message stringlengths 2 67k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 463 109k | cluster float64 19 19 | __index_level_0__ int64 926 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,969 | 19 | 201,938 |
"Correct Solution:
```
N, K = map(int, input().split())
V = list(map(int, input().split()))
ans = 0
KN = min(K,N)
for i in range(1,KN+1):
for j in range(i+1):
rec = V[:i-j]+V[N-j:]
rec.sort(reverse=True)
away = max(min(i, K-i),0)
for l in range(away+1):
recsum = sum(rec[:i-l])
if recsum > ans:
ans=recsum
print(ans)
``` | output | 1 | 100,969 | 19 | 201,939 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,970 | 19 | 201,940 |
"Correct Solution:
```
n, k = map(int, input().split())
vs = list(map(int, input().split()))
m = 0
r = min(n, k)
for a in range(r+1):
for b in range(r-a+1):
cd = k - (a + b)
if b == 0:
hold = sorted(vs[:a])
else:
hold = sorted(vs[:a] + vs[-b:])
ap = hold[min(len([1 for x in hold if x < 0]), cd):]
s = sum(ap)
m = max(s, m)
print(m)
``` | output | 1 | 100,970 | 19 | 201,941 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,971 | 19 | 201,942 |
"Correct Solution:
```
n,k=map(int,input().split())
v=list(map(int,input().split()))
ans=0
for a in range(50):
for b in range(50):
if a+b <= min(n,k):
li=v[:a]+v[n-b:]
sumv = sum(li)
li.sort()
for i in range(min(k-a-b,len(li))):
if li[i] < 0:
li[i] = 0
else:
break
ans=max(ans,sum(li))
print(ans)
``` | output | 1 | 100,971 | 19 | 201,943 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,972 | 19 | 201,944 |
"Correct Solution:
```
N, K = [int(i) for i in input().split()]
V = [int(i) for i in input().split()]
M = min(N, K)
ma = 0
for l in range(M + 1):
for r in range(M - l + 1):
G = V[:l] + V[N - r:]
G.sort()
for i in range(K - l - r + 1):
ma = max(ma, sum(G[i:]))
print(ma)
``` | output | 1 | 100,972 | 19 | 201,945 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,973 | 19 | 201,946 |
"Correct Solution:
```
n,k = map(int,input().split())
v = list(map(int,input().split()))
ans = 0
for i in range(1,n+1):
vt = v[i:][::-1] + v[:i]
# print(vt)
for j in range(0, min(n,k)+1):
tmp = sorted(vt[:j])
tmp_n = sum(tmp)
for l in range(min(k-j, j)):
if(tmp[l] < 0):
tmp_n -= tmp[l]
ans = max(ans, tmp_n)
print(ans)
``` | output | 1 | 100,973 | 19 | 201,947 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,974 | 19 | 201,948 |
"Correct Solution:
```
N, K = map(int, input().split())
V = list(map(int, input().split()))
ans = float('-inf')
for l in range(min(N+1, K+1)):
for r in range(min(N-l+1, K-l+1)):
arr = V[:l] + V[N-r:]
t = sum(arr)
arr.sort()
for i in range(min(l+r, K-l-r)):
a = arr[i]
if a >= 0:
break
t -= a
ans = ans if ans > t else t
print(ans)
``` | output | 1 | 100,974 | 19 | 201,949 |
Provide a correct Python 3 solution for this coding contest problem.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0 | instruction | 0 | 100,975 | 19 | 201,950 |
"Correct Solution:
```
n,k = map(int,input().split())
v = list(map(int,input().split()))
ma = 0
kaisu = min(n,k)
for a in range(kaisu+1):
for b in range(kaisu-a+1):
push = k-(a+b)
li = v[:a]+v[n-b:]
li.sort(reverse=True)
for c in range(push):
if li:
if li[-1] < 0:
li.pop()
ma = max(ma,sum(li))
print(ma)
``` | output | 1 | 100,975 | 19 | 201,951 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k=map(int,input().split())
v=list(map(int,input().split()))
ans=-10**18
for l in range(min(n,k)+1):
for r in range(min(n,k)+1):
if l+r>n or l+r>k:
continue
trg=v[:l]+v[n-r:]
trg.sort(reverse=True)
for i in range(k-l-r+1):
ans=max(ans,sum(trg[:len(trg)-i]))
print(ans)
``` | instruction | 0 | 100,976 | 19 | 201,952 |
Yes | output | 1 | 100,976 | 19 | 201,953 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k,*v=map(int,open(0).read().split())
m=0
r=range
for i in r(n+1):
for j in r(min(k,n)-i+1):
t=sorted(v[:i]+v[-j:]*(j>0))[::-1]
while t and(k-i-j)*t[-1]<0:t.pop();j+=1
m=max(m,sum(t))
print(m)
``` | instruction | 0 | 100,977 | 19 | 201,954 |
Yes | output | 1 | 100,977 | 19 | 201,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k = map(int,input().split())
lst = list(map(int,input().split()))
ans = 0
m = min(n,k)
for a in range(m+1):
for b in range(m+1-a):
jewel = lst[:a] + lst[n-b:]
jewel.sort()
for c in range(k+1-a-b):
ans = max(ans, sum(jewel[c:]))
print(ans)
``` | instruction | 0 | 100,978 | 19 | 201,956 |
Yes | output | 1 | 100,978 | 19 | 201,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k = map(int,input().split())
