message stringlengths 2 22.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 145 109k | cluster float64 9 9 | __index_level_0__ int64 290 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.
Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.
Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.
Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
Input
The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.
Output
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
Examples
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Note
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.
In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
Submitted Solution:
```
m,n = map(int , input().split())
a = [int(i) for i in input().split()]
b = [int(i) for i in input().split()]
def f(x):
por = 0
for i in range(m):
por += x*a[i] - b[i]
return por
l = 0
r = 1000*(m+1)
while r - l >1 :
mid = (l+r)//2
if f(mid) <= n:
l = mid
else:
r = mid
print(l)
``` | instruction | 0 | 59,312 | 9 | 118,624 |
No | output | 1 | 59,312 | 9 | 118,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.
Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.
Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.
Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
Input
The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.
Output
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
Examples
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Note
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.
In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
Submitted Solution:
```
import sys
n, k = list(map(int, sys.stdin.readline().split()))
lst = list(map(int, sys.stdin.readline().split()))
ingr = list(map(int, sys.stdin.readline().split()))
ans = [[0,0,0]] * n
for i in range(n):
ans[i][0] = ingr[i]//lst[i]
ans[i][1] = ingr[i]%lst[i]
ans[i][2] = i
ans.sort(key = lambda x: [x[0],x[1]])
minn = ans[0][0]
while (k > 0):
ans[0][1] += 1
k -= 1
if (ans[0][1] >= lst[ans[0][2]]):
ans[0][0] += 1
ans[0][1] = 0
ans.sort(key = lambda x: [x[0],x[1]])
if (ans[0][0] > minn):
minn = ans[0][0]
print(minn)
``` | instruction | 0 | 59,313 | 9 | 118,626 |
No | output | 1 | 59,313 | 9 | 118,627 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is given in two versions that differ only by constraints. If you can solve this problem in large constraints, then you can just write a single solution to the both versions. If you find the problem too difficult in large constraints, you can write solution to the simplified version only.
Waking up in the morning, Apollinaria decided to bake cookies. To bake one cookie, she needs n ingredients, and for each ingredient she knows the value ai — how many grams of this ingredient one needs to bake a cookie. To prepare one cookie Apollinaria needs to use all n ingredients.
Apollinaria has bi gram of the i-th ingredient. Also she has k grams of a magic powder. Each gram of magic powder can be turned to exactly 1 gram of any of the n ingredients and can be used for baking cookies.
Your task is to determine the maximum number of cookies, which Apollinaria is able to bake using the ingredients that she has and the magic powder.
Input
The first line of the input contains two positive integers n and k (1 ≤ n, k ≤ 1000) — the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, needed to bake one cookie.
The third line contains the sequence b1, b2, ..., bn (1 ≤ bi ≤ 1000), where the i-th number is equal to the number of grams of the i-th ingredient, which Apollinaria has.
Output
Print the maximum number of cookies, which Apollinaria will be able to bake using the ingredients that she has and the magic powder.
Examples
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Note
In the first sample it is profitably for Apollinaria to make the existing 1 gram of her magic powder to ingredient with the index 2, then Apollinaria will be able to bake 4 cookies.
In the second sample Apollinaria should turn 1 gram of magic powder to ingredient with the index 1 and 1 gram of magic powder to ingredient with the index 3. Then Apollinaria will be able to bake 3 cookies. The remaining 1 gram of the magic powder can be left, because it can't be used to increase the answer.
Submitted Solution:
```
def magicPowder():
ingredients, mps = map(int, input().split())
need = [int(s) for s in input().split()]
have = [int(s) for s in input().split()]
low = 0
high = 1000
a = 0
total = 0
while(low <= high):
mid = low + (high - low) // 2
#print("low, high: ", low, high)
#print("mid: ", mid)
#print("need, have: ", need, have)
#print("mps: ", mps)
for i in range(ingredients):
total += need[i] * mid - have[i]
#print("total: ", total)
if total <= mps:
low = mid + 1
a = mid
else:
high = mid - 1
total = 0
#cookies = list(map(lambda n, h : int(h) // int(n), need, have))
print(a)
magicPowder()
``` | instruction | 0 | 59,314 | 9 | 118,628 |
No | output | 1 | 59,314 | 9 | 118,629 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,399 | 9 | 120,798 |
"Correct Solution:
```
N=int(input())
from collections import defaultdict
import math
d = defaultdict(int)
MOD=10**9+7
for i in range(N):
a,b=map(int,input().split())
x=math.gcd(abs(a),abs(b))
if a==0:
if b==0:
k=(0,0)
else:
k=(0,1)
elif b==0:
k=(1,0)
else:
if a<0:
a,b=-a,-b
if b<0:
a,b=a//x,-((-b)//x)
else:
a,b=a//x,b//x
k=(a,b)
d[k]+=1
ind=defaultdict(int)
index=1
_ans=[1]
for a,b in list(d.keys()):
if a==b==0:
continue
if b<=0:
_k=(-b,a)
else:
_k=(b,-a)
if d[_k]==0:
_ans[0]*=pow(2,d[(a,b)])
else:
if ind[_k]==0:
_ans.append(pow(2,d[(a,b)],MOD))
ind[(a,b)]=index
index+=1
else:
_ans[ind[_k]]+=pow(2,d[(a,b)],MOD)-1
ans=1
for a in _ans:
ans=ans*a%MOD
print((d[(0,0)]+ans-1+MOD)%MOD)
``` | output | 1 | 60,399 | 9 | 120,799 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,400 | 9 | 120,800 |
"Correct Solution:
```
