375,
-0.014739990234375,
-0.75390625,
-0.35888671875,
-0.135498046875,
0.077880859375,
0.08697509765625,
0.94873046875,
0.46337890625,
0.006717681884765625,
0.3876953125,
-0.68798828125,
-0.237060546875,
-0.134765625,
-0.78564453125,
-0.8540... | 11 |
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)
```
No
| 105,684 | [
0.349365234375,
-0.28173828125,
0.14990234375,
0.00379180908203125,
-0.744140625,
-0.31591796875,
-0.0943603515625,
0.1160888671875,
0.025360107421875,
0.94873046875,
0.53662109375,
0.0078277587890625,
0.373046875,
-0.70263671875,
-0.282470703125,
-0.09771728515625,
-0.70458984375,
... | 11 |
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)
```
No
| 105,685 | [
0.45166015625,
-0.35693359375,
0.152099609375,
-0.04937744140625,
-0.63720703125,
-0.388427734375,
-0.11083984375,
-0.03955078125,
0.1895751953125,
0.92431640625,
0.58447265625,
0.058746337890625,
0.25927734375,
-0.6162109375,
-0.1951904296875,
-0.06402587890625,
-0.75,
-0.97314453... | 11 |
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)
```
No
| 105,686 | [
0.4072265625,
-0.29052734375,
0.125244140625,
-0.026458740234375,
-0.74951171875,
-0.337646484375,
-0.1455078125,
0.0253143310546875,
0.1240234375,
0.96875,
0.6337890625,
-0.0277557373046875,
0.388671875,
-0.76904296875,
-0.36328125,
-0.2080078125,
-0.6875,
-0.888671875,
-0.74267... | 11 |
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:
```
import re
s = input()
p1 = re.compile('[A-Z]*AB[A-Z]*BA[A-Z]*')
p2 = re.compile('[A-Z]*BA[A-Z]*AB[A-Z]*')
print('YES' if p1.match(s) or p2.match(s) else 'NO')
```
No
| 105,687 | [
0.40380859375,
-0.286376953125,
0.1773681640625,
-0.07183837890625,
-0.77783203125,
-0.288330078125,
-0.16943359375,
0.07659912109375,
0.061676025390625,
0.84716796875,
0.5302734375,
-0.07977294921875,
0.420166015625,
-0.77099609375,
-0.223876953125,
-0.1761474609375,
-0.76025390625,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 β€ x β€ 3). The following conditions must hold:
* For each x (1 β€ x β€ 3) cntx > 0;
* For any two degrees x and y cntx β€ 2Β·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 β€ x β€ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 β€ n β€ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 β€ ai β€ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
print("-1 3 2 1")
```
No
| 105,825 | [
0.48193359375,
-0.053680419921875,
-0.01503753662109375,
0.3056640625,
-0.5478515625,
-0.24365234375,
-0.0159912109375,
0.51708984375,
-0.294189453125,
1.037109375,
0.372802734375,
0.12353515625,
0.26416015625,
-0.55419921875,
-0.4130859375,
0.1080322265625,
-0.68212890625,
-0.9663... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 β€ x β€ 3). The following conditions must hold:
* For each x (1 β€ x β€ 3) cntx > 0;
* For any two degrees x and y cntx β€ 2Β·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 β€ x β€ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 β€ n β€ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 β€ ai β€ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip1 = {}
dip2 = {}
dip3 = {}
ndip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
if ndip.__contains__(b[i]):
ndip[b[i]] += 1
else:
ndip[b[i]] = 1
for i in range(third_index, second_index):
if dip3.__contains__(b[i]):
dip3[b[i]] += 1
else:
dip3[b[i]] = 1
for i in range(second_index, first_index):
if dip2.__contains__(b[i]):
dip2[b[i]] += 1
else:
dip2[b[i]] = 1
for i in range(first_index, n):
if dip1.__contains__(b[i]):
dip1[b[i]] += 1
else:
dip1[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for x in a:
if ndip.__contains__(x) and ndip[x]:
print('-1', end=' ')
ndip[x] -= 1
elif dip1.__contains__(x) and dip1[x]:
print('1', end=' ')
dip1[x] -= 1
elif dip2.__contains__(x) and dip2[x]:
print('2', end=' ')
dip2[x] -= 1
elif dip3.__contains__(x) and dip3[x]:
print('3', end=' ')
dip3[x] -= 1
```
No
| 105,826 | [
0.48193359375,
-0.053680419921875,
-0.01503753662109375,
0.3056640625,
-0.5478515625,
-0.24365234375,
-0.0159912109375,
0.51708984375,
-0.294189453125,
1.037109375,
0.372802734375,
0.12353515625,
0.26416015625,
-0.55419921875,
-0.4130859375,
0.1080322265625,
-0.68212890625,
-0.9663... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 β€ x β€ 3). The following conditions must hold:
* For each x (1 β€ x β€ 3) cntx > 0;
* For any two degrees x and y cntx β€ 2Β·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 β€ x β€ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 β€ n β€ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 β€ ai β€ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
dip[b[i]] = -1
for i in range(third_index, second_index):
dip[b[i]] = 3
for i in range(second_index, first_index):
dip[b[i]] = 2
for i in range(first_index, n):
dip[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for x in a:
print(dip[x], end=' ')
```
No
| 105,827 | [
0.48193359375,
-0.053680419921875,
-0.01503753662109375,
0.3056640625,
-0.5478515625,
-0.24365234375,
-0.0159912109375,
0.51708984375,
-0.294189453125,
1.037109375,
0.372802734375,
0.12353515625,
0.26416015625,
-0.55419921875,
-0.4130859375,
0.1080322265625,
-0.68212890625,
-0.9663... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alexey recently held a programming contest for students from Berland. n students participated in a contest, i-th of them solved ai problems. Now he wants to award some contestants. Alexey can award the students with diplomas of three different degrees. Each student either will receive one diploma of some degree, or won't receive any diplomas at all. Let cntx be the number of students that are awarded with diplomas of degree x (1 β€ x β€ 3). The following conditions must hold:
* For each x (1 β€ x β€ 3) cntx > 0;
* For any two degrees x and y cntx β€ 2Β·cnty.
Of course, there are a lot of ways to distribute the diplomas. Let bi be the degree of diploma i-th student will receive (or - 1 if i-th student won't receive any diplomas). Also for any x such that 1 β€ x β€ 3 let cx be the maximum number of problems solved by a student that receives a diploma of degree x, and dx be the minimum number of problems solved by a student that receives a diploma of degree x. Alexey wants to distribute the diplomas in such a way that:
1. If student i solved more problems than student j, then he has to be awarded not worse than student j (it's impossible that student j receives a diploma and i doesn't receive any, and also it's impossible that both of them receive a diploma, but bj < bi);
2. d1 - c2 is maximum possible;
3. Among all ways that maximize the previous expression, d2 - c3 is maximum possible;
4. Among all ways that correspond to the two previous conditions, d3 - c - 1 is maximum possible, where c - 1 is the maximum number of problems solved by a student that doesn't receive any diploma (or 0 if each student is awarded with some diploma).
Help Alexey to find a way to award the contestants!