v = [int(x) for x in input().split()]
ans = 0
for l in range(n+1):
for r in range(l,max(n,l+1)):
m = v[:l]+v[r+1:]
if len(m) > k:
continue
m.sort()
i=0
while i < len(m) and i < k - len(m) and m[i] < 0:
i+=1
ans = max(ans,sum(m[i:]))
print(ans)
``` | instruction | 0 | 100,979 | 19 | 201,958 |
Yes | output | 1 | 100,979 | 19 | 201,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k=map(int,input().split())
v=list(map(int,input().split()))
#k回行える
nk=min(n,k)
ll=[-10**9]*max(n,k)
rr=[-10**9]*max(n,k)
#左側から確認
mm=[]
pp=[]
import heapq # heapqライブラリのimport
#heapq.heapify(mm) # リストを優先度付きキューへ
#print(heapq.heappop(a)) # 最小値の取り出し
#heapq.heappush(a, -2)
import copy
for i in range(nk):
#print("i#########",i)
cnt=i
if v[i]>=0:
pp.append(v[i])
else:
mm.append(v[i])
heapMm = copy.deepcopy(mm)
heapq.heapify(heapMm)
hoge=sum(pp)+sum(mm)
#一発入れてみる
ll[cnt]=max(ll[cnt],hoge)
#print(ll)
#print(heapMm)
while len(list(heapMm))>0:
tmp=heapq.heappop(heapMm)
cnt+=1
if cnt>=k:
break
hoge-=tmp
ll[cnt]=max(ll[cnt],hoge)
#print(ll)
for i in range(1,len(ll)):
if ll[i-1]>ll[i]:
ll[i]=ll[i-1]
#print(ll)
##右側から
pp=[]
mm=[]
v=v[::-1]
for i in range(nk):
#print("i#########",i)
cnt=i
if v[i]>=0:
pp.append(v[i])
else:
mm.append(v[i])
heapMm = copy.deepcopy(mm)
heapq.heapify(heapMm)
hoge=sum(pp)+sum(mm)
#一発入れてみる
rr[cnt]=max(rr[cnt],hoge)
#print(rr)
#print(heapMm)
while len(list(heapMm))>0:
tmp=heapq.heappop(heapMm)
cnt+=1
if cnt>=k:
break
hoge-=tmp
rr[cnt]=max(rr[cnt],hoge)
#print(rr)
for i in range(1,len(rr)):
if rr[i-1]>rr[i]:
rr[i]=rr[i-1]
ll=[0]+ll
rr=[0]+rr
result=-10**9
for i in range(nk+1):
result=max(result,ll[i]+rr[nk-i])
print(max(result,0))
``` | instruction | 0 | 100,980 | 19 | 201,960 |
No | output | 1 | 100,980 | 19 | 201,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sun Jun 2 14:23:16 2019
@author: hisaki
"""
#宝石問題を得くプログラム
decision = []
def func(recursive_num, myList, ValueList):
global Sum
global K
#得点計算
temp = sum(myList)
if temp > Sum:
Sum = temp
if recursive_num == K:
return
if ValueList:
save_myList = myList[:]
save_ValueList = ValueList[:]
# myList.append(ValueList.pop(0))
save_myList.append(save_ValueList.pop(0))
func(recursive_num+1,save_myList,save_ValueList)
# ValueList.insert(0,myList.pop(-1))#もとに戻す
if ValueList:
# myList.append(ValueList.pop(-1))
save_myList = myList[:]
save_ValueList = ValueList[:]
save_myList.append(save_ValueList.pop(-1))
func(recursive_num+1, save_myList, save_ValueList)
# ValueList.insert(-1,myList.pop(-1))#もとに戻す
if myList:
temp_index = myList.index(min(myList))
# ValueList.insert(0,myList.pop(temp_index))
save_myList = myList[:]
save_ValueList = ValueList[:]
save_ValueList.insert(0,save_myList.pop(temp_index))
func(recursive_num+1, save_myList, save_ValueList)
# myList.append(ValueList.pop(0))
if myList:
temp_index = myList.index(min(myList))
# ValueList.insert(-1,myList.pop(temp_index))
save_myList = myList[:]
save_ValueList = ValueList[:]
save_ValueList.insert(-1,save_myList.pop(temp_index))
func(recursive_num+1, save_myList, save_ValueList)
# myList.append(ValueList.pop(-1))
return
N, K = map(int, input().split())
ValueList = list(map(int, input().split()))
recursive_num = 0
myList=[]
Sum=0
func(recursive_num, myList, ValueList)
print (Sum)
``` | instruction | 0 | 100,981 | 19 | 201,962 |
No | output | 1 | 100,981 | 19 | 201,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
n,k = map(int,readline().split())
lst1 = list(map(int,readline().split()))
ans = 0
for a in range(n+1):
for b in range(n+1):
if a+b >= min(n,k):
continue
left = lst1[:a]
right = lst1[n-b:]
res = left + right
rest = k - (a + b)
if rest > 0:
res.sort()
while rest:
if len(res) == 0:
break
if res[0] < 0:
del res[0]
rest -=1
else:
break
ans = max(ans,sum(res))
print(ans)
``` | instruction | 0 | 100,982 | 19 | 201,964 |
No | output | 1 | 100,982 | 19 | 201,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Your friend gave you a dequeue D as a birthday present.
D is a horizontal cylinder that contains a row of N jewels.
The values of the jewels are V_1, V_2, ..., V_N from left to right. There may be jewels with negative values.
In the beginning, you have no jewel in your hands.
You can perform at most K operations on D, chosen from the following, at most K times (possibly zero):
* Operation A: Take out the leftmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation B: Take out the rightmost jewel contained in D and have it in your hand. You cannot do this operation when D is empty.