from math import gcd
from collections import defaultdict
N=int(input())
mod = 10**9+7
d_1 = defaultdict(int)
d_2 = defaultdict(int)
zeros = 0
def to_irreducible(a,b):
GCD = gcd(a,b)
return list(map(lambda x:x//GCD,[a,b]))
for i in range(N):
a,b=map(int,input().split(' '))
if a==b==0:
zeros += 1
elif a*b >= 0 and b!=0:
a,b = to_irreducible(abs(a),abs(b))
d_1[(a,b)] +=1
else:
a,b = to_irreducible(abs(b),abs(a))
d_1[(a,b)] +=0
d_2[(a,b)] +=1
ans = 1
for k,v_1 in d_1.items():
v_2 = d_2[k]
ans *= (pow(2,v_1,mod) + pow(2,v_2,mod) -1)
ans %= mod
print((ans+zeros-1)%mod)
``` | output | 1 | 60,400 | 9 | 120,801 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,401 | 9 | 120,802 |
"Correct Solution:
```
mod = 1000000007
from collections import defaultdict
def gcd(a, b):
while b:
a, b = b, a % b
return a
N = int(input())
AB = [list(map(int,input().split())) for _ in range(N)]
d1 = defaultdict(lambda:0)
d2 = defaultdict(lambda:0)
ca = 0
cb = 0
cc = 0
for i in range(N):
a,b = AB[i]
if a==0 and b!=0:
ca += 1
continue
elif a!=0 and b==0:
cb += 1
continue
elif a==0 and b==0:
cc += 1
continue
a_ = abs(a)
b_ = abs(b)
d = gcd(a,b)
if d != 0:
a //= d
b //= d
if a < 0:
a *= -1
b *= -1
if b>0:
d1[(a,b)] += 1
else:
d2[(a,b)] += 1
d1[(-b,a)] += 0
ans = pow(2,ca)+pow(2,cb)-1
for a,b in d1:
x = d1[(a,b)]
y = d2[(b,-a)]
ans *= pow(2,x,mod)+pow(2,y,mod)-1
ans %= mod
ans -= 1
ans += cc
ans %= mod
print(ans)
``` | output | 1 | 60,401 | 9 | 120,803 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,402 | 9 | 120,804 |
"Correct Solution:
```
from math import gcd
MOD = 10**9 + 7
n = int(input())
cnt = {}
zero_cnt = 0
for i in range(n):
a, b = map(int, input().split())
if (a,b) == (0,0):
zero_cnt += 1
continue
g = gcd(abs(a), abs(b))
a //= g
b //= g
rotate = 0
while not (a >= 0 and b > 0):
a,b = -b,a
rotate += 1
if (a,b) not in cnt:
cnt[(a,b)] = [0,0]
cnt[(a,b)][rotate % 2] += 1
ans = 1
for key in cnt:
A, B = cnt[key]
ans *= 1 + pow(2, A, MOD) - 1 + pow(2, B, MOD) - 1
ans %= MOD
ans += zero_cnt
ans -= 1
print(ans % MOD)
``` | output | 1 | 60,402 | 9 | 120,805 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,403 | 9 | 120,806 |
"Correct Solution:
```
from collections import Counter
import math
n = int(input())
pp = []
a0 = 0
b0 = 0
z = 0
for i in range(n):
a,b = map(int,input().split())
if a==0 and b==0:
z += 1
elif a==0:
a0 += 1
elif b==0:
b0 += 1
else:
gomi = math.gcd(a,b)
if a < 0:
a *= -1
b *= -1
pp.append((a//gomi,b//gomi))
mod = 10**9+7
a0 %= mod
b0 %= mod
z %= mod
p = Counter(pp)
r = 1
s = set()
for i,j in p.keys():
if (i,j) in s:
continue
ans = p[(i,j)]
if j < 0:
f = (-j,i)
else:
f = (j,-i)
chk = p[f]
r *= (pow(2,ans,mod)+pow(2,chk,mod)-1)%mod
r %= mod
s.add(f)
r *= (pow(2,a0,mod)+pow(2,b0,mod)-1)%mod
r %= mod
print((r+z-1)%mod)
``` | output | 1 | 60,403 | 9 | 120,807 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,404 | 9 | 120,808 |
"Correct Solution:
```
from math import gcd
_, *e = [[*map(int, t.split())] for t in open(0)]
ans = 1
mod = 10 ** 9 + 7
slope_dict = {}
zeros = 0
for x, y in e:
if x == y == 0:
zeros += 1
else:
d = gcd(x, y)
x //= d
y //= d
if x < 0 or x == 0 < y:
x, y = -x, -y
s = 0
if y < 0:
x, y, s = -y, x, 1
if (x, y) not in slope_dict:
slope_dict[(x, y)] = [0, 0]
slope_dict[(x, y)][s] += 1
for k in slope_dict:
ans = ans * (pow(2, slope_dict[k][0], mod) +
pow(2, slope_dict[k][1], mod) - 1) % mod
print((ans + zeros - 1) % mod)
``` | output | 1 | 60,404 | 9 | 120,809 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,405 | 9 | 120,810 |
"Correct Solution:
```
from math import gcd
n = int(input())
XY = [tuple(map(int, input().split())) for i in range(n)]
C = {}
z = 0
for x, y in XY:
if x == 0 and y == 0:
z += 1
continue
g = gcd(x, y)
x, y = x // g, y // g
if y < 0:
x, y = -x, -y
try:
C[(x, y)] += 1
except KeyError:
C[(x, y)] = 1
ans = 1
p = 10**9 + 7
for x, y in C:
if C[(x, y)] == 0:
continue
a = C[(x, y)]
if x > 0:
x0, y0 = -y, x
else:
x0, y0 = y, -x
try:
b = C[(x0, y0)]
C[(x0, y0)] = 0
except KeyError:
b = 0
ans *= pow(2, a, p) + pow(2, b, p) - 1
ans %= p
C[(x, y)] = 0
print((ans + z - 1) % p)
``` | output | 1 | 60,405 | 9 | 120,811 |
Provide a correct Python 3 solution for this coding contest problem.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479 | instruction | 0 | 60,406 | 9 | 120,812 |
"Correct Solution:
```
from math import gcd
from collections import defaultdict
n = int(input())
a, b = [], []
for i in range(n):
data = list(map(int, input().split()))
a.append(data[0])
b.append(data[1])
d = defaultdict(lambda: [0, 0])
zeros = 0
for i in range(n):
if a[i] * b[i] != 0:
g = gcd(a[i], b[i])
a[i] //= g
b[i] //= g
if a[i]*b[i] > 0:
d[(abs(a[i]), abs(b[i]))][0] += 1
elif a[i]*b[i] < 0:
d[(abs(b[i]), abs(a[i]))][1] += 1
elif a[i] == 0 and b[i] != 0:
d[(0, 1)][0] += 1
elif b[i] == 0 and a[i] != 0:
d[(0, 1)][1] += 1
elif a[i] == b[i] == 0:
zeros += 1
mod = 1000000007
cnt = 1
for pair in d.values():
cnt *= pow(2, pair[0], mod) + pow(2, pair[1], mod) - 1
cnt %= mod
cnt += zeros - 1
cnt %= mod
print(cnt)
``` | output | 1 | 60,406 | 9 | 120,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
from math import gcd
MOD = 10**9+7
N = int(input())
d = {}
zero = 0
for i in range(N):
a, b = map(int, input().split())
if a == 0 and b == 0:
zero += 1
continue
flip = 0
while not (a > 0 and b >= 0):
a, b = -b, a
flip ^= 1
g = gcd(a, b)
a //= g
b //= g
if not (a, b) in d:
d[(a, b)] = [0, 0]
d[(a, b)][flip] += 1
ans = 1
for (v1, v2) in d.values():
ans *= pow(2, v1, MOD) + pow(2, v2, MOD)-1
ans %= MOD
ans -= 1
ans += zero
ans %= MOD
print(ans)
``` | instruction | 0 | 60,407 | 9 | 120,814 |
Yes | output | 1 | 60,407 | 9 | 120,815 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
import math
import sys
from collections import defaultdict
input = sys.stdin.readline
mod = 1000000007
n = int(input())
n2 = [1]
for _ in range(n):