Input
The first line contains one integer number n (3 β€ n β€ 3000).
The second line contains n integer numbers a1, a2, ..., an (1 β€ ai β€ 5000).
Output
Output n numbers. i-th number must be equal to the degree of diploma i-th contestant will receive (or - 1 if he doesn't receive any diploma).
If there are multiple optimal solutions, print any of them. It is guaranteed that the answer always exists.
Examples
Input
4
1 2 3 4
Output
3 3 2 1
Input
6
1 4 3 1 1 2
Output
-1 1 2 -1 -1 3
Submitted Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
b = sorted(a)
dip1 = {}
dip2 = {}
dip3 = {}
ndip = {}
first_value = 0
first_index = 2
for i in range(first_index, n):
if b[i] - b[i-1] >= first_value:
first_value = b[i] - b[i-1]
first_index = i
second_value = 0
second_index = 1
for i in range(second_index, first_index):
if b[i] - b[i-1] >= second_value:
second_value = b[i] - b[i-1]
second_index = i
third_value = 0
third_index = 0
for i in range(third_index, second_index):
if i == 0:
continue
if b[i] - b[i-1] >= third_value:
third_value = b[i] - b[i-1]
third_index = i
for i in range(0, third_index):
if ndip.__contains__(b[i]):
ndip[b[i]] += 1
else:
ndip[b[i]] = 1
for i in range(third_index, second_index):
if dip3.__contains__(b[i]):
dip3[b[i]] += 1
else:
dip3[b[i]] = 1
for i in range(second_index, first_index):
if dip2.__contains__(b[i]):
dip2[b[i]] += 1
else:
dip2[b[i]] = 1
for i in range(first_index, n):
if dip1.__contains__(b[i]):
dip1[b[i]] += 1
else:
dip1[b[i]] = 1
# print(first_index, second_index, third_index, )
# print(dip)
#
# print(a)
for i in range(n):
if ndip.__contains__(a[i]) and ndip[a[i]]:
print('-1', end=' ')
ndip[a[i]] -= 1
elif dip1.__contains__(a[i]) and dip1[a[i]]:
print('1', end=' ')
dip1[a[i]] -= 1
elif dip2.__contains__(a[i]) and dip2[a[i]]:
print('2', end=' ')
dip2[a[i]] -= 1
elif dip3.__contains__(a[i]) and dip3[a[i]]:
print('3', end=' ')
dip3[a[i]] -= 1
```
No
| 105,828 | [
0.48193359375,
-0.053680419921875,
-0.01503753662109375,
0.3056640625,
-0.5478515625,
-0.24365234375,
-0.0159912109375,
0.51708984375,
-0.294189453125,
1.037109375,
0.372802734375,
0.12353515625,
0.26416015625,
-0.55419921875,
-0.4130859375,
0.1080322265625,
-0.68212890625,
-0.9663... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence a1, a2, ..., an of one-dimensional segments numbered 1 through n. Your task is to find two distinct indices i and j such that segment ai lies within segment aj.
Segment [l1, r1] lies within segment [l2, r2] iff l1 β₯ l2 and r1 β€ r2.
Print indices i and j. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Input
The first line contains one integer n (1 β€ n β€ 3Β·105) β the number of segments.
Each of the next n lines contains two integers li and ri (1 β€ li β€ ri β€ 109) β the i-th segment.
Output
Print two distinct indices i and j such that segment ai lies within segment aj. If there are multiple answers, print any of them. If no answer exists, print -1 -1.
Examples
Input
5
1 10
2 9
3 9
2 3
2 9
Output
2 1
Input
3
1 5
2 6
6 20
Output
-1 -1
Note
In the first example the following pairs are considered correct:
* (2, 1), (3, 1), (4, 1), (5, 1) β not even touching borders;
* (3, 2), (4, 2), (3, 5), (4, 5) β touch one border;
* (5, 2), (2, 5) β match exactly.
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input().strip())
ais = [tuple(map(int, input().strip().split())) for _ in range(n)]
def solve(ais):
bis = [(l, r, i + 1) for i, (l, r) in enumerate(ais)]
bis.sort(key=lambda t: (t[0], -t[1]))
rr = bis[0][1] - 1
ir = bis[0][2]
for l, r, i in bis:
if r <= rr:
return (i, ir)
else:
rr = r
ir = i
return (-1, -1)
i, j = solve(ais)
print (i, j)
```
Yes
| 105,874 | [
0.1639404296875,
0.53369140625,
-0.035675048828125,
0.1282958984375,
-0.54931640625,
-0.350341796875,
-0.28369140625,
-0.06915283203125,
0.1590576171875,
0.92822265625,
0.17626953125,
0.33544921875,
-0.229736328125,
-1.0888671875,
-0.4912109375,
0.068115234375,
-0.52197265625,
-0.4... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a set of n weights. You know that their masses are a_1, a_2, ..., a_n grams, but you don't know which of them has which mass. You can't distinguish the weights.
However, your friend does know the mass of each weight. You can ask your friend to give you exactly k weights with the total mass m (both parameters k and m are chosen by you), and your friend will point to any valid subset of weights, if it is possible.
You are allowed to make this query only once. Find the maximum possible number of weights you can reveal after this query.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of weights.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 100) β the masses of the weights.
Output
Print the maximum number of weights you can learn the masses for after making a single query.
Examples
Input
4
1 4 2 2
Output
2
Input
6
1 2 4 4 4 9
Output
2
Note
In the first example we can ask for a subset of two weights with total mass being equal to 4, and the only option is to get \{2, 2\}.
Another way to obtain the same result is to ask for a subset of two weights with the total mass of 5 and get \{1, 4\}. It is easy to see that the two remaining weights have mass of 2 grams each.
In the second example we can ask for a subset of two weights with total mass being 8, and the only answer is \{4, 4\}. We can prove it is not possible to learn masses for three weights in one query, but we won't put the proof here.
Submitted Solution:
```
n = int(input())
geers = [int(x) for x in input().split()]
if len(set(geers)) <= 2:
print(n)
else:
print(2)
```
No
| 106,148 | [
0.3525390625,
0.0047454833984375,
-0.0733642578125,
0.260498046875,
-0.84375,
-0.212158203125,
-0.035736083984375,
0.293212890625,
0.284912109375,
0.78125,
0.89208984375,
-0.38623046875,
0.2449951171875,
-0.91650390625,
-0.5185546875,
0.47021484375,
-0.85791015625,
-1.0830078125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a set of n weights. You know that their masses are a_1, a_2, ..., a_n grams, but you don't know which of them has which mass. You can't distinguish the weights.
However, your friend does know the mass of each weight. You can ask your friend to give you exactly k weights with the total mass m (both parameters k and m are chosen by you), and your friend will point to any valid subset of weights, if it is possible.
You are allowed to make this query only once. Find the maximum possible number of weights you can reveal after this query.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of weights.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 100) β the masses of the weights.
Output
Print the maximum number of weights you can learn the masses for after making a single query.
Examples
Input
4
1 4 2 2
Output
2
Input
6
1 2 4 4 4 9
Output
2
Note
In the first example we can ask for a subset of two weights with total mass being equal to 4, and the only option is to get \{2, 2\}.