* Operation C: Choose a jewel in your hands and insert it to the left end of D. You cannot do this operation when you have no jewel in your hand.
* Operation D: Choose a jewel in your hands and insert it to the right end of D. You cannot do this operation when you have no jewel in your hand.
Find the maximum possible sum of the values of jewels in your hands after the operations.
Constraints
* All values in input are integers.
* 1 \leq N \leq 50
* 1 \leq K \leq 100
* -10^7 \leq V_i \leq 10^7
Input
Input is given from Standard Input in the following format:
N K
V_1 V_2 ... V_N
Output
Print the maximum possible sum of the values of jewels in your hands after the operations.
Examples
Input
6 4
-10 8 2 1 2 6
Output
14
Input
6 4
-6 -100 50 -2 -5 -3
Output
44
Input
6 3
-6 -100 50 -2 -5 -3
Output
0
Submitted Solution:
```
n,k=map(int,input().split())
v=list(map(int,input().split()))
q=[[0,0,n-1,[]]]
qq=[]
ans=0
a=min(k,n-1)
while len(q)>0:
#print(q)
qi=q.pop()
#print(qi)
if qi[0]==a or qi[1]==qi[2]+1:
continue
q+=[[qi[0]+1,qi[1]+1,qi[2],qi[3]+[v[qi[1]]]]]
q+=[[qi[0]+1,qi[1],qi[2]-1,qi[3]+[v[qi[2]]]]]
qq+=[[qi[0]+1,qi[1]+1,qi[2],qi[3]+[v[qi[1]]]]]
qq+=[[qi[0]+1,qi[1],qi[2]-1,qi[3]+[v[qi[2]]]]]
if k>=n:
qq+=[[n,v]]
while len(qq)>0:
qi=qq.pop()
ans=max(ans,sum(qi[3]))
if qi[0]==k:
continue
if len(qi[3])==0:
continue
if len(qi[3])==1:
continue
#print(qi)
qi[3].remove(min(qi[3]))
qq+=[[qi[0]+1,qi[1],qi[2],qi[3]]]
#print(qi)
print(ans)
``` | instruction | 0 | 100,983 | 19 | 201,966 |
No | output | 1 | 100,983 | 19 | 201,967 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,792 | 19 | 203,584 |
"Correct Solution:
```
a, b, c = map(int, input().split())
k = int(input())
while b <= a:
b = b * 2
k -= 1
while c <= b:
c = c * 2
k -= 1
print('Yes' if k >= 0 else 'No')
``` | output | 1 | 101,792 | 19 | 203,585 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,793 | 19 | 203,586 |
"Correct Solution:
```
A,B,C = list(map(int,input().split()))
K = int(input())
while A>=B:
B*=2
K-=1
while B>=C:
C*=2
K-=1
if K>=0:
print("Yes")
else:
print("No")
``` | output | 1 | 101,793 | 19 | 203,587 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,794 | 19 | 203,588 |
"Correct Solution:
```
a,b,c=map(int,input().split())
k=int(input())
ans=0
while(b<=a):
b<<=1
ans+=1
while(c<=b):
c<<=1
ans+=1
if ans<=k:
print("Yes")
else:
print("No")
``` | output | 1 | 101,794 | 19 | 203,589 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,795 | 19 | 203,590 |
"Correct Solution:
```
a,b,c=map(int,input().split())
k=int(input())
cnt=0
while a>=b:
b*=2
cnt+=1
while b>=c:
c*=2
cnt+=1
if cnt<=k:
print("Yes")
else:
print("No")
``` | output | 1 | 101,795 | 19 | 203,591 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,796 | 19 | 203,592 |
"Correct Solution:
```
a, b, c = map(int, input().split())
k = int(input())
while b <= a:
k -= 1
b *= 2
while c <= b:
k -= 1
c *= 2
print("Yes" if k>=0 else "No")
``` | output | 1 | 101,796 | 19 | 203,593 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,797 | 19 | 203,594 |
"Correct Solution:
```
a,b,c=map(int,input().split())
k=int(input())
x=0
while a>=b:
b=b*2
x=x+1
while b>=c:
c=c*2
x=x+1
if k>=x:
print('Yes')
else:
print('No')
``` | output | 1 | 101,797 | 19 | 203,595 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,798 | 19 | 203,596 |
"Correct Solution:
```
a,b,c=map(int,input().split())
k=int(input())
while b<=a:
b*=2
k-=1
while c<=b:
c*=2
k-=1
print("No")if k<0 else print("Yes")
``` | output | 1 | 101,798 | 19 | 203,597 |
Provide a correct Python 3 solution for this coding contest problem.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No | instruction | 0 | 101,799 | 19 | 203,598 |
"Correct Solution:
```
a,b,c=map(int,input().split())
k=int(input())
n=0
while a>=b:
b*=2
n+=1
while b>=c:
c*=2
n+=1
print('YNeos'[n>k::2])
``` | output | 1 | 101,799 | 19 | 203,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
a,b,c = map(int, input().split())
k = int(input())
t = 0
while(a>=b):
b = 2*b
t +=1
while(b>=c):
c = 2*c
t +=1
if(t<=k):
print("Yes")
else:
print("No")
``` | instruction | 0 | 101,800 | 19 | 203,600 |
Yes | output | 1 | 101,800 | 19 | 203,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
a, b, c = map(int, input().split())
k = int(input())
for i in range(k):
if(a >= b): b *= 2
elif(b >= c): c *= 2
if(a < b < c): print("Yes")
else: print("No")
``` | instruction | 0 | 101,801 | 19 | 203,602 |
Yes | output | 1 | 101,801 | 19 | 203,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
R,G,B = map(int, input().split())