n2.append(n2[-1] * 2 % mod)
zero = 0
plus = defaultdict(int)
minus = defaultdict(int)
for _ in range(n):
a, b = map(int, input().split())
if a == 0 and b == 0:
zero += 1
continue
elif a == 0:
plus[(1, 0)] += 1
elif b == 0:
minus[(1, 0)] += 1
else:
p = math.gcd(a, b)
a //= p
b //= p
if a * b > 0:
plus[(abs(a), abs(b))] += 1
else:
minus[(abs(b), abs(a))] += 1
n -= zero
ans = 1
for k, v in plus.items():
if minus[k] > 0:
ans *= (n2[v] + n2[minus[k]] - 1)
ans %= mod
n -= v + minus[k]
ans *= n2[n]
ans %= mod
ans += zero - 1
ans %= mod
print(ans)
``` | instruction | 0 | 60,408 | 9 | 120,816 |
Yes | output | 1 | 60,408 | 9 | 120,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
def resolve():
import math
from fractions import Fraction
mod = 10**9 + 7
n = int(input())
enpty = 0
dic1 = {}
for i in range(n):
a, b = map(int, input().split())
if a == 0 and b == 0:
enpty += 1
else:
c = math.gcd(a,b)
a = a//c
b = b//c
if b < 0 or (b==0 and a<0):
a = -a
b = -b
if (a, b) in dic1:
dic1[(a, b)] += 1
else:
dic1[(a,b)] = 1
ans = 1
for (a, b), num in dic1.items():
if (-b, a) in dic1:
ans = (ans*(pow(2, num, mod)+pow(2, dic1[(-b, a)], mod)-1))%mod
elif not (b, -a) in dic1:
ans = (ans*pow(2, num, mod))%mod
print((ans-1+enpty)%mod)
resolve()
``` | instruction | 0 | 60,409 | 9 | 120,818 |
Yes | output | 1 | 60,409 | 9 | 120,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
import math,sys
input = sys.stdin.readline
from collections import defaultdict
n = int(input())
MOD = 1000000007
d = defaultdict(int)
l,r = 0,0
count = 0
for i in range(n):
a,b = map(int,input().split())
if a == b == 0:
count += 1
continue
elif a == 0 or b == 0:
if a == 0:
l += 1
else:
r += 1
else:
g = math.gcd(a,b)
a,b = a//g,b//g
if a<0:
a,b = -a,-b
d[str(a)+':'+str(b)] += 1
c = 1
for i in d.keys():
if d[i] != 0:
cur = pow(2,d[i],MOD)
if '-' in i:
ind = i.index(':')
ci = i[ind+2:] + ':' + i[:ind]
else:
ind = i.index(':')
ci = i[ind+1:] + ':-' + i[:ind]
if ci in d.keys():
cur += pow(2,d[ci],MOD)-1
d[ci] = 0
c = (c*cur)%MOD
c = (c*(pow(2,l,MOD)+pow(2,r,MOD)-1)-1+count)%MOD
print(c)
``` | instruction | 0 | 60,410 | 9 | 120,820 |
Yes | output | 1 | 60,410 | 9 | 120,821 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor, cos, radians, pi, sin, gcd
from operator import mul
from functools import reduce
from operator import mul
sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007
n = I()
D = defaultdict(int)
A = []
B = []
F = []
pow2 = [1] * (n + 1)
for i in range(1, n + 1):
pow2[i] = pow2[i - 1] * 2 % mod
ans = 1
cnt1 = 0
cnt2 = 0
for i in range(n):
a, b = LI()
flg = 0
if a * b < 0:
flg = 1
if a == 0 and b == 0:
continue
a = abs(a)
b = abs(b)
g = gcd(a, b)
a = a // g
b = b // g
A += [a]
B += [b]
F += [flg]
if a == 0:
cnt1 += 1
elif b == 0:
cnt2 += 1
else:
D[(flg, a, b)] += 1
visited = defaultdict(int)
for j in range(n):
fi, ai, bi = F[j], A[j], B[j]
if ai == 0 or bi == 0:
continue
if visited[(fi, ai, bi)]:
continue
visited[(fi, ai, bi)] = 1
visited[(fi ^ 1, bi, ai)] = 1
ans = ans * (pow2[D[(fi, ai, bi)]] + pow2[D[(fi ^ 1, bi, ai)]] - 1) % mod
print(ans * (pow2[cnt1] + pow2[cnt2] - 1) % mod - 1)
``` | instruction | 0 | 60,411 | 9 | 120,822 |
No | output | 1 | 60,411 | 9 | 120,823 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
from collections import Counter
def gcd(a, b):
while b: a, b = b, a % b
return abs(a)
MOD = 1000000007
def main():
N = int(input())
d = {}
d[(1,0)] = (0,0)
z = 0
for i in range(N):
a,b = (int(x) for x in input().split())
g = gcd(a,b)
if g != 0:
a = a//g
b = b//g
if a*b<0: a = -abs(a)
elif a*b>0: a = abs(a)
b = abs(b)
if a*b==0:
if a==0 and b==0 :z+=1
elif b==0: d[(1,0)] = (d[(1,0)][0]+1,d[(1,0)][1])
elif a==0: d[(1,0)] = (d[(1,0)][0], d[(1,0)][1]+1)
else:
if (a,b) in d: d[(a,b)] = (d[(a,b)][0]+1,d[(a,b)][1])
elif (b, -a) in d: d[(b,-a)] = (d[(b,-a)][0], d[(b,-a)][1]+1)
elif (-b, a) in d: d[(-b,a)] = (d[(-b,a)][0], d[(-b,a)][1]+1)
else: d[(a,b)] = (1,0)
ans = 1
for i, j in d.values():
ans *= (pow(2,i,MOD) + pow(2,j,MOD) - 1)
ans %= MOD
print(ans-1+z)
if __name__ == '__main__':
main()
``` | instruction | 0 | 60,412 | 9 | 120,824 |
No | output | 1 | 60,412 | 9 | 120,825 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
#!/usr/bin/env python3
#E
import sys
import math
from bisect import bisect_right as br
from bisect import bisect_left as bl
sys.setrecursionlimit(2147483647)
from heapq import heappush, heappop,heappushpop
from collections import defaultdict
from itertools import accumulate
from collections import Counter
from collections import deque
from operator import itemgetter
from itertools import permutations
mod = 10**9 + 7
inf = float('inf')
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def comb(n, r):
return fact[n]*pow(fact[n-r], mod-2, mod)*pow(fact[r], mod-2, mod) % mod
n = I()
ab = [LI() for _ in range(n)]
ans = pow(2, n, mod) - 1
fact = [1]*(n+1)
for i in range(1, n+1):
fact[i] = i * fact[i-1]
fact[i] %= mod
a_zero = 0
b_zero = 0
AB = []
lst = []
for a, b in ab:
if a == 0 and b == 0:
ans -= pow(2, n-1, mod)
ans %= mod
n -= 1
elif a == 0:
a_zero += 1
elif b == 0:
if a_zero:
lst.append(b_zero)
b_zero += 1
else:
AB.append([a, b])
x = [a / b for a, b in AB]
y = [- b / a for a, b in AB]
j = 0
if a_zero:
while j + 1 < b_zero:
ans -= comb(b_zero, j + 2) * pow(2, n - (b_zero - (j+2)) - (j+2), mod)
ans %= mod
j += 1
z = defaultdict(lambda : 0)
for i,j in enumerate(x):
if z[j]:
lst.append(z[j]+1)
z[y[i]] += 1
for i in lst:
j = 0
while j + 1 < i:
if n - (i-(j+2)) - (j+2) >= 0:
ans -= comb(i, j+2) * pow(2, n - (i-(j+2)) - (j+2), mod)
ans %= mod
j += 1
n -= i
print(ans)
``` | instruction | 0 | 60,413 | 9 | 120,826 |
No | output | 1 | 60,413 | 9 | 120,827 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We have caught N sardines. The deliciousness and fragrantness of the i-th sardine is A_i and B_i, respectively.