Another way to obtain the same result is to ask for a subset of two weights with the total mass of 5 and get \{1, 4\}. It is easy to see that the two remaining weights have mass of 2 grams each.
In the second example we can ask for a subset of two weights with total mass being 8, and the only answer is \{4, 4\}. We can prove it is not possible to learn masses for three weights in one query, but we won't put the proof here.
Submitted Solution:
```
from sys import stdin,stdout
import copy
from collections import Counter
n=int(stdin.readline())
arr=list(map(int,stdin.readline().strip().split(' ')))
temp=Counter(arr)
if len(temp)<=2:
stdout.write(str(len(arr)))
else:
m=sum(arr)
np=[True for i in range(m+1)]
pna=[False for i in range(m+1)]
pc=[{} for i in range(m+1)]
for ele in arr:
npc=copy.deepcopy(np)
pnac=copy.deepcopy(pna)
pcc=copy.deepcopy(pc)
# print(ele,"START")
# print(np)
# print(pc)
# print(pna)
for i in range(m+1):
if i==ele:
if npc[i]==True:
npc[i]=False
pcc[i][ele]=1
else:
if pnac[i]:
pnac[i+ele]=True
npc[i+ele]=False
for k in pc[i]:
npc[i+ele]=False
if k==ele:
pcc[i+ele][ele]=pcc[i][ele]+1
else:
pnac[i+ele]=True
continue
if npc[i]:
continue
else:
if pna[i]:
pnac[i+ele]=True
for k in pc[i]:
npc[i+ele]=False
if k==ele:
pcc[i+ele][ele]=pcc[i][ele]+1
else:
pnac[i+ele]=True
np=npc
pc=pcc
pna=pnac
# print(ele,"END")
# print(np)
# print(pc)
# print(pna)
ans=-1
for i in range(m+1):
if not pna[i]:
for j in pc[i]:
ans=max(ans,pc[i][j])
stdout.write(str(ans))
# 4
# 1 4 2 2
```
No
| 106,149 | [
0.30419921875,
0.047271728515625,
0.001155853271484375,
0.324462890625,
-0.8408203125,
-0.2403564453125,
-0.202392578125,
0.291015625,
0.259521484375,
0.86181640625,
0.83251953125,
-0.410888671875,
0.2978515625,
-0.8388671875,
-0.63525390625,
0.54345703125,
-0.7763671875,
-1.043945... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a set of n weights. You know that their masses are a_1, a_2, ..., a_n grams, but you don't know which of them has which mass. You can't distinguish the weights.
However, your friend does know the mass of each weight. You can ask your friend to give you exactly k weights with the total mass m (both parameters k and m are chosen by you), and your friend will point to any valid subset of weights, if it is possible.
You are allowed to make this query only once. Find the maximum possible number of weights you can reveal after this query.
Input
The first line contains a single integer n (1 β€ n β€ 100) β the number of weights.
The second line contains n integers a_1, a_2, β¦, a_n (1 β€ a_i β€ 100) β the masses of the weights.
Output
Print the maximum number of weights you can learn the masses for after making a single query.
Examples
Input
4
1 4 2 2
Output
2
Input
6
1 2 4 4 4 9
Output
2
Note
In the first example we can ask for a subset of two weights with total mass being equal to 4, and the only option is to get \{2, 2\}.
Another way to obtain the same result is to ask for a subset of two weights with the total mass of 5 and get \{1, 4\}. It is easy to see that the two remaining weights have mass of 2 grams each.
In the second example we can ask for a subset of two weights with total mass being 8, and the only answer is \{4, 4\}. We can prove it is not possible to learn masses for three weights in one query, but we won't put the proof here.
Submitted Solution:
```
n = int(input())
num = n
a = input().split()
up = []
for i in range(n):
a[i] = int(a[i])
for i in range(n):
k = a.count(a[i])
up.append(k)
l1 = []
l2 = []
for i in range(n):
if up[i] == 1:
l1.append(a[i])
else:
l2.append(a[i])
if sum(l1) != sum(l2):
ans = len(l1)
else:
ch = []
for i in range(len(l1)):
ch.append(0)
while len(l2) > 1:
l = []
l1.append(l2[0])
del(l2[0])
n = len(ch)
while ch[0] != 2:
for i in range(n - 1, 0, -1):
if ch[i] == 2:
ch[i] = 0
ch[i - 1] += 1
if sum(ch) < 2:
ch[n - 1] += 1
continue
s = 0
for i in range(n):
if ch[i] == 1:
s += l1[i]
if s != 0:
l.append(s)
ch[n - 1] += 1
flag = True
for i in range(len(l1)):
if l[i] == sum(l2):
flag = False
if flag == True:
ans = len(l2)
break
del(ch[0])
if ans == num:
ans = 1
print(ans)
```
No
| 106,151 | [
0.33837890625,
-0.03192138671875,
-0.02288818359375,
0.248291015625,
-0.794921875,
-0.23388671875,
-0.08612060546875,
0.334716796875,
0.3505859375,
0.86328125,
0.89501953125,
-0.3544921875,
0.2763671875,
-0.85205078125,
-0.55615234375,
0.60986328125,
-0.796875,
-1.001953125,
-0.3... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
from sys import stdin
input = stdin.readline
n, m, k = list(map(int, input().split()))
if min(m, k) >= n:
print('Yes')
else:
print('No')
```
Yes
| 106,197 | [
0.485595703125,
-0.17236328125,
-0.0850830078125,
0.1962890625,
-0.822265625,
-0.2332763671875,
-0.28173828125,
0.42724609375,
-0.1505126953125,
0.59375,
0.263671875,
0.203369140625,
0.2274169921875,
-0.2060546875,
-0.49169921875,
-0.21044921875,
-0.6689453125,
-0.90869140625,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
t=list(map(int,(input().split())))
if t[1]>=t[0] and t[2]>=t[0]:
print ("YES")
else :
print("NO")
```
Yes
| 106,198 | [
0.488037109375,
-0.154296875,
-0.10791015625,
0.1849365234375,
-0.80615234375,
-0.25146484375,
-0.27392578125,
0.413818359375,
-0.1435546875,
0.60791015625,
0.291015625,
0.21875,
0.2325439453125,
-0.2318115234375,
-0.460693359375,
-0.199462890625,
-0.67626953125,
-0.9052734375,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
n,m,k = [int(x) for x in input().split()]
if(n<=m and n<=k):
print("Yes")
else:
print("No")
```
Yes
| 106,199 | [
0.478515625,
-0.145263671875,
-0.0955810546875,
0.1932373046875,
-0.8193359375,
-0.26708984375,
-0.26416015625,
0.395263671875,
-0.1356201171875,
0.6103515625,
0.317626953125,
0.218017578125,
0.2158203125,
-0.260498046875,
-0.487060546875,
-0.1954345703125,
-0.6767578125,
-0.922363... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
string = input()
li = string.split(" ")
n = int(li[0])
m = int(li[1])
k = int(li[2])
#print(n,m,k)
if (k>=n) and (m>=n):
print("Yes")
else:
print("No")
```
Yes
| 106,200 | [
0.466796875,
-0.177490234375,
-0.0927734375,
0.19482421875,
-0.82861328125,
-0.287109375,
-0.2230224609375,
0.398681640625,
-0.09588623046875,
0.58154296875,
0.34228515625,
0.2169189453125,
0.2236328125,
-0.258544921875,
-0.48779296875,
-0.196044921875,
-0.6845703125,