K = int(input())
while R >= G:
G *= 2
K -= 1
while G >= B:
B *= 2
K -= 1
print('Yes' if K >= 0 else 'No')
``` | instruction | 0 | 101,802 | 19 | 203,604 |
Yes | output | 1 | 101,802 | 19 | 203,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
A,B,C=map(int,input().split())
K=int(input())
for i in range(K):
if A>=B:
B*=2
continue
C*=2
if A<B<C:
print('Yes')
else:
print('No')
``` | instruction | 0 | 101,803 | 19 | 203,606 |
Yes | output | 1 | 101,803 | 19 | 203,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
a, b, c = map(int, input().split())#赤midoriao
k = int(input())
#赤＜緑＜青
if a < b and b<c:
print("Yes")
else:
while a>=b:
b = b*2
k-=1
#print(b)
while b>=c or k>0:
c = c*2
k-=1
if a < b and b<c:
print("Yes")
else:
print("No")
``` | instruction | 0 | 101,804 | 19 | 203,608 |
No | output | 1 | 101,804 | 19 | 203,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
A,B,C = list(map(int,input().split()))
K = int(input())
flag= 0
for i in range(K):
if A>=B:
B *=2
if C <= B:
C *=2
if C >B and B>A:
flag = 1
break
if flag == 0:
print("No")
else:
print("Yes")
``` | instruction | 0 | 101,805 | 19 | 203,610 |
No | output | 1 | 101,805 | 19 | 203,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
R,G,B = map(int,input().split())
K = int(input())
for i in range(K):
if G < R:
G *= 2
elif B <= G:
B *= 2
# print('Green: ' + str(G) + ' blue:' + str(B))
if R < G and G < B:
print('Yes')
else:
print('No')
``` | instruction | 0 | 101,806 | 19 | 203,612 |
No | output | 1 | 101,806 | 19 | 203,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
M-kun has the following three cards:
* A red card with the integer A.
* A green card with the integer B.
* A blue card with the integer C.
He is a genius magician who can do the following operation at most K times:
* Choose one of the three cards and multiply the written integer by 2.
His magic is successful if both of the following conditions are satisfied after the operations:
* The integer on the green card is strictly greater than the integer on the red card.
* The integer on the blue card is strictly greater than the integer on the green card.
Determine whether the magic can be successful.
Constraints
* 1 \leq A, B, C \leq 7
* 1 \leq K \leq 7
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C
K
Output
If the magic can be successful, print `Yes`; otherwise, print `No`.
Examples
Input
7 2 5
3
Output
Yes
Input
7 4 2
3
Output
No
Submitted Solution:
```
import itertools
abc = list(map(int, input().split()))
K = int(input())
tmp = [0, 1, 2, 3]
ans = 'No'
for i in itertools.permutations(tmp, 3):
if sum(i) == K:
tmpabc = []
for j in range(3):
tmpabc.append(abc[j] * (2 ** i[j]))
if tmpabc[0] < tmpabc[1] and tmpabc[1] < tmpabc[2]:
ans = 'Yes'
break
print(ans)
``` | instruction | 0 | 101,807 | 19 | 203,614 |
No | output | 1 | 101,807 | 19 | 203,615 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,936 | 19 | 203,872 |
"Correct Solution:
```
from collections import Counter
X,M = map(int, input().split())
A = list(map(int, input().split()))
C = Counter(A)
P = [0] * M # mod iの数
Q = [0] * M # mod iの同一数字のペアの数
for k,v in C.items():
P[k % M] += v
Q[k % M] += v//2
#print(P)
#print(Q)
ans = 0
for i in range(1,(M+1)//2):
p1, p2 = P[i], P[M-i]
q1, q2 = Q[i], Q[M-i]
if p1 > p2:
p1,p2 = p2,p1
q1,q2 = q2,q1
#print(ans, p1,p2,q1,q2)
ans += p1
ans += min((p2-p1)//2, q2)
ans += P[0] // 2
if M%2 == 0:
ans += P[M//2] // 2
print(ans)
``` | output | 1 | 101,936 | 19 | 203,873 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,937 | 19 | 203,874 |
"Correct Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inp(): return int(input())
def inpl(): return list(map(int, input().split()))
def inpls(): return list(input().split())
N,M = inpl()
xx = inpl()
nums = [0]*M
Pairs = [0]*M
ed = defaultdict(int)
for x in xx:
nums[x%M] += 1
if ed[x] == 0:
ed[x] = 1
else:
Pairs[x%M]+=1
ed[x] = 0
ans = 0
ans += nums[0]//2
if M%2 == 0:
ans += nums[M//2]//2
for i in range(1,M//2):
a = nums[i]
b = nums[M-i]
if a < b:
ans += a + min((b-a)//2,Pairs[M-i])
else:#b < a:
ans += b + min((a-b)//2,Pairs[i])
else:
for i in range(1,M//2+1):
a = nums[i]
b = nums[M-i]
if a < b:
ans += a + min((b-a)//2,Pairs[M-i])
else:#b < a:
ans += b + min((a-b)//2,Pairs[i])
print(ans)
``` | output | 1 | 101,937 | 19 | 203,875 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,938 | 19 | 203,876 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import time,random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def main():
n,m = LI()