We will choose one or more of these sardines and put them into a cooler. However, two sardines on bad terms cannot be chosen at the same time.
The i-th and j-th sardines (i \neq j) are on bad terms if and only if A_i \cdot A_j + B_i \cdot B_j = 0.
In how many ways can we choose the set of sardines to put into the cooler? Since the count can be enormous, print it modulo 1000000007.
Constraints
* All values in input are integers.
* 1 \leq N \leq 2 \times 10^5
* -10^{18} \leq A_i, B_i \leq 10^{18}
Input
Input is given from Standard Input in the following format:
N
A_1 B_1
:
A_N B_N
Output
Print the count modulo 1000000007.
Examples
Input
3
1 2
-1 1
2 -1
Output
5
Input
10
3 2
3 2
-1 1
2 -1
-3 -9
-8 12
7 7
8 1
8 2
8 4
Output
479
Submitted Solution:
```
import sys
def input(itr=iter(sys.stdin.readlines())): # pylint: disable= redefined-builtin
return next(itr).rstrip('\n\r')
###############################################################################
import collections
import math as m
from fractions import Fraction
MOD = 1000000007
N = int(input())
Sardine = collections.namedtuple("Sardine", "A B")
sardines = [Sardine(*map(int, input().split())) for _ in range(N)]
def slope(sardine):
if sardine.B == 0:
return m.inf
else:
return Fraction(sardine.A, sardine.B)
def perpendicular(slope):
if slope == 0:
return m.inf
elif slope == m.inf:
return 0
else:
return -1 / slope
groups = collections.Counter()
for s in sardines:
groups[slope(s)] += 1
ways = 1
while groups:
slope, count = groups.popitem()
perp_slope = perpendicular(slope)
perp_count = groups[perp_slope]
ways *= (2**count + 2**perp_count - 1) % MOD
ways %= MOD
del groups[perp_slope]
print((ways-1)%MOD)
``` | instruction | 0 | 60,414 | 9 | 120,828 |
No | output | 1 | 60,414 | 9 | 120,829 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.
Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
Constraints
* 1 ≤ A, B, C ≤ 5000
* 1 ≤ X, Y ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C X Y
Output
Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.
Examples
Input
1500 2000 1600 3 2
Output
7900
Input
1500 2000 1900 3 2
Output
8500
Input
1500 2000 500 90000 100000
Output
100000000
Submitted Solution:
```
def getint(): return int(input())
def getints(): return list(map(int, input().split()))
a,b,c,x,y=getints()
res = 1 << 29
for c_count in range(0, max(a,b)*2 + 1):
x_need = max(0, x - c_count // 2)
y_need = max(0, y - c_count // 2)
tmp = x_need * a + y_need * b + c_count * c
res = min(res, tmp)
print(res)
``` | instruction | 0 | 60,491 | 9 | 120,982 |
No | output | 1 | 60,491 | 9 | 120,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
"Pizza At", a fast food chain, offers three kinds of pizza: "A-pizza", "B-pizza" and "AB-pizza". A-pizza and B-pizza are completely different pizzas, and AB-pizza is one half of A-pizza and one half of B-pizza combined together. The prices of one A-pizza, B-pizza and AB-pizza are A yen, B yen and C yen (yen is the currency of Japan), respectively.
Nakahashi needs to prepare X A-pizzas and Y B-pizzas for a party tonight. He can only obtain these pizzas by directly buying A-pizzas and B-pizzas, or buying two AB-pizzas and then rearrange them into one A-pizza and one B-pizza. At least how much money does he need for this? It is fine to have more pizzas than necessary by rearranging pizzas.
Constraints
* 1 ≤ A, B, C ≤ 5000
* 1 ≤ X, Y ≤ 10^5
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B C X Y
Output
Print the minimum amount of money required to prepare X A-pizzas and Y B-pizzas.
Examples
Input
1500 2000 1600 3 2
Output
7900
Input
1500 2000 1900 3 2
Output
8500
Input
1500 2000 500 90000 100000
Output
100000000
Submitted Solution:
```
A,B,C,X,Y = map(int,input().split())
m = min(X,Y)
m2 = max(X,Y)
if (A+B)/2 > C and m > 1:#できるだけCを買う方がお得
ans = C*2*m
if X>Y:
ans += A*(X-Y)
else:
ans += B*(Y-X)
if ans > C*2*m2:
print(C*2*m2)
else:
print(ans)
else:
print(A*X+B*Y)
``` | instruction | 0 | 60,494 | 9 | 120,988 |
No | output | 1 | 60,494 | 9 | 120,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,721 | 9 | 121,442 |
Tags: greedy, math
Correct Solution:
```
import math
n , k = map(int,input().split())
ls = []
for i in range(n):
a = int(input())
ls.append(a)
l = list(set(ls))
l.sort()
for i in range(len(l)):
l[i] = ls.count(l[i])
sm = 0
for i in range(len(l)):
sm += l[i] % 2
#print(sm)
print(n - sm // 2 )
``` | output | 1 | 60,721 | 9 | 121,443 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,722 | 9 | 121,444 |
Tags: greedy, math
Correct Solution:
```
a,b=map(int,input().split())
if a%2==0:
sets=a//2
else:
sets=a//2+1
dic={}
for i in range(a):
m=int(input())
if m in dic:
dic[m]+=1
else:
dic[m]=1
l=[]
for i in dic:
l.append(dic[i])
l.sort()
i=len(l)-1
z=0
count=0
while i>=0 and sets>0:
if l[i]>=2:
c=l[i]//2
if sets>=c:
count+=(l[i]//2)*2
l[i]=l[i]%2
z+=l[i]
sets-=c
else:
count+=sets*2
l[i]=l[i]-sets*2
z+=l[i]
sets=0
else:
z+=l[i]
i-=1
if sets>0 and z>0:
count+=min(z,sets)
print(count)
``` | output | 1 | 60,722 | 9 | 121,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,723 | 9 | 121,446 |
Tags: greedy, math
Correct Solution:
```
import math
ceil = math.ceil
l1 = [int(x) for x in input().split()]
n,k = map(int,l1)
c=[]
d=[0]*k
for i in range(n):
temp = int(input())
d[temp-1]+=1
d.sort(reverse=True)
#print(d)
for j in range(int(n/2)+n%2):
if d[0]==1:
d.remove(1)
else:
d[0]-=2
if d[0]==0:
d.remove(0)
d.sort(reverse=True)
print(n-sum(d))
``` | output | 1 | 60,723 | 9 | 121,447 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,724 | 9 | 121,448 |
Tags: greedy, math
Correct Solution:
```
n, k = map(int,input().split())
cnt = []
for i in range(k):
cnt.append(0)
ans = 0
for i in range(n):
x = input()
x = int(x)
cnt[x-1]+=1
cnt.sort(reverse=True)
ans = 0
tmp = 0
for i in range(k):
ans+=cnt[i]//2*2
cnt[i]%=2
if cnt[i] is not 0:
tmp+=1
print(ans+tmp//2+tmp%2)
``` | output | 1 | 60,724 | 9 | 121,449 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,725 | 9 | 121,450 |
Tags: greedy, math
Correct Solution:
```
Input=lambda:map(int,input().split())
# This code written during the contest
# Codeforces Round #574 (Div. 2)
n, k = Input()