-0.943359375,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
n,p,m=map(int,input().split())
if p>=n and m>=p:
print('Yes')
else:
print('No')
```
No
| 106,201 | [
0.5009765625,
-0.1485595703125,
-0.1134033203125,
0.1890869140625,
-0.80029296875,
-0.2568359375,
-0.270751953125,
0.405517578125,
-0.1463623046875,
0.58251953125,
0.333740234375,
0.211181640625,
0.2318115234375,
-0.22314453125,
-0.460205078125,
-0.205078125,
-0.66455078125,
-0.899... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
string = input()
li = string.split(" ")
n = li[0]
m = li[1]
k = li[2]
if (k>=n) and (m>=n):
print("Yes")
else:
print("No")
```
No
| 106,202 | [
0.482666015625,
-0.168701171875,
-0.0875244140625,
0.1837158203125,
-0.83837890625,
-0.2900390625,
-0.239501953125,
0.412353515625,
-0.0986328125,
0.58349609375,
0.334228515625,
0.2181396484375,
0.225341796875,
-0.2275390625,
-0.48291015625,
-0.22705078125,
-0.66796875,
-0.90136718... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
q,w,e = input().split()
q = int(q)
w = int(w)
e = int(e)
def check(a,s,d):
if a/s >= 1 :
if a/d >= 1 :
print("yes")
else:
print("no")
else:
print("no")
check(q,w,e)
```
No
| 106,203 | [
0.466064453125,
-0.1580810546875,
-0.10223388671875,
0.172607421875,
-0.802734375,
-0.252197265625,
-0.2139892578125,
0.4208984375,
-0.12115478515625,
0.6416015625,
0.325439453125,
0.2008056640625,
0.1927490234375,
-0.281982421875,
-0.51220703125,
-0.1790771484375,
-0.7294921875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vus the [Cossack](https://en.wikipedia.org/wiki/Cossacks) holds a programming competition, in which n people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly m pens and k notebooks.
Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.
Input
The first line contains three integers n, m, and k (1 β€ n, m, k β€ 100) β the number of participants, the number of pens, and the number of notebooks respectively.
Output
Print "Yes" if it possible to reward all the participants. Otherwise, print "No".
You can print each letter in any case (upper or lower).
Examples
Input
5 8 6
Output
Yes
Input
3 9 3
Output
Yes
Input
8 5 20
Output
No
Note
In the first example, there are 5 participants. The Cossack has 8 pens and 6 notebooks. Therefore, he has enough pens and notebooks.
In the second example, there are 3 participants. The Cossack has 9 pens and 3 notebooks. He has more than enough pens but only the minimum needed number of notebooks.
In the third example, there are 8 participants but only 5 pens. Since the Cossack does not have enough pens, the answer is "No".
Submitted Solution:
```
j=input().split()
n,m,k=j[0],j[1],j[2]
if n<=m or n<=k:
print("Yes")
else:
print("No")
```
No
| 106,204 | [
0.505859375,
-0.1326904296875,
-0.10552978515625,
0.1737060546875,
-0.8330078125,
-0.283935546875,
-0.260498046875,
0.422607421875,
-0.1077880859375,
0.5966796875,
0.3115234375,
0.2200927734375,
0.205322265625,
-0.238037109375,
-0.47705078125,
-0.2198486328125,
-0.67236328125,
-0.9... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
def solve():
arange=int(input())
if(arange%2==0):
print("NO")
return
diapasone=list(range(1,(arange*2)+1))
answer=[]
i=0
j=len(diapasone)-1
while(len(answer)<len(diapasone)):
temp_i=i
temp_j=j
counter=arange
while(counter>0):
answer.append(diapasone[temp_i])
counter-=1
if(counter>0):
answer.append(diapasone[temp_j])
counter-=1
temp_i+=2
temp_j-=2
i+=1
j-=1
answer_final=''
for i in answer:
answer_final+=str(i)
answer_final+=' '
print("YES")
print(answer_final[:-1])
solve()
```
Yes
| 106,213 | [
0.292724609375,
0.1173095703125,
-0.1300048828125,
0.12432861328125,
-0.60595703125,
-0.2171630859375,
-0.23974609375,
0.298828125,
0.297119140625,
0.998046875,
0.57275390625,
-0.0947265625,
-0.0594482421875,
-0.380126953125,
-0.61767578125,
0.205322265625,
-0.931640625,
-0.8388671... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
n = int(input())
if n % 2 == 0: print("NO")
elif n == 1:
print("YES")
print("1 2")
else:
print("YES")
circleTop = []
circleBottom = []
for x in range(n):
if x % 2 == 0:
circleTop.append(x*2+1)
circleBottom.append(x*2+2)
else:
circleTop.append(x*2+2)
circleBottom.append(x*2+1)
# works
print(" ".join(map(str, circleTop+circleBottom)))
```
Yes
| 106,214 | [
0.460205078125,
-0.0104827880859375,
-0.02008056640625,
-0.16943359375,
-0.483642578125,
-0.279052734375,
-0.0799560546875,
0.1971435546875,
0.27685546875,
0.94775390625,
0.68017578125,
0.049224853515625,
0.103271484375,
-0.50390625,
-0.53466796875,
0.07513427734375,
-0.91455078125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
def find(n):
ans=[0]*(2*n)
if n%2==0:
return []
temp=0
for i in range(1,n+1):
left=2*i-1
right=2*i
if temp%2==0:
ans[i-1]=left
ans[i+n-1]=right
else:
ans[i-1]=right
ans[i+n-1]=left
temp+=1
return ans
n=int(input())
ans=find(n)
if ans==[]:
print("NO")
else:
print("YES")
print(*ans)
```
Yes
| 106,215 | [
0.42724609375,
0.040557861328125,
-0.0631103515625,
-0.1534423828125,
-0.53076171875,
-0.31689453125,
-0.11212158203125,
0.138671875,
0.262939453125,
1.0595703125,
0.6806640625,
-0.01340484619140625,
0.131103515625,
-0.5009765625,
-0.55029296875,
0.07586669921875,
-0.85888671875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
size1=int(input())
ar1=[]
ar2=[]
count=1
if size1%2==0:
print('NO')
else:
for i in range((size1)):
if i%2==0:
ar1.append(count)
count+=1
ar2.append(count)
count+=1
else:
ar2.append(count)
count += 1
ar1.append(count)
count+=1
print('YES')
ar1=ar1+ar2
for i in ar1:
print(i, end=" ")
print()
```
Yes
| 106,216 | [
0.459228515625,
0.034027099609375,
-0.0223846435546875,
-0.171142578125,
-0.57666015625,
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-0.0814208984375,
0.10308837890625,
0.2325439453125,
1,
0.77294921875,
-0.01537322998046875,
0.1044921875,
-0.471923828125,
-0.55859375,
0.1148681640625,
-0.93310546875,
-0.860... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
n = int(input())
if n&1 == 0:
print("NO")
else:
j = [1]
for y in range(n+1,2*n):
j.append(y)
for y in range(2,n+1):
j.append(y)
j.append(2*n)
print("YES")
print(*j)
```
No
| 106,217 | [
0.42138671875,
0.03436279296875,
-0.031524658203125,
-0.151123046875,
-0.5888671875,
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-0.07855224609375,
0.13525390625,
0.2392578125,