xc = collections.Counter(LI())
c = collections.defaultdict(int)
d = collections.defaultdict(int)
for x,v in xc.items():
if v % 2 == 1:
c[x % m] += 1
d[x % m] += v
r = d[0] // 2
if m % 2 == 0:
r += d[m // 2] // 2
for i in range(1, (m+1)//2):
t = min(d[i], d[m-i])
r += t
r += (d[i] - max(c[i], t)) // 2
r += (d[m-i] - max(c[m-i], t)) // 2
return r
print(main())
``` | output | 1 | 101,938 | 19 | 203,877 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,939 | 19 | 203,878 |
"Correct Solution:
```
import collections
n,m=map(int,input().split())
arr=list(map(int,input().split()))
cnt=collections.Counter(arr)
cnt1=collections.defaultdict(int)
cnt2=collections.defaultdict(int)
for key in cnt.keys():
tmp=key%m
val=cnt[key]
if val%2==0:
cnt2[tmp]+=val
else:
cnt1[tmp]+=1
cnt2[tmp]+=val-1
ans=0
for i in range(m):
l=i
r=(m-i)%m
if l>r:
break
elif l==r:
ans+=(cnt1[l]+cnt2[l])//2
else:
ans+=min(cnt1[l],cnt1[r])
if cnt1[l]==cnt1[r]:
ans+=(cnt2[l]+cnt2[r])//2
elif cnt1[l]<cnt1[r]:
cnt1[r]-=cnt1[l]
ans+=min(cnt2[l],cnt1[r])
cnt2[l]=max(0,cnt2[l]-cnt1[r])
ans+=(cnt2[l]+cnt2[r])//2
else:
cnt1[l]-=cnt1[r]
ans+=min(cnt1[l],cnt2[r])
cnt2[r]=max(0,cnt2[r]-cnt1[l])
ans+=(cnt2[l]+cnt2[r])//2
print(ans)
``` | output | 1 | 101,939 | 19 | 203,879 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,940 | 19 | 203,880 |
"Correct Solution:
```
import math
N, M = map(int, input().split())
X = list(map(int, input().split()))
mod_arr = [{} for i in range(M)]
for x in X:
d = mod_arr[x % M]
if x in d:
d[x] += 1
else:
d[x] = 1
ans = 0
if M == 1:
print(N // 2)
exit()
def calc_only_one(d):
sum_v = sum(d.values())
return sum_v // 2
ans += calc_only_one(mod_arr[0])
# even pattern
if M % 2 == 0:
ans += calc_only_one(mod_arr[M // 2])
def calc_two(d_S, d_T):
res = 0
# print(d1, d2)
"""
if len(d_S) == 0:
for v in d_S.values():
res += v // 2
return res
elif len(d_T) == 0:
for v in d_T.values():
res += v // 2
return res
"""
if sum(d_S.values()) < sum(d_T.values()):
d_S, d_T = d_T, d_S
cnt_S = sum(d_S.values())
cnt_T = sum(d_T.values())
remain_for_pair = cnt_S - cnt_T
max_pair_cnt = sum([v // 2 for v in d_S.values()])
pair_cnt = min(remain_for_pair // 2 , max_pair_cnt)
res = cnt_T + pair_cnt
# print(d_S, d_T)
# print(remain_for_pair, max_pair_cnt, pair_cnt, res)
return res
for i in range(1, math.ceil(M / 2)):
ans += calc_two(mod_arr[i], mod_arr[M - i])
print(ans)
``` | output | 1 | 101,940 | 19 | 203,881 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,941 | 19 | 203,882 |
"Correct Solution:
```
from collections import Counter
N, M = map(int, input().split())
X = list(map(int, input().split()))
S = [0] * M
MM = [0] * M
for x, n in Counter(X).items():
MM[x%M] += n//2
S[x%M] += n
ans = 0
ans += S[0] // 2
if M%2 == 0:
ans += S[M//2] // 2
for i1 in range(1, 1000000):
i2 = M-i1
if i1 >= i2:
break
s1, s2 = S[i1], S[i2]
if s1 < s2:
ans += s1 + min((s2-s1)//2, MM[i2])
else:
ans += s2 + min((s1-s2)//2, MM[i1])
print(ans)
``` | output | 1 | 101,941 | 19 | 203,883 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,942 | 19 | 203,884 |
"Correct Solution:
```
import math
N, M = map(int, input().split())
X = list(map(int, input().split()))
mod_arr = [{} for i in range(M)]
for x in X:
d = mod_arr[x % M]
if x in d:
d[x] += 1
else:
d[x] = 1
ans = 0
if M == 1:
print(N // 2)
exit()
def calc_only_one(d):
sum_v = sum(d.values())
return sum_v // 2
ans += calc_only_one(mod_arr[0])
# even pattern
if M % 2 == 0:
ans += calc_only_one(mod_arr[M // 2])
def calc_two(d_S, d_T):
res = 0
if len(d_S) == 0:
for v in d_T.values():
res += v // 2
return res
elif len(d_T) == 0:
for v in d_S.values():
res += v // 2
return res
if sum(d_S.values()) < sum(d_T.values()):
d_S, d_T = d_T, d_S
cnt_S = sum(d_S.values())
cnt_T = sum(d_T.values())
remain_for_pair = cnt_S - cnt_T
max_pair_cnt = sum([v // 2 for v in d_S.values()])
pair_cnt = min(remain_for_pair // 2 , max_pair_cnt)
res = cnt_T + pair_cnt
# print(d_S, d_T)
# print(remain_for_pair, max_pair_cnt, pair_cnt, res)
return res
for i in range(1, math.ceil(M / 2)):
ans += calc_two(mod_arr[i], mod_arr[M - i])
print(ans)
``` | output | 1 | 101,942 | 19 | 203,885 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6 | instruction | 0 | 101,943 | 19 | 203,886 |
"Correct Solution:
```
from collections import Counter
n, m = map(int,input().split())
x = list(map(int,input().split()))
data = [[] for i in range(m)]
def mod(x):
return x%m - 1
def insert(x):
y = mod(x)
data[y].append(x)
for i in x:
insert(i)
ans = len(data[m-1])//2
#print(data)
for i in range(m//2):