arr = [0]*k
for i in range(n):
a = int(input()) - 1
arr[a]+=1
odd = 0
for i in range(k):
odd+=(arr[i]%2)
print(n-(odd//2))
``` | output | 1 | 60,725 | 9 | 121,451 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,726 | 9 | 121,452 |
Tags: greedy, math
Correct Solution:
```
n,k=map(int,input().split())
l1=[0]*k
for i in range(n):
x=int(input())
l1[x-1]+=1
ans=0
odds=0
for item in l1:
if item%2==0:
ans+=item
else :
ans+=(item-1)
odds+=1
if n%2==0:
print(ans+odds//2)
else :
print(ans+(odds-1)//2+1)
``` | output | 1 | 60,726 | 9 | 121,453 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,727 | 9 | 121,454 |
Tags: greedy, math
Correct Solution:
```
import math
n,k=map(int,input().split())
a=[]
c=[]
for i in range (n):
b=int(input())
a.append(b)
z=n/2
z=math.ceil(z)
for i in range(k):
c.append(0)
for i in range(n):
r=a[i]-1
c[r]=c[r]+1
c.sort(reverse=True)
q=0
for i in c:
if(i%2==1):
q+=1
print(n-q//2)
``` | output | 1 | 60,727 | 9 | 121,455 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth). | instruction | 0 | 60,728 | 9 | 121,456 |
Tags: greedy, math
Correct Solution:
```
n, k = map(int, input().split())
counters = [0] * k
for i in range(n):
drink = int(input())
counters[drink - 1] += 1
result = 0
rest = 0
for c in counters:
result += (c // 2) * 2
rest += c % 2
result += (rest + 1) // 2
print(result)
``` | output | 1 | 60,728 | 9 | 121,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
from math import ceil
n,k = map(int,input().split())
d = {}
for _ in range(n):
num = int(input())
d[num] = d.get(num,0)+1
ans = 0
count = 0
for x in d.values():
if x%2==0:
ans+=x
else:
ans+=x-1
count+=1
print(ans+ceil(count/2))
# for x in d:
# d[x] = ceil(d[x]/2)
# n = ceil(n/2)
# vals = sorted(d.values(),reverse = True)
# print(vals)
``` | instruction | 0 | 60,729 | 9 | 121,458 |
Yes | output | 1 | 60,729 | 9 | 121,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
N,K=[int(x) for x in input().split()]
L=[0]*(K+1)
for i in range(N):
L[int(input())]+=1
#print(L)
x=0
for i in range(1,K+1):
#print((L[i])//2)
x+=L[i]//2
L[i]=L[i]%2
#print(L)
Sum=sum(L)
#print(Sum)
if N%2==1:
print(N-(Sum-((N+1)//2-x)))
else:
print(N-(Sum-((N)//2-x)))
``` | instruction | 0 | 60,730 | 9 | 121,460 |
Yes | output | 1 | 60,730 | 9 | 121,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
[n,k]=list(map(int,input().strip().split()))
l=[]
happy=0
if n%2==0:
cangive=n/2
else:
cangive=(n//2)+1
for i in range(n):
inputi=int(input().strip())
if inputi in l:
happy+=2
cangive-=1
l.remove(inputi)
else:
l.append(inputi)
if cangive==0:
break
if cangive!=0 and len(l)!=0:
happy+=min(len(l),cangive)
print(int(happy))
``` | instruction | 0 | 60,731 | 9 | 121,462 |
Yes | output | 1 | 60,731 | 9 | 121,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
import collections as cl
n, k = map(int, input().split())
a = list(sorted(cl.Counter([int(input()) for _ in range(n)]).values(), reverse=True))
v = sum([a[i] - a[i] % 2 for i in range(len(a))])
print(v + ((n + 1) // 2 - v // 2))
``` | instruction | 0 | 60,732 | 9 | 121,464 |
Yes | output | 1 | 60,732 | 9 | 121,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
n, k = map(int, input().split())
s = []
kind = list([0]*k)
MAX = 0
for i in range(n):
s.append(int(input()))
kind[s[i]-1] += 1
if n % 2 == 1:
n+=1
while len(kind) > 0:
if n - 2*((s.count(kind.index(max(kind))+1)+1)//2) >= 0:
MAX += s.count(kind.index(max(kind))+1)
n -= 2*((s.count(kind.index(max(kind))+1)+1)//2)
kind[kind.index(max(kind))] = 0
elif n >= 0:
MAX += n
n -= n
break
print(MAX)
``` | instruction | 0 | 60,733 | 9 | 121,466 |
No | output | 1 | 60,733 | 9 | 121,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
#for i in range(1,1001):print(1)
n,k=map(int,input().split());dp=[0]*(1004)
for i in range(n):
a=int(input());dp[a]+=1
odd=[];ans=0;sett=(n+1)//2;cnt=0
for i in dp:
if sett>0 and i%2==0:
ans+=i
sett-=i//2
else:ans+=(i-1);cnt+=1;sett-=(i)//2
#print(sett,cnt)
print(min(n,ans+min(cnt,sett)))
``` | instruction | 0 | 60,734 | 9 | 121,468 |
No | output | 1 | 60,734 | 9 | 121,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
from math import ceil
from collections import defaultdict
hash = defaultdict(int)
n,k = map(int,input().split())
for i in range(n):
a = int(input())
hash[a]+=1
ans = 0
l = [hash[i] for i in hash.keys()]
l.sort()
l.reverse()
k = 0
i = 0
while True:
ans+=ceil(l[k])
i+=ceil(l[k]/2)
if i == ceil(n/2):
break
elif i>ceil(n/2):
ans+=(ceil(n/2)-i)
break
k+=1
print(ans)
``` | instruction | 0 | 60,735 | 9 | 121,470 |
No | output | 1 | 60,735 | 9 | 121,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Old timers of Summer Informatics School can remember previous camps in which each student was given a drink of his choice on the vechorka (late-evening meal). Or may be the story was more complicated?
There are n students living in a building, and for each of them the favorite drink a_i is known. So you know n integers a_1, a_2, ..., a_n, where a_i (1 ≤ a_i ≤ k) is the type of the favorite drink of the i-th student. The drink types are numbered from 1 to k.
There are infinite number of drink sets. Each set consists of exactly two portions of the same drink. In other words, there are k types of drink sets, the j-th type contains two portions of the drink j. The available number of sets of each of the k types is infinite.
You know that students will receive the minimum possible number of sets to give all students exactly one drink. Obviously, the number of sets will be exactly ⌈ n/2 ⌉, where ⌈ x ⌉ is x rounded up.
After students receive the sets, they will distribute their portions by their choice: each student will get exactly one portion. Note, that if n is odd then one portion will remain unused and the students' teacher will drink it.
What is the maximum number of students that can get their favorite drink if ⌈ n/2 ⌉ sets will be chosen optimally and students will distribute portions between themselves optimally?
Input
The first line of the input contains two integers n and k (1 ≤ n, k ≤ 1 000) — the number of students in the building and the number of different drinks.