1.025390625,
0.66015625,
0.02581787109375,
0.126220703125,
-0.51416015625,
-0.57275390625,
0.0684814453125,
-0.9501953125,
-0.916015... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
n=int(input())
if(n%2==1):
print("YES")
else:
print("NO")
```
No
| 106,218 | [
0.4462890625,
0.0391845703125,
-0.0660400390625,
-0.222900390625,
-0.52978515625,
-0.3828125,
-0.0789794921875,
0.1446533203125,
0.2261962890625,
1.013671875,
0.6943359375,
0.048370361328125,
0.141845703125,
-0.465087890625,
-0.60107421875,
0.0306396484375,
-0.91943359375,
-0.89062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
n = int(input())
c = 2*n
res = list(range(1,c+1))
res2=[]
print(res)
for i in range(1,int(len(res)/2)):
b = res[i]
res[i] = res[i+2]
res[i+2] = b
print(res)
for i in range(0,len(res)-2):
d = res[i]+res[i+1]+res[i+2]
res2.append(d)
for i in range(0,2):
d = res[i-2]+res[i-1]+res[i]
res2.append(d)
for i in range(0,len(res2)//n,2):
e = res2[i] -res2[i+1]
if e == -1 or e == 1:
f = 'True'
else:
print('NO')
quit()
print('YES')
print(res)
```
No
| 106,219 | [
0.428955078125,
0.0291900634765625,
-0.039459228515625,
-0.194091796875,
-0.54638671875,
-0.31103515625,
-0.12139892578125,
0.0867919921875,
0.19287109375,
1.048828125,
0.71142578125,
-0.01739501953125,
0.138916015625,
-0.4619140625,
-0.548828125,
0.0899658203125,
-0.94873046875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given integer n. You have to arrange numbers from 1 to 2n, using each of them exactly once, on the circle, so that the following condition would be satisfied:
For every n consecutive numbers on the circle write their sum on the blackboard. Then any two of written on the blackboard 2n numbers differ not more than by 1.
For example, choose n = 3. On the left you can see an example of a valid arrangement: 1 + 4 + 5 = 10, 4 + 5 + 2 = 11, 5 + 2 + 3 = 10, 2 + 3 + 6 = 11, 3 + 6 + 1 = 10, 6 + 1 + 4 = 11, any two numbers differ by at most 1. On the right you can see an invalid arrangement: for example, 5 + 1 + 6 = 12, and 3 + 2 + 4 = 9, 9 and 12 differ more than by 1.
<image>
Input
The first and the only line contain one integer n (1 β€ n β€ 10^5).
Output
If there is no solution, output "NO" in the first line.
If there is a solution, output "YES" in the first line. In the second line output 2n numbers β numbers from 1 to 2n in the order they will stay in the circle. Each number should appear only once. If there are several solutions, you can output any of them.
Examples
Input
3
Output
YES
1 4 5 2 3 6
Input
4
Output
NO
Note
Example from the statement is shown for the first example.
It can be proved that there is no solution in the second example.
Submitted Solution:
```
n = int(input());
if n % 2 == 0:
print("NO");
else:
ans = [];
for i in range(2*n):
ans.append(0);
current = 1;
for i in range(n):
ans[i] = current if i % 2 == 0 else current+1;
ans[n+i] = current+1 if i % 2 == 0 else current;
current += 2;
print("YES");
print(ans);
```
No
| 106,220 | [
0.410888671875,
0.02850341796875,
-0.073486328125,
-0.1912841796875,
-0.50634765625,
-0.335205078125,
-0.044281005859375,
0.11505126953125,
0.2498779296875,
0.9990234375,
0.72705078125,
-0.008056640625,
0.125732421875,
-0.51025390625,
-0.591796875,
0.035980224609375,
-0.9423828125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
import sys
from math import ceil
tests = int(input())
for _ in range(tests):
n, d = list(map(int, sys.stdin.readline().split()))
ok = False
for i in range(n, -1 ,-1):
result = i + ceil(d/(i + 1))
if result <= n:
print("YES")
ok = True
break
if not ok:
print("NO")
```
Yes
| 106,265 | [
0.5068359375,
0.205322265625,
-0.21484375,
0.3857421875,
-0.3515625,
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0.0172119140625,
0.0496826171875,
0.10296630859375,
0.83251953125,
0.53466796875,
-0.49169921875,
0.223876953125,
-0.876953125,
-0.57763671875,
-0.174072265625,
-0.53662109375,
-0.71142578125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
def func(x, d):
#d_div_x = d // (x + 1) if d % (x + 1) == 0 else d // (x + 1) + 1
return x + d / (x + 1)
# 14
def ternary_search_min(l, r, d):
while r - l >= 3:
m1 = int(r - 2 * (r - l) / 3)
m2 = int(r - (r - l) / 3)
if func(m1, d) <= func(m2, d):
r = m2
else:
l = m1
remain_args = [i for i in range(l, r + 1)]
func_values = list(map(lambda x: func(x, d), remain_args))
return min(func_values)
test_size = int(input())
for i in range(test_size):
n, d = map(int, input().split())
min_days = ternary_search_min(0, n, d)
if min_days <= n or d <= n:
print("YES")
else:
print("NO")
```
Yes
| 106,266 | [
0.5478515625,
0.2127685546875,
-0.190673828125,
0.34375,
-0.311279296875,
-0.1839599609375,
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0.036407470703125,
0.08087158203125,
0.83837890625,
0.53515625,
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0.225830078125,
-0.84716796875,
-0.5166015625,
-0.162353515625,
-0.59716796875,
-0.6904296... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
from math import sqrt
for w in range(int(input())):
n,d = map(int,input().split())
print('yes') if 2*sqrt(d) <= n+1 else print('no')
```
Yes
| 106,267 | [
0.52392578125,
0.19091796875,
-0.244140625,
0.386474609375,
-0.3232421875,
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0.02191162109375,
0.00521087646484375,
0.09716796875,
0.87548828125,
0.5751953125,
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0.2078857421875,
-0.89990234375,
-0.54296875,
-0.1483154296875,
-0.5830078125,
-0.696777343... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
import math
t=int(input())
for _ in range(t):
n,d=map(int,input().split())
x=n//2
y=math.ceil(d/(x+1))
if(x+y<=n):
print('YES')
else:
print('NO')
```
Yes
| 106,268 | [
0.55712890625,
0.1611328125,
-0.225341796875,
0.346435546875,
-0.312255859375,
-0.1448974609375,
0.0216064453125,
0.01715087890625,
0.10504150390625,
0.87548828125,
0.54638671875,
-0.4912109375,
0.193359375,
-0.8798828125,
-0.52978515625,
-0.1710205078125,
-0.556640625,
-0.68164062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
for t in range(int(input())):
n, d = [int(i) for i in input().split()]
if d <= n:
print("YES")
elif (n-1)**2 + 4*(n-d) > 0:
print("YES")
else:
print("NO")
```
No
| 106,269 | [
0.5458984375,
0.18603515625,
-0.188720703125,
0.3525390625,
-0.313720703125,