p = len(data[i])
q = len(data[m-i-2])
if p == q:
if i != m-i-2:
ans += p
else:
ans += p//2
elif p > q:
ans += q
r = (p-q)//2
same = 0
set_i = Counter(data[i])
for keys in set_i:
same +=set_i[keys]//2
#print(set_i)
ans += min(r,same)
else:
ans += p
r = (q-p)//2
set_j = Counter(data[m-i-2])
same = 0
for keys in set_j:
same += set_j[keys] // 2
#print(set_j)
ans += min(r,same)
print(ans)
``` | output | 1 | 101,943 | 19 | 203,887 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
from collections import Counter
N, M = map(int, input().split())
Xs = list(map(int, input().split()))
cnt = Counter(Xs)
nums = [0] * M
numSames = [0] * M
for key, num in cnt.items():
k = key%M
nums[k] += num
numSames[k] += num//2
ans = 0
for k in range(M//2+1):
k2 = M-k
if k == 0 or k == k2:
numPair = nums[k]//2
ans += numPair
nums[k] -= numPair*2
else:
numPair = min(nums[k], nums[k2])
ans += numPair
nums[k] -= numPair
nums[k2] -= numPair
for k in range(M):
ans += min(numSames[k], nums[k]//2)
print(ans)
``` | instruction | 0 | 101,944 | 19 | 203,888 |
Yes | output | 1 | 101,944 | 19 | 203,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
from collections import Counter
N,M = map(int,input().split())
src = list(map(int,input().split()))
if M == 1:
print(N // 2)
exit()
hist = [0] * M
ctrs = [Counter() for i in range(M)]
for a in src:
hist[a%M] += 1
ctrs[a%M].update([a])
ans = 0
for i in range(1, (M+1)//2):
j = M - i
if hist[j] < hist[i]: i,j = j,i
ans += hist[i]
rem = (hist[j] - hist[i]) // 2
n = sum([v//2 for v in ctrs[j].values()])
ans += min(rem, n)
ans += sum(ctrs[0].values()) // 2
if M%2 == 0:
ans += sum(ctrs[M//2].values()) // 2
print(ans)
``` | instruction | 0 | 101,945 | 19 | 203,890 |
Yes | output | 1 | 101,945 | 19 | 203,891 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)
N,M = map(int,input().split())
X = [int(x) for x in input().split()]
counter = [0] * M
pair_counter = [0] * M
se = set()
for x in X:
r = x%M
counter[r] += 1
if x in se:
pair_counter[r] += 1
se.remove(x)
else:
se.add(x)
answer = 0
for r in range(1,(M+1)//2):
s = M - r
cr,cs = counter[r], counter[s]
pr,ps = pair_counter[r], pair_counter[s]
if cr < cs:
cr,cs = cs,cr
pr,ps = ps,pr
x = cs + min(pr,(cr-cs)//2)
answer += x
# 0とM//2
rs = [0] if M&1 else [0,M//2]
answer += sum(counter[r]//2 for r in rs)
print(answer)
``` | instruction | 0 | 101,946 | 19 | 203,892 |
Yes | output | 1 | 101,946 | 19 | 203,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**10
mod = 10**9 + 7
def f():
n,m = list(map(int, input().split()))
b = [[] for _ in range(m)]
for c in map(int, input().split()):
b[c%m].append(c)
t = 0
for i in range(m//2):
if i+1 > m-i-1:
break
if i+1 == m-i-1:
l = len(b[i+1])
t += l//2
break
l = len(b[i+1])
r = len(b[m-i-1])
if l > r:
t += r
bt = b[i+1]
sa = l-r
else:
t += l
bt = b[m-i-1]
sa = r-l
bt.sort()
lt = 0
i = 0
sa = sa // 2
while i < l-1 and lt < sa:
if bt[i] == bt[i+1]:
lt += 1
i += 2
else:
i += 1
t += lt
l = len(b[0])
t += l//2
return t
print(f())
``` | instruction | 0 | 101,947 | 19 | 203,894 |
No | output | 1 | 101,947 | 19 | 203,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**10
mod = 10**9 + 7
def f():
n,m = list(map(int, input().split()))
b = [[] for _ in range(m)]
for c in map(int, input().split()):
b[c%m].append(c)
t = 0
for i in range(m):
if i+1 > m-i-1:
break
if i+1 == m-i-1:
l = len(b[i+1])
t += l//2
break
l = len(b[i+1])
r = len(b[m-i-1])
if l > r:
t += r
bt = b[i+1]
sa = l-r
else:
t += l
bt = b[m-i-1]
sa = r-l
bt.sort()
lt = 0
i = 0
sa = sa // 2
while i < l-1 and lt < sa:
if bt[i] == bt[i+1]:
lt += 1
i += 2
else:
i += 1
t += lt
l = len(b[0])
t += l//2
return t
print(f())
``` | instruction | 0 | 101,948 | 19 | 203,896 |
No | output | 1 | 101,948 | 19 | 203,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**10
mod = 10**9 + 7
def f():
n,m = list(map(int, input().split()))
b = [[] for _ in range(m)]
for c in map(int, input().split()):
b[c%m].append(c)
t = 0
for i in range(m//2):
if i+1 == m-i-1:
l = len(b[i+1])
t += l//2
break
l = len(b[i+1])
r = len(b[m-i-1])
if l > r:
t += r
bt = b[i+1]
bt.sort()
sa = l-r
lt = 0
i = 0
while i < l-1 and lt < sa:
if bt[i] == bt[i+1]:
lt += 1
i += 2
else:
i += 1
t += lt
else:
t += l
bt = b[m-i-1]
bt.sort()
sa = r-l
lt = 0
i = 0
while i < r-1 and lt < sa:
if bt[i] == bt[i+1]:
lt += 1
i += 2
else:
i += 1
t += lt
l = len(b[0])
t += l//2
return t
print(f())
``` | instruction | 0 | 101,949 | 19 | 203,898 |
No | output | 1 | 101,949 | 19 | 203,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi is playing with N cards.