The next n lines contain student's favorite drinks. The i-th line contains a single integer from 1 to k — the type of the favorite drink of the i-th student.
Output
Print exactly one integer — the maximum number of students that can get a favorite drink.
Examples
Input
5 3
1
3
1
1
2
Output
4
Input
10 3
2
1
3
2
3
3
1
3
1
2
Output
9
Note
In the first example, students could choose three sets with drinks 1, 1 and 2 (so they will have two sets with two drinks of the type 1 each and one set with two drinks of the type 2, so portions will be 1, 1, 1, 1, 2, 2). This way all students except the second one will get their favorite drinks.
Another possible answer is sets with drinks 1, 2 and 3. In this case the portions will be 1, 1, 2, 2, 3, 3. Then all the students except one will gain their favorite drinks. The only student that will not gain the favorite drink will be a student with a_i = 1 (i.e. the first, the third or the fourth).
Submitted Solution:
```
n,k=map(int,input().split());dp=[0]*(1001)
for i in range(n):
a=int(input());dp[a]+=1
odd=[];ans=0;sett=(n+1)//2;flag=True
for i in dp:
if i%2==0:
ans+=i
sett-=i//2
else:odd.append(i)
if sett==0:print(ans)
else:
#print(odd);
odd=sorted(odd)[::-1]
for i in odd:
if sett>0:
if sett>=(i+1)//2:ans+=i;sett-=(i+1)//2
elif sett<(i+1)//2:ans+=2*sett;sett=0
print(ans)
``` | instruction | 0 | 60,736 | 9 | 121,472 |
No | output | 1 | 60,736 | 9 | 121,473 |
Provide a correct Python 3 solution for this coding contest problem.
Taro Aizu's company has a boss who hates being indivisible. When Taro goes out to eat with his boss, he pays by splitting the bill, but when the payment amount is not divisible by the number of participants, his boss always pays for it.
One day, Taro became the secretary of the dinner party. Mr. Taro, who has little money, wondered if he could invite his boss to get a treat. I have to place an order with a restaurant, but I don't know how many people will participate yet, so it seems that I want to place an order so that any number of people can participate. At the same time as Mr. Taro, you who are also planning to attend the dinner party decided to cooperate with Mr. Taro to calculate the amount that is less than the budget amount and is not divisible by any number of people.
Create a program that inputs the type of dish, the price of each dish, and the budget amount, and outputs the total amount (excluding 1 and the total amount) that is not divisible by any number that is less than or equal to the budget amount. You can order multiple dishes of each type, but you do not have to order all types of dishes. However, if there is no such total amount, output NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n x
v1
v2
::
vn
On the first line, the type of dish n (1 ≤ n ≤ 30) and the budget amount x (1 ≤ x ≤ 1000000) are given, separated by blanks. The next n lines are given the integer vi (1 ≤ vi ≤ 1000000), which represents the amount of the i-type dish.
The number of datasets does not exceed 100.
Output
For each input dataset, print the total amount or NA closest to the budget amount on one line.
Example
Input
4 15000
305
260
129
500
3 400
10
20
30
3 200909
5
9
12
0 0
Output
14983
NA
200909 | instruction | 0 | 61,392 | 9 | 122,784 |
"Correct Solution:
```
def solve(price, ps, is_prime):
dp = [0] * ((price >> 5) + 1)
for i in range(0, price + 1,ps[0]):
dp[i >> 5] |= 1 << (i & 31)
dp[0] = 1
for i in range(1, len(ps)):
cur_p = ps[i]
r = cur_p & 31
rest = 0
if cur_p >= 32:
for p in range(cur_p, price + 1, 32):
i1 = (p - cur_p) >> 5
i2 = p >> 5
v = dp[i1]
dp[i2] |= ((v << r)&(2**33-1)) | rest
rest = (v >> (32 - r)) & ((1 << r) - 1)
else:
for p in range(cur_p, price + 1):
dp[p >> 5] |= (dp[(p - cur_p) >> 5] >> ((p - cur_p) & 31) & 1) << (p & 31)
p = -1
for i in range(price+1, 0,-1):
if is_prime[i] and (dp[i >> 5] >> (i & 31) & 1):
p = i
break
if p == -1:
print("NA")
else:
print(p)
def main():
N = 1000001
temp = [True]*(N+1)
temp[0] = temp[1] = False
for i in range(2, int((N+1)**0.5)+1):
if temp[i]:
temp[i+i::i] = [False]*(len(temp[i+i::i]))
while True:
n, price = map(int,input().split())
if n == 0 and price == 0:
return
ps = []
for i in range(n):
ps.append(int(input()))
solve(price, ps, temp)
main()
``` | output | 1 | 61,392 | 9 | 122,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro Aizu's company has a boss who hates being indivisible. When Taro goes out to eat with his boss, he pays by splitting the bill, but when the payment amount is not divisible by the number of participants, his boss always pays for it.
One day, Taro became the secretary of the dinner party. Mr. Taro, who has little money, wondered if he could invite his boss to get a treat. I have to place an order with a restaurant, but I don't know how many people will participate yet, so it seems that I want to place an order so that any number of people can participate. At the same time as Mr. Taro, you who are also planning to attend the dinner party decided to cooperate with Mr. Taro to calculate the amount that is less than the budget amount and is not divisible by any number of people.
Create a program that inputs the type of dish, the price of each dish, and the budget amount, and outputs the total amount (excluding 1 and the total amount) that is not divisible by any number that is less than or equal to the budget amount. You can order multiple dishes of each type, but you do not have to order all types of dishes. However, if there is no such total amount, output NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n x
v1
v2
::
vn
On the first line, the type of dish n (1 ≤ n ≤ 30) and the budget amount x (1 ≤ x ≤ 1000000) are given, separated by blanks. The next n lines are given the integer vi (1 ≤ vi ≤ 1000000), which represents the amount of the i-type dish.
The number of datasets does not exceed 100.
Output
For each input dataset, print the total amount or NA closest to the budget amount on one line.
Example
Input
4 15000
305
260
129
500
3 400
10
20
30
3 200909
5
9
12
0 0
Output
14983
NA
200909
Submitted Solution:
```
isPrime = [False,False]+[True]*(1000000-2)
isPrime[4::2] = [False]*len(isPrime[4::2])
for i in range(3,1000,2):
if isPrime[i]:
isPrime[2*i::i] = [False]*len(isPrime[2*i::i])
while True:
n,x = [int(i) for i in input().split()]
if n==0: break
prices = [int(input()) for i in range(n)]
isPayable = [True]+[False]*x
for pri in prices:
for i in range(0,x-pri+1):
if isPayable[i]:
isPayable[i+pri] = True
li = [i for i in range(x+1) if isPayable[i] and isPrime[i]]
print(max(li) if li else 'NA')
``` | instruction | 0 | 61,393 | 9 | 122,786 |
No | output | 1 | 61,393 | 9 | 122,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro Aizu's company has a boss who hates being indivisible. When Taro goes out to eat with his boss, he pays by splitting the bill, but when the payment amount is not divisible by the number of participants, his boss always pays for it.