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0.0033130645751953125,
0.02447509765625,
0.08856201171875,
0.86865234375,
0.552734375,
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0.1900634765625,
-0.88818359375,
-0.515625,
-0.1614990234375,
-0.5625,
-0.7080078125,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
import math
t=int(input())
for i in range(t):
n,d=map(int,input().split())
if d<=n:
print("YES")
else:
m=int(pow(d,0.5))
z=m+math.ceil(d/(m+1))
#print(m)
if z<=n:
print("YES")
else:
print("NO")
```
No
| 106,270 | [
0.56201171875,
0.166015625,
-0.2281494140625,
0.342529296875,
-0.31591796875,
-0.14697265625,
0.01910400390625,
0.01666259765625,
0.10467529296875,
0.87451171875,
0.5458984375,
-0.484130859375,
0.195556640625,
-0.87158203125,
-0.52587890625,
-0.1727294921875,
-0.5625,
-0.6879882812... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
import math
for _ in range(int(input())):
n,d=map(int,input().split())
if n>=d:
print("YES")
else:
g=1
for i in range(1,n+1):
if math.ceil((d/(i+1)))+i<=n:
k=0
print("YES")
break
if g==1:
print("NO")
```
No
| 106,271 | [
0.5439453125,
0.1676025390625,
-0.230712890625,
0.36083984375,
-0.306396484375,
-0.156005859375,
0.038116455078125,
0.0236663818359375,
0.10186767578125,
0.8681640625,
0.56787109375,
-0.50439453125,
0.22021484375,
-0.90673828125,
-0.5419921875,
-0.155029296875,
-0.5546875,
-0.67480... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Adilbek was assigned to a special project. For Adilbek it means that he has n days to run a special program and provide its results. But there is a problem: the program needs to run for d days to calculate the results.
Fortunately, Adilbek can optimize the program. If he spends x (x is a non-negative integer) days optimizing the program, he will make the program run in \leftβ (d)/(x + 1) \rightβ days (\leftβ a \rightβ is the ceiling function: \leftβ 2.4 \rightβ = 3, \leftβ 2 \rightβ = 2). The program cannot be run and optimized simultaneously, so the total number of days he will spend is equal to x + \leftβ (d)/(x + 1) \rightβ.
Will Adilbek be able to provide the generated results in no more than n days?
Input
The first line contains a single integer T (1 β€ T β€ 50) β the number of test cases.
The next T lines contain test cases β one per line. Each line contains two integers n and d (1 β€ n β€ 10^9, 1 β€ d β€ 10^9) β the number of days before the deadline and the number of days the program runs.
Output
Print T answers β one per test case. For each test case print YES (case insensitive) if Adilbek can fit in n days or NO (case insensitive) otherwise.
Example
Input
3
1 1
4 5
5 11
Output
YES
YES
NO
Note
In the first test case, Adilbek decides not to optimize the program at all, since d β€ n.
In the second test case, Adilbek can spend 1 day optimizing the program and it will run \leftβ 5/2 \rightβ = 3 days. In total, he will spend 4 days and will fit in the limit.
In the third test case, it's impossible to fit in the limit. For example, if Adilbek will optimize the program 2 days, it'll still work \leftβ (11)/(2+1) \rightβ = 4 days.
Submitted Solution:
```
import math
t = int(input())
while t:
t-=1
n,d = map(int,input().split())
if d<=n or 1+math.ceil(d/(2))<=n or 2+math.ceil(d/3)<=n:
print("YES")
else:
print("NO")
```
No
| 106,272 | [
0.57275390625,
0.1922607421875,
-0.2353515625,
0.327392578125,
-0.31103515625,
-0.1436767578125,
0.019683837890625,
-0.00048470497131347656,
0.101318359375,
0.84912109375,
0.5283203125,
-0.472900390625,
0.2042236328125,
-0.8642578125,
-0.5341796875,
-0.1778564453125,
-0.55029296875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
2 5 4
5 7
4 8
Output
8 7
Submitted Solution:
```
N, A, B = map(int, input().split())
P = [list(map(int, input().split())) for i in range(N)]
C = 153
memo = {}
def dfs(i, turn, A, B):
if (i, turn, A, B) in memo:
return memo[i, turn, A, B]
if i == N:
return 0
r, s = P[i]
res = 0
if turn == 1:
# B
res = dfs(i+1, 0, A, min(B + r, C))
if B:
res = min(res, dfs(i, 0, A, B-1))
else:
# A
res = dfs(i+1, 1, min(A + r, C), B) + s
if A:
res = max(res, dfs(i, 1, A-1, B))
memo[i, turn, A, B] = res
return res
S = sum(s for r, s in P)
res = dfs(0, 0, min(A, C), min(B, C))
print(res, S-res)
```
No
| 106,973 | [
0.438720703125,
0.007015228271484375,
0.0936279296875,
-0.107177734375,
-0.9052734375,
-0.265380859375,
0.01345062255859375,
0.382568359375,
-0.06671142578125,
1.041015625,
0.33349609375,
0.009063720703125,
0.277099609375,
-0.5302734375,
-0.60595703125,
-0.1614990234375,
-0.597167968... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
T = int(input())
inputs = [''] * T
for i in range(T):
inputs[i] = input()
for s in inputs:
n, x = map(int, s.split())
print(2 * x)
```
Yes
| 107,119 | [
0.190185546875,
0.1221923828125,
-0.152099609375,
0.251953125,
-0.52978515625,
-0.47802734375,
-0.09576416015625,
-0.01549530029296875,
0.384521484375,
0.8720703125,
0.333251953125,
0.1885986328125,
-0.1357421875,
-0.65869140625,
-0.498291015625,
0.2332763671875,
-0.426513671875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
T = int(input())
for _ in range(T):
n, x = map(int,input().split())
print(2*x)
```
Yes
| 107,120 | [
0.1717529296875,
0.10198974609375,
-0.1497802734375,
0.267822265625,
-0.51123046875,
-0.4921875,
-0.06976318359375,
-0.0042572021484375,
0.376953125,
0.8916015625,
0.336181640625,
0.175537109375,
-0.142822265625,
-0.64794921875,
-0.501953125,
0.23974609375,
-0.426025390625,
-0.8442... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
for _ in range(t):
l = list(map(int,input().split()))
print(l[1]<<1)
```
Yes
| 107,121 | [
0.1724853515625,
0.0960693359375,
-0.134521484375,
0.26123046875,
-0.5224609375,
-0.494384765625,
-0.07965087890625,
-0.003570556640625,
0.387451171875,
0.8662109375,
0.313232421875,
0.172607421875,
-0.1243896484375,
-0.62109375,
-0.52001953125,
0.23876953125,
-0.44189453125,
-0.84... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
for i in range(t):
n, x = list(map(int, input().split(" ")))
prox = x * 2
print(prox)
```
Yes
| 107,122 | [
0.2359619140625,
0.12060546875,
-0.1353759765625,
0.304443359375,
-0.53125,
-0.5654296875,
-0.1124267578125,
0.0201263427734375,
0.42724609375,
0.8720703125,
0.351806640625,
0.176025390625,
-0.1314697265625,
-0.6435546875,