The i-th card has an integer X_i on it.
Takahashi is trying to create as many pairs of cards as possible satisfying one of the following conditions:
* The integers on the two cards are the same.
* The sum of the integers on the two cards is a multiple of M.
Find the maximum number of pairs that can be created.
Note that a card cannot be used in more than one pair.
Constraints
* 2≦N≦10^5
* 1≦M≦10^5
* 1≦X_i≦10^5
Input
The input is given from Standard Input in the following format:
N M
X_1 X_2 ... X_N
Output
Print the maximum number of pairs that can be created.
Examples
Input
7 5
3 1 4 1 5 9 2
Output
3
Input
15 10
1 5 6 10 11 11 11 20 21 25 25 26 99 99 99
Output
6
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time
sys.setrecursionlimit(10**7)
inf = 10**10
mod = 10**9 + 7
def f():
n,m = list(map(int, input().split()))
b = [[] for _ in range(m)]
for c in map(int, input().split()):
b[c%m].append(c)
t = 0
for i in range(1,m):
if i > m-i:
break
if i == m-i:
l = len(b[i])
t += l//2
break
l = len(b[i])
r = len(b[m-i])
if l > r:
t += r
bt = b[i]
sa = l-r
else:
t += l
bt = b[m-i]
sa = r-l
bt.sort()
lt = 0
j = 0
sa = sa // 2
while j < l-1 and lt < sa:
if bt[j] == bt[j+1]:
lt += 1
j += 2
else:
k += 1
t += lt
l = len(b[0])
t += l//2
return t
print(f())
``` | instruction | 0 | 101,950 | 19 | 203,900 |
No | output | 1 | 101,950 | 19 | 203,901 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play the game using a bag containing several cards with integers written on it. In each game, participants first declare one of their favorite number n. Then, take out an appropriate number of cards from the bag at a time, and if the sum of the numbers written on those cards is equal to n, you will receive a luxurious prize. After each game, the cards will be returned to the bag.
Create a program that inputs the information of m types of cards in the bag and the number declared by the participants in g games, and outputs how many combinations of cards can receive luxury products in each game. please.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
m
a1 b1
a2 b2
::
am bm
g
n1
n2
::
ng
The number of card types m (1 ≤ m ≤ 7) on the first line, the integer ai (1 ≤ ai ≤ 100) written on the i-type card on the following m line, and the number bi (1 ≤ bi ≤ 10) Are given with a space delimiter.
The next line is given the number of games g (1 ≤ g ≤ 10), and the next g line is given the integer ni (1 ≤ ni ≤ 1,000) declared in game i.
The number of datasets does not exceed 100.
Output
For each input dataset, the i line prints the number of card combinations that will give you a gorgeous prize in Game i.
Example
Input
5
1 10
5 3
10 3
25 2
50 2
4
120
500
100
168
7
1 10
3 10
5 10
10 10
25 10
50 10
100 10
3
452
574
787
0
Output
16
0
12
7
9789
13658
17466 | instruction | 0 | 101,967 | 19 | 203,934 |
"Correct Solution:
```
while 1:
n=int(input())
if n==0:break
dp=[1]+[0]*3000
for i in range(n):
a,b=map(int,input().split())
for j in range(1001,-1,-1):
for k in range(1,b+1):
dp[j+a*k]+=dp[j]
for _ in range(int(input())):
print(dp[int(input())])
``` | output | 1 | 101,967 | 19 | 203,935 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play the game using a bag containing several cards with integers written on it. In each game, participants first declare one of their favorite number n. Then, take out an appropriate number of cards from the bag at a time, and if the sum of the numbers written on those cards is equal to n, you will receive a luxurious prize. After each game, the cards will be returned to the bag.
Create a program that inputs the information of m types of cards in the bag and the number declared by the participants in g games, and outputs how many combinations of cards can receive luxury products in each game. please.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
m
a1 b1
a2 b2
::
am bm
g
n1
n2
::
ng
The number of card types m (1 ≤ m ≤ 7) on the first line, the integer ai (1 ≤ ai ≤ 100) written on the i-type card on the following m line, and the number bi (1 ≤ bi ≤ 10) Are given with a space delimiter.
The next line is given the number of games g (1 ≤ g ≤ 10), and the next g line is given the integer ni (1 ≤ ni ≤ 1,000) declared in game i.
The number of datasets does not exceed 100.
Output
For each input dataset, the i line prints the number of card combinations that will give you a gorgeous prize in Game i.