One day, Taro became the secretary of the dinner party. Mr. Taro, who has little money, wondered if he could invite his boss to get a treat. I have to place an order with a restaurant, but I don't know how many people will participate yet, so it seems that I want to place an order so that any number of people can participate. At the same time as Mr. Taro, you who are also planning to attend the dinner party decided to cooperate with Mr. Taro to calculate the amount that is less than the budget amount and is not divisible by any number of people.
Create a program that inputs the type of dish, the price of each dish, and the budget amount, and outputs the total amount (excluding 1 and the total amount) that is not divisible by any number that is less than or equal to the budget amount. You can order multiple dishes of each type, but you do not have to order all types of dishes. However, if there is no such total amount, output NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n x
v1
v2
::
vn
On the first line, the type of dish n (1 ≤ n ≤ 30) and the budget amount x (1 ≤ x ≤ 1000000) are given, separated by blanks. The next n lines are given the integer vi (1 ≤ vi ≤ 1000000), which represents the amount of the i-type dish.
The number of datasets does not exceed 100.
Output
For each input dataset, print the total amount or NA closest to the budget amount on one line.
Example
Input
4 15000
305
260
129
500
3 400
10
20
30
3 200909
5
9
12
0 0
Output
14983
NA
200909
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys,math
import pdb
#Debug=True
Debug=False
nmax=1000000
unit_cost = []
total_cost=0
pmax=-1
dic = {2:True,3:True}
def prime3(n):
global dic
if n in dic:
return(dic[n])
else:
if n % 2==0:
return False
while True:
for i in range(3,math.ceil(math.sqrt(n))+1,2):
if n % i == 0:
dic[n]=False
return(False)
dic[n]=True
return(True)
def cost_gen1(money, lvl=0, lst=[]):
try:
uc = unit_cost[lvl]
if lvl==len(unit_cost)-1:
yield uc * (money//uc)
else:
for i in range(money//uc,-1,-1):
tot_uc = uc*i
rem = money - tot_uc
if rem < unit_cost[lvl+1]:
yield tot_uc
else:
for j in cost_gen1(rem, lvl+1):
yield j + tot_uc
except Exception:
print('NA')
def check_mutually_prime(v):
s = min(v)
for i in range(2,s+1):
if sum(map(lambda x:x%i, v))==0:
return False
return True
def main():
global unit_cost, total_cost, pmax
if not Debug:
fh=sys.stdin
else:
#fh=open('vol2_0202b.txt', 'r')
fh=open('vol2_0202.txt', 'r')
#pdb.set_trace()
while True:
m, total_cost = list(map(int, fh.readline().strip().split()))
if m==0:
break
v=[]
for _ in range(m):
v.append(int(fh.readline()))
unit_cost = sorted(v, reverse=True) # 大きい順
# 互いに素(relative prime)、でないときの処理
if not check_mutually_prime(unit_cost):
print('NA')
continue
#t0=time.time()
# print('NA')
# continue
p=[]
pmax=-1
gen=cost_gen1(total_cost, 0)
for i in gen:
if i > pmax and prime3(i):
p.append(i)
pmax=max(p)
if pmax==total_cost:
break
print(pmax)
#if Debug:
# print('t=', time.time()-t0)
fh.close()
if __name__ == "__main__":
main()
``` | instruction | 0 | 61,394 | 9 | 122,788 |
No | output | 1 | 61,394 | 9 | 122,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro Aizu's company has a boss who hates being indivisible. When Taro goes out to eat with his boss, he pays by splitting the bill, but when the payment amount is not divisible by the number of participants, his boss always pays for it.
One day, Taro became the secretary of the dinner party. Mr. Taro, who has little money, wondered if he could invite his boss to get a treat. I have to place an order with a restaurant, but I don't know how many people will participate yet, so it seems that I want to place an order so that any number of people can participate. At the same time as Mr. Taro, you who are also planning to attend the dinner party decided to cooperate with Mr. Taro to calculate the amount that is less than the budget amount and is not divisible by any number of people.
Create a program that inputs the type of dish, the price of each dish, and the budget amount, and outputs the total amount (excluding 1 and the total amount) that is not divisible by any number that is less than or equal to the budget amount. You can order multiple dishes of each type, but you do not have to order all types of dishes. However, if there is no such total amount, output NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n x
v1
v2
::
vn
On the first line, the type of dish n (1 ≤ n ≤ 30) and the budget amount x (1 ≤ x ≤ 1000000) are given, separated by blanks. The next n lines are given the integer vi (1 ≤ vi ≤ 1000000), which represents the amount of the i-type dish.
The number of datasets does not exceed 100.
Output
For each input dataset, print the total amount or NA closest to the budget amount on one line.
Example
Input
4 15000
305
260
129
500
3 400
10
20
30
3 200909
5
9
12
0 0
Output
14983
NA
200909
Submitted Solution:
```
isPrime = [False,False]+[True]*(1000000-1) # 1,000,001
isPrime[4::2] = [False]*len(isPrime[4::2])
for i in range(3,1000,2):
if isPrime[i]:
isPrime[2*i::i] = [False]*len(isPrime[2*i::i])
while True:
# n,x = [int(i) for i in input().split()]
n,x = list(map(int, input().split()))
if n==0 and x==0: break
prices = [int(input()) for i in range(n)] # n
isPayable = [False]*(x+1) # x+1
for pri in prices:
isPayable[pri] = True
for i in range(0,x-pri+1):
if isPayable[i]:
isPayable[i+pri] = True
for i in reversed(range(x+1)):
if isPayable[i] and isPrime[i]:
print(i)
break
if i == 0:
print('NA')
break
``` | instruction | 0 | 61,395 | 9 | 122,790 |
No | output | 1 | 61,395 | 9 | 122,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro Aizu's company has a boss who hates being indivisible. When Taro goes out to eat with his boss, he pays by splitting the bill, but when the payment amount is not divisible by the number of participants, his boss always pays for it.
One day, Taro became the secretary of the dinner party. Mr. Taro, who has little money, wondered if he could invite his boss to get a treat. I have to place an order with a restaurant, but I don't know how many people will participate yet, so it seems that I want to place an order so that any number of people can participate. At the same time as Mr. Taro, you who are also planning to attend the dinner party decided to cooperate with Mr. Taro to calculate the amount that is less than the budget amount and is not divisible by any number of people.
Create a program that inputs the type of dish, the price of each dish, and the budget amount, and outputs the total amount (excluding 1 and the total amount) that is not divisible by any number that is less than or equal to the budget amount. You can order multiple dishes of each type, but you do not have to order all types of dishes. However, if there is no such total amount, output NA.
Input
A sequence of multiple datasets is given as input. The end of the input is indicated by two lines of zeros. Each dataset is given in the following format:
n x
v1
v2
::
vn
On the first line, the type of dish n (1 ≤ n ≤ 30) and the budget amount x (1 ≤ x ≤ 1000000) are given, separated by blanks. The next n lines are given the integer vi (1 ≤ vi ≤ 1000000), which represents the amount of the i-type dish.
The number of datasets does not exceed 100.
Output
For each input dataset, print the total amount or NA closest to the budget amount on one line.