-0.50048828125,
0.265380859375,
-0.40625,
-0.84716796875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
def main_function():
t = int(input())
for i in range(t):
a, b = [int(i) for i in input().split(" ")]
return b * 2
print(main_function())
```
No
| 107,123 | [
0.1871337890625,
0.08441162109375,
-0.134765625,
0.2568359375,
-0.5185546875,
-0.492431640625,
-0.0858154296875,
-0.0178070068359375,
0.3935546875,
0.837890625,
0.33642578125,
0.1533203125,
-0.16796875,
-0.62109375,
-0.541015625,
0.245849609375,
-0.47802734375,
-0.8701171875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
noOfTestCases = int(input())
number, position = map(int, input().split(" "))
print(position*2)
```
No
| 107,124 | [
0.17138671875,
0.1270751953125,
-0.11993408203125,
0.266845703125,
-0.464599609375,
-0.669921875,
-0.1248779296875,
0.0296173095703125,
0.2978515625,
0.9013671875,
0.403564453125,
0.03424072265625,
-0.037109375,
-0.6142578125,
-0.52001953125,
0.160400390625,
-0.369140625,
-0.811035... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
t = int(input())
[n, x] = [int(v) for v in input().split()]
print(2 * x)
```
No
| 107,125 | [
0.20068359375,
0.10626220703125,
-0.142822265625,
0.273681640625,
-0.53955078125,
-0.5048828125,
-0.056671142578125,
0.017974853515625,
0.395751953125,
0.8896484375,
0.336181640625,
0.165283203125,
-0.176513671875,
-0.64794921875,
-0.52490234375,
0.22119140625,
-0.41748046875,
-0.8... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a list of numbers from 1 to n written from left to right on the blackboard.
You perform an algorithm consisting of several steps (steps are 1-indexed). On the i-th step you wipe the i-th number (considering only remaining numbers). You wipe the whole number (not one digit).
<image>
When there are less than i numbers remaining, you stop your algorithm.
Now you wonder: what is the value of the x-th remaining number after the algorithm is stopped?
Input
The first line contains one integer T (1 β€ T β€ 100) β the number of queries. The next T lines contain queries β one per line. All queries are independent.
Each line contains two space-separated integers n and x (1 β€ x < n β€ 10^{9}) β the length of the list and the position we wonder about. It's guaranteed that after the algorithm ends, the list will still contain at least x numbers.
Output
Print T integers (one per query) β the values of the x-th number after performing the algorithm for the corresponding queries.
Example
Input
3
3 1
4 2
69 6
Output
2
4
12
Submitted Solution:
```
try:
T=int(input())
l1=[]
for i in range(T):
n,x=map(int,input().split())
l=list(range(1,n+1))
if n>x:
for i in range(n):
if i<len(l):
l.remove(l[i])
else:
break
l1.append(l[x-1])
else:
pass
for i in l1:
print(i)
except:
pow
```
No
| 107,126 | [
0.10101318359375,
0.0268096923828125,
-0.06292724609375,
0.311767578125,
-0.56787109375,
-0.42626953125,
-0.129638671875,
0.0225372314453125,
0.434814453125,
0.8427734375,
0.357666015625,
0.1590576171875,
-0.1346435546875,
-0.6826171875,
-0.53369140625,
0.2293701171875,
-0.4592285156... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.
Iahub asks Iahubina: can you build a rooted tree, such that
* each internal node (a node with at least one son) has at least two sons;
* node i has ci nodes in its subtree?
Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.
Input
The first line of the input contains integer n (1 β€ n β€ 24). Next line contains n positive integers: the i-th number represents ci (1 β€ ci β€ n).
Output
Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).
Examples
Input
4
1 1 1 4
Output
YES
Input
5
1 1 5 2 1
Output
NO
Submitted Solution:
```
import fileinput
def solve(n, v):
v.sort()
if v[-1] != n: return 0
f = [0 for _ in range(n)]
for i in range(n):
if v[i] == 1: continue
k, c = v[i]-1, 0
for j in reversed(range(i)):
if not f[j] and v[j] <= k:
if not c and v[j] == k: continue
f[j] = 1
k -= v[j]
c += 1
if not k: break
if k: return 0
return 1
f = fileinput.input()
n = int(f.readline())
v = list(map(int, f.readline().rstrip().split()))
print("YES" if solve(n, v) else "NO")
```
No
| 107,332 | [
0.386962890625,
0.30859375,
-0.0188140869140625,
-0.1890869140625,
-0.1771240234375,
-0.0034351348876953125,
-0.4306640625,
0.71630859375,
0.4169921875,
0.7822265625,
0.46533203125,
-0.09521484375,
0.279541015625,
-0.71923828125,
-0.409423828125,
0.019683837890625,
-0.492431640625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.
Iahub asks Iahubina: can you build a rooted tree, such that
* each internal node (a node with at least one son) has at least two sons;
* node i has ci nodes in its subtree?
Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.
Input
The first line of the input contains integer n (1 β€ n β€ 24). Next line contains n positive integers: the i-th number represents ci (1 β€ ci β€ n).
Output
Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).
Examples
Input
4
1 1 1 4
Output
YES
Input
5
1 1 5 2 1
Output
NO
Submitted Solution:
```
def DFS(x):
for i in range(x):
if(Seen[i]):
continue
if(Rem[i]>=C[x]):
if(Rem[i]==C[x] and len(Children[i])==0):
continue
Rem[i]-=C[x]
Parent[x]=i
Children[i].append(x)
return True
for i in range(x):
if(Seen[i]):
continue
Y=[]
Seen[i]=True
for j in range(len(Children[i])):
child=Children[i][j]
Parent[child]=-1
Rem[i]+=C[child]
if(DFS(child)):
continue
Parent[child]=i
Rem[i]-=C[child]
Y.append(child)
Seen[i]=False
Children[i]=list(Y)
if(Rem[i]>=C[x]):
if(Rem[i]==C[x] and len(Children[i])==0):
continue
Rem[i]-=C[x]
Children[i].append(x)
Parent[x]=i
return True
return False
n=int(input())
C=list(map(int,input().split()))
Rem=[-1]*n
Parent=[-1]*n
Children=[]
Seen=[False]*n
C.sort(reverse=True)
if(C[0]!=n or C.count(2)>0):
print("NO")
else:
for i in range(n):
Rem[i]=C[i]-1
Children.append([])
Parent[0]=0
Ans="YES"
for i in range(1,n):
if(DFS(i)==False):
Ans="NO"
break
for i in range(n):
if(Rem[i]!=0 and C[i]!=1):
Ans="NO"
break
print(Ans)
```
No
| 107,333 | [
0.401611328125,
0.27392578125,
-0.05194091796875,
-0.16162109375,
-0.189208984375,
0.0271148681640625,
-0.374755859375,
0.7490234375,
0.420166015625,
0.7529296875,
0.489501953125,
-0.110107421875,
0.281982421875,
-0.63818359375,
-0.432861328125,
0.0279998779296875,
-0.487060546875,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.