Example
Input
5
1 10
5 3
10 3
25 2
50 2
4
120
500
100
168
7
1 10
3 10
5 10
10 10
25 10
50 10
100 10
3
452
574
787
0
Output
16
0
12
7
9789
13658
17466 | instruction | 0 | 101,968 | 19 | 203,936 |
"Correct Solution:
```
while True:
m = int(input())
if m == 0:
break
nums = []
for _ in range(m):
a, b = map(int, input().split())
lst = [a * i for i in range(b + 1)]
nums.append(lst)
"""
dp[x][y] ... x種類目まででyを作る場合の数
dp[x][y] = sum(dp[x - 1][y - wk] for wk in nums[x])
"""
def solve(n):
dp = [[0] * (n + 1) for _ in range(m + 1)]
for i in range(m + 1):
dp[i][0] = 1
for x in range(1, m + 1):
for y in range(1, n + 1):
dp[x][y] = sum([dp[x - 1][y - v] for v in nums[x - 1] if v <= y])
print(dp[m][n])
g = int(input())
for _ in range(g):
solve(int(input()))
``` | output | 1 | 101,968 | 19 | 203,937 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play the game using a bag containing several cards with integers written on it. In each game, participants first declare one of their favorite number n. Then, take out an appropriate number of cards from the bag at a time, and if the sum of the numbers written on those cards is equal to n, you will receive a luxurious prize. After each game, the cards will be returned to the bag.
Create a program that inputs the information of m types of cards in the bag and the number declared by the participants in g games, and outputs how many combinations of cards can receive luxury products in each game. please.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
m
a1 b1
a2 b2
::
am bm
g
n1
n2
::
ng
The number of card types m (1 ≤ m ≤ 7) on the first line, the integer ai (1 ≤ ai ≤ 100) written on the i-type card on the following m line, and the number bi (1 ≤ bi ≤ 10) Are given with a space delimiter.
The next line is given the number of games g (1 ≤ g ≤ 10), and the next g line is given the integer ni (1 ≤ ni ≤ 1,000) declared in game i.
The number of datasets does not exceed 100.
Output
For each input dataset, the i line prints the number of card combinations that will give you a gorgeous prize in Game i.
Example
Input
5
1 10
5 3
10 3
25 2
50 2
4
120
500
100
168
7
1 10
3 10
5 10
10 10
25 10
50 10
100 10
3
452
574
787
0
Output
16
0
12
7
9789
13658
17466 | instruction | 0 | 101,969 | 19 | 203,938 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0154
"""
import sys
from sys import stdin
input = stdin.readline
def solve(m, cards, g, guesses):
ans = []
cards.append([0, 0])
cards.sort()
W = max(guesses) + 1
dp = [0] * W
dp[0] = 1
for i in range(1, m+1):
card_num = cards[i][0]
card_rem = cards[i][1]
for j in range(W-1, 0, -1):
for k in range(1, card_rem+1):
if card_num*k <= j:
dp[j] += dp[j-card_num*k]
for gg in guesses:
ans.append(dp[gg])
return ans
def main(args):
# m = 5
# cards = [[1, 10], [5, 3], [10, 3], [25, 2], [50, 2]]
# g = 4
# guesses = [120, 500, 100, 168]
while True:
m = int(input())
if m == 0:
break
cards = []
for _ in range(m):
a, b = map(int, input().split())
cards.append([a, b])
g = int(input())
guesses = [int(input()) for _ in range(g)]
ans = solve(m, cards, g, guesses)
print(*ans, sep='\n')
if __name__ == '__main__':
main(sys.argv[1:])
``` | output | 1 | 101,969 | 19 | 203,939 |
Provide a correct Python 3 solution for this coding contest problem.
Let's play the game using a bag containing several cards with integers written on it. In each game, participants first declare one of their favorite number n. Then, take out an appropriate number of cards from the bag at a time, and if the sum of the numbers written on those cards is equal to n, you will receive a luxurious prize. After each game, the cards will be returned to the bag.
Create a program that inputs the information of m types of cards in the bag and the number declared by the participants in g games, and outputs how many combinations of cards can receive luxury products in each game. please.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by a single line of zeros. Each dataset is given in the following format:
m
a1 b1
a2 b2
::
am bm
g
n1
n2
::
ng
The number of card types m (1 ≤ m ≤ 7) on the first line, the integer ai (1 ≤ ai ≤ 100) written on the i-type card on the following m line, and the number bi (1 ≤ bi ≤ 10) Are given with a space delimiter.
The next line is given the number of games g (1 ≤ g ≤ 10), and the next g line is given the integer ni (1 ≤ ni ≤ 1,000) declared in game i.
The number of datasets does not exceed 100.
Output
For each input dataset, the i line prints the number of card combinations that will give you a gorgeous prize in Game i.
Example
Input
5
1 10
5 3
10 3
25 2
50 2
4
120
500
100
168
7
1 10
3 10
5 10
10 10
25 10
50 10
100 10
3
452
574
787
0
Output
16
0
12
7
9789
13658
17466 | instruction | 0 | 101,970 | 19 | 203,940 |
"Correct Solution:
```
# AOJ 0154 Sum of Cards
# Python3 2018.6.19 bal4u
MAX = 1001
while True:
m = int(input())
if m == 0: break
ab = [0]
for i in range(m):
ab.append(list(map(int, input().split())))
dp = [[0 for j in range(MAX)] for r in range(8)]
dp[0][0] = 1
for i in range(1, m+1):
for j in range(MAX):
if dp[i-1][j] == 0: continue;
s = j
for k in range(ab[i][1], -1, -1):
if s < MAX: dp[i][s] += dp[i-1][j]
s += ab[i][0]
for i in range(int(input())):
print(dp[m][int(input())])
``` | output | 1 | 101,970 | 19 | 203,941 |
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