Example
Input
4 15000
305
260
129
500
3 400
10
20
30
3 200909
5
9
12
0 0
Output
14983
NA
200909
Submitted Solution:
```
from sys import stdin
readline = stdin.readline
import math
def sieve(n):
p = [1] * (n + 1)
p[0] = p[1] = 0
for i in range(math.ceil((n + 1) ** 0.5)):
if p[i]:
for j in range(2 * i, len(p), i):
p[j] = 0
return p
prime = sieve(1000000)
def solve(price, x):
dp = [0] * (x + 1)
dp[0] = 1
price.sort()
for pi in price:
for i in range(0, len(dp), pi):
dp[i] = 1
for pi in price[1:]:
for i in range(pi, len(dp)):
if not dp[i] and dp[i - pi]:
dp[i] = 1
for i in reversed(range(2, len(dp))):
if dp[i] and prime[i]:
return i
return None
def main():
while True:
n, x = map(int, readline().split())
if n == 0:
break
price = [int(readline()) for _ in range(n)]
amount = solve(price, x)
print(amount if amount else 'NA')
main()
``` | instruction | 0 | 61,396 | 9 | 122,792 |
No | output | 1 | 61,396 | 9 | 122,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,635 | 9 | 123,270 |
Tags: greedy, implementation, math
Correct Solution:
```
q = int(input())
for qq in range(q):
s = input().split()
a = int(s[0])
b = int(s[1])
n = int(s[2])
m = int(s[3])
if m <= min(a, b) and m + n <= a + b:
print('Yes')
else:
print('No')
``` | output | 1 | 61,635 | 9 | 123,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,636 | 9 | 123,272 |
Tags: greedy, implementation, math
Correct Solution:
```
t = int(input())
for i in range(t):
a, b, n, m = input().split()
a = int(a)
b = int(b)
n = int(n)
m = int(m)
if a < b:
a, b = b, a
if b >= m and a + b >= n + m:
print("Yes")
else:
print("No")
``` | output | 1 | 61,636 | 9 | 123,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,637 | 9 | 123,274 |
Tags: greedy, implementation, math
Correct Solution:
```
#
import sys
readline = lambda: sys.stdin.readline().strip('\n')
readlist = lambda: list(map(int,readline().split()))
#problem C
for i in range(int(input())):
a,b,n,m = readlist()
if m > min(a,b) or (a+b)<(n+m):
print('No')
else:
print('Yes')
quit()
#problem B
for i in range(int(input())):
n,r = readlist()
s = 0
if n>r:
print(r*(r+1)//2)
else:
print(n*(n-1)//2 + 1)
#problem A
for i in range(int(input())):
n = int(input())
print(n//2 + n%2)
a = 1
``` | output | 1 | 61,637 | 9 | 123,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,638 | 9 | 123,276 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import deque
from fractions import Fraction as f
from copy import *
from bisect import *
from heapq import *
from math import *
from itertools import permutations
def eprint(*args):
print(*args, file=sys.stderr)
zz=1
#sys.setrecursionlimit(10**6)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
def li():
return [int(x) for x in input().split()]
def gi():
return [x for x in input().split()]
def fi():
return int(input())
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
def bo(i):
return ord(i)-ord('a')
t=fi()
while t>0:
t-=1
a,b,n,m=mi()
ans=["Yes","No"]
if a+b<n+m:
print(ans[1])
continue
if min(a,b)>=m:
print(ans[0])
else:
print(ans[1])
``` | output | 1 | 61,638 | 9 | 123,277 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,639 | 9 | 123,278 |
Tags: greedy, implementation, math
Correct Solution:
```
t = int(input())
for case in range(t):
a, b, n, m = list(map(int, input().split(' ')))
if (a + b) < (n + m):
print('No')
else:
if m > min(a, b):
print('No')
else:
print('Yes')
``` | output | 1 | 61,639 | 9 | 123,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,640 | 9 | 123,280 |
Tags: greedy, implementation, math
Correct Solution:
```
t = input()
t = int(t)
while(t > 0):
t = int(t)-1
# print(t)
a, b, n, m = input().split()
a = int(a)
b = int(b)
n = int(n)
m = int(m)
if(a+b < n+m):
print("NO")
continue
# cout<<a-m;
if(min(a, b) < m):
print("NO")
continue
print("YES")
``` | output | 1 | 61,640 | 9 | 123,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,641 | 9 | 123,282 |
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
a,b,n,m = map(int,input().split())
print("No") if min(a,b) < m or (n+m) > (a+b) else print("Yes")
``` | output | 1 | 61,641 | 9 | 123,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Anna is a girl so brave that she is loved by everyone in the city and citizens love her cookies. She is planning to hold a party with cookies. Now she has a vanilla cookies and b chocolate cookies for the party.
She invited n guests of the first type and m guests of the second type to the party. They will come to the party in some order. After coming to the party, each guest will choose the type of cookie (vanilla or chocolate) to eat. There is a difference in the way how they choose that type:
If there are v vanilla cookies and c chocolate cookies at the moment, when the guest comes, then
* if the guest of the first type: if v>c the guest selects a vanilla cookie. Otherwise, the guest selects a chocolate cookie.
* if the guest of the second type: if v>c the guest selects a chocolate cookie. Otherwise, the guest selects a vanilla cookie.
After that:
* If there is at least one cookie of the selected type, the guest eats one.
* Otherwise (there are no cookies of the selected type), the guest gets angry and returns to home.
Anna wants to know if there exists some order of guests, such that no one guest gets angry. Your task is to answer her question.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains four integers a, b, n, m (0 ≤ a,b,n,m ≤ 10^{18}, n+m ≠ 0).
Output
For each test case, print the answer in one line. If there exists at least one valid order, print "Yes". Otherwise, print "No".
You can print each letter in any case (upper or lower).
Example
Input
6
2 2 1 2
0 100 0 1
12 13 25 1
27 83 14 25
0 0 1 0
1000000000000000000 1000000000000000000 1000000000000000000 1000000000000000000
Output
Yes
No
No
Yes
No
Yes
Note
In the first test case, let's consider the order \{1, 2, 2\} of types of guests. Then:
* The first guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 1 chocolate cookie.
* The second guest eats a chocolate cookie. After that, there are 2 vanilla cookies and 0 chocolate cookies.
* The last guest selects a chocolate cookie, but there are no chocolate cookies. So, the guest gets angry.
So, this order can't be chosen by Anna.
Let's consider the order \{2, 2, 1\} of types of guests. Then:
* The first guest eats a vanilla cookie. After that, there is 1 vanilla cookie and 2 chocolate cookies.
* The second guest eats a vanilla cookie. After that, there are 0 vanilla cookies and 2 chocolate cookies.
* The last guest eats a chocolate cookie. After that, there are 0 vanilla cookies and 1 chocolate cookie.
So, the answer to this test case is "Yes".
In the fifth test case, it is illustrated, that the number of cookies (a + b) can be equal to zero, but the number of guests (n + m) can't be equal to zero.
In the sixth test case, be careful about the overflow of 32-bit integer type. | instruction | 0 | 61,642 | 9 | 123,284 |
Tags: greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
a,b,n,m = [int(x) for x in input().split()]
if a+b<n+m:
print("No")
continue
if m>0 and min(a,b)==0:
print("No")
continue
else:
if min(a,b)<m:
print("No")
else:
print("Yes")
``` | output | 1 | 61,642 | 9 | 123,285 |
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