Iahub asks Iahubina: can you build a rooted tree, such that
* each internal node (a node with at least one son) has at least two sons;
* node i has ci nodes in its subtree?
Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.
Input
The first line of the input contains integer n (1 β€ n β€ 24). Next line contains n positive integers: the i-th number represents ci (1 β€ ci β€ n).
Output
Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).
Examples
Input
4
1 1 1 4
Output
YES
Input
5
1 1 5 2 1
Output
NO
Submitted Solution:
```
def DFS(x):
for i in range(x):
if(Seen[i]):
continue
if(Rem[i]>=C[x]):
Rem[i]-=C[x]
Parent[x]=i
Children[i].append(x)
return True
for i in range(x):
if(Seen[i]):
continue
Y=[]
Seen[i]=True
for j in range(len(Children[i])):
child=Children[i][j]
Parent[child]=-1
Rem[i]+=C[child]
if(DFS(child)):
continue
Parent[child]=i
Rem[i]-=C[child]
Y.append(child)
Seen[i]=False
Children[i]=list(Y)
if(Rem[i]>=C[x]):
Rem[i]-=C[x]
Children[i].append(x)
Parent[x]=i
return True
return False
n=int(input())
C=list(map(int,input().split()))
Rem=[-1]*n
Parent=[-1]*n
Children=[]
Seen=[False]*n
C.sort(reverse=True)
if(C[0]!=n or C.count(2)>0):
print("NO")
else:
for i in range(n):
Rem[i]=C[i]-1
Children.append([])
Parent[0]=0
Ans="YES"
for i in range(1,n):
if(DFS(i)==False):
Ans="NO"
break
for i in range(n):
if(Rem[i]!=0 and C[i]!=1):
Ans="NO"
break
print(Ans)
```
No
| 107,334 | [
0.388427734375,
0.268798828125,
-0.03472900390625,
-0.179931640625,
-0.1998291015625,
0.021575927734375,
-0.38330078125,
0.73388671875,
0.439697265625,
0.75341796875,
0.496826171875,
-0.11932373046875,
0.29345703125,
-0.638671875,
-0.442626953125,
0.0146331787109375,
-0.48583984375,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Iahub and Iahubina went to a picnic in a forest full of trees. Less than 5 minutes passed before Iahub remembered of trees from programming. Moreover, he invented a new problem and Iahubina has to solve it, otherwise Iahub won't give her the food.
Iahub asks Iahubina: can you build a rooted tree, such that
* each internal node (a node with at least one son) has at least two sons;
* node i has ci nodes in its subtree?
Iahubina has to guess the tree. Being a smart girl, she realized that it's possible no tree can follow Iahub's restrictions. In this way, Iahub will eat all the food. You need to help Iahubina: determine if there's at least one tree following Iahub's restrictions. The required tree must contain n nodes.
Input
The first line of the input contains integer n (1 β€ n β€ 24). Next line contains n positive integers: the i-th number represents ci (1 β€ ci β€ n).
Output
Output on the first line "YES" (without quotes) if there exist at least one tree following Iahub's restrictions, otherwise output "NO" (without quotes).
Examples
Input
4
1 1 1 4
Output
YES
Input
5
1 1 5 2 1
Output
NO
Submitted Solution:
```
def can_be_a_tree(A, n):
for i in A:
if i == 2:
return False
return sum(A) == 2 * n - 1
n = int(input())
A = map(int, input().split())
if can_be_a_tree(A, n):
print("YES")
else:
print("NO")
```
No
| 107,335 | [
0.456787109375,
0.29833984375,
-0.1783447265625,
-0.1685791015625,
-0.1815185546875,
-0.01503753662109375,
-0.348876953125,
0.74365234375,
0.42822265625,
0.7802734375,
0.4765625,
-0.08843994140625,
0.340576171875,
-0.68310546875,
-0.485107421875,
0.06988525390625,
-0.51904296875,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
a,b = list(map(int,input().split()))
if a*2 <= b <= a*4 and b%2 == 0:
print("Yes")
else:
print("No")
```
Yes
| 107,630 | [
0.409912109375,
0.2802734375,
-0.041046142578125,
0.08502197265625,
-0.63427734375,
-0.407470703125,
0.10198974609375,
0.623046875,
0.366455078125,
0.61962890625,
0.544921875,
-0.12744140625,
0.056915283203125,
-0.292236328125,
-0.48828125,
-0.08551025390625,
-0.54833984375,
-0.508... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
X,Y=map(int,input().split())
if(Y%2==0 and X*4>=Y and X*2<=Y):
print('Yes')
else:
print('No')
```
Yes
| 107,631 | [
0.414306640625,
0.313232421875,
-0.0364990234375,
0.08099365234375,
-0.61474609375,
-0.43017578125,
0.09173583984375,
0.60546875,
0.385986328125,
0.61767578125,
0.564453125,
-0.10498046875,
0.054779052734375,
-0.305908203125,
-0.453369140625,
-0.1416015625,
-0.544921875,
-0.5068359... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
#b
x,y=map(int, input().split())
if x*2 <= y <=x*4 and y%2 == 0:
print("Yes")
else:
print("No")
```
Yes
| 107,632 | [
0.420166015625,
0.324462890625,
-0.0240936279296875,
0.0784912109375,
-0.61279296875,
-0.416748046875,
0.0986328125,
0.6171875,
0.36181640625,
0.642578125,
0.5625,
-0.10174560546875,
0.061126708984375,
-0.316650390625,
-0.448974609375,
-0.10943603515625,
-0.5185546875,
-0.520019531... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are some animals in a garden. Each of them is a crane with two legs or a turtle with four legs.
Takahashi says: "there are X animals in total in the garden, and they have Y legs in total." Determine whether there is a combination of numbers of cranes and turtles in which this statement is correct.
Constraints
* 1 \leq X \leq 100
* 1 \leq Y \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
X Y
Output
If there is a combination of numbers of cranes and turtles in which the statement is correct, print `Yes`; otherwise, print `No`.
Examples
Input
3 8
Output
Yes
Input
2 100
Output
No
Input
1 2
Output
Yes
Submitted Solution:
```
a, b = map(int, input().split())
print("Yes" if b%2==0 and 2*a<=b<=4*a else "No")
```
Yes
| 107,633 | [
0.4375,
0.299072265625,
-0.0367431640625,
0.06683349609375,
-0.6142578125,
-0.420166015625,
0.104736328125,
0.6162109375,
0.345458984375,
0.6162109375,
0.56884765625,
-0.1153564453125,
0.025360107421875,
-0.298583984375,
-0.480224609375,
-0.11572265625,
-0.55322265625,
-0.516113281... | 11 |
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