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.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
q=int(input())
for i in range(q):
a,b,c=map(int,input().split())
result = (a+b+c)//2
print(result)
``` | instruction | 0 | 10,014 | 9 | 20,028 |
Yes | output | 1 | 10,014 | 9 | 20,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
n = int(input())
sum0 = []
for i in range(0, n):
pile1, pile2, pile3 = input().split()
intpile1 = int(pile1)
intpile2 = int(pile2)
intpile3 = int(pile3)
sum0.append((intpile1+intpile2+intpile3)//2)
for i in range(len(sum0)):
print(sum0[i])
``` | instruction | 0 | 10,015 | 9 | 20,030 |
Yes | output | 1 | 10,015 | 9 | 20,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
q=int(input(""))
for i in range(q):
l=[int(x) for x in input("").split()]
f=l[0]
s=l[1]
t=l[2]
print((f+s+t)//2)
``` | instruction | 0 | 10,016 | 9 | 20,032 |
Yes | output | 1 | 10,016 | 9 | 20,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
q = int(input())
for i in range(q):
a, b, c = list(map(int, input().split()))
if a + b == c or a + c == b or b + c == a:
print(max(a, b, c))
elif a == b == c:
print(a * 1.5)
else:
x, y, z = 0, 0, 0
if a > b and a > c:
x, y, z = a, b, c
elif b > a and b > c:
x, y, z = b, a, c
else:
x, y, z = c, a, b
a = (x - abs(y - z)) // 2
c = abs(z - y)
z += a
y += a
b = min(z, y)
d = max(z, y)
b += c
print(int(min(b, d)))
``` | instruction | 0 | 10,017 | 9 | 20,034 |
No | output | 1 | 10,017 | 9 | 20,035 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
n = input();
n = int(n);
candy1 = [];
candy2 = [];
for i in range(n):
candy1.append(0);
candy2.append(0);
a,b,c = input().split();
a = int(a);
b = int(b);
c = int(c);
q = (a+b+c)//2;
great = max(a,b,c);
if a == great:
x = a//2;
y = a/2;
if x==y:
candy1[i] = b+x;
candy2[i] = c+x;
else:
candy1[i] = b+x;
candy2[i] = c+x+1;
if candy1[i]>candy2[i]:
v = candy1[i]-candy2[i];
if v>1:
candy1[i] = candy1[i]-(v/2);
candy2[i] = candy2[i] +(v/2);
v = candy1[i] - candy2[i];
if v==1:
if candy1[i]>candy2[i]:
candy1[i] = candy1[i] - 1;
else:
candy2[i] = candy2[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
else:
v = candy2[i]-candy1[i];
if v>1:
candy2[i] = candy2[i]-(v/2);
candy1[i] = candy1[i]+(v/2);
v = candy2[i]-candy1[i];
if v==1:
if candy2[i]>candy1[i]:
candy2[i] = candy2[i]-1;
else:
candy1[i] = candy1[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
elif b == great:
x = b//2;
y = b/2;
if x==y:
candy1[i] = a+x;
candy2[i] = c+x;
else:
candy1[i] = a+x;
candy2[i] = c+x+1;
if candy1[i]>candy2[i]:
v = candy1[i]-candy2[i];
if v>1:
candy1[i] = candy1[i]-(v/2);
candy2[i] = candy2[i] +(v/2);
v = candy1[i] - candy2[i];
if v==1:
if candy1[i]>candy2[i]:
candy1[i] = candy1[i] - 1;
else:
candy2[i] = candy2[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
else:
v = candy2[i]-candy1[i];
if v>1:
candy2[i] = candy2[i]-(v/2);
candy1[i] = candy1[i]+(v/2);
v = candy2[i]-candy1[i];
if v==1:
if candy2[i]>candy1[i]:
candy2[i] = candy2[i]-1;
else:
candy1[i] = candy1[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
elif c == great:
x = c//2;
y = c/2;
if x==y:
candy1[i] = b+x;
candy2[i] = a+x;
else:
candy1[i] = b+x;
candy2[i] = a+x+1;
if candy1[i] > candy2[i]:
v = candy1[i]-candy2[i];
if v>1:
candy1[i] = candy1[i]-(v/2);
candy2[i] = candy2[i] +(v/2);
v = candy1[i] - candy2[i];
if v==1:
if candy1[i]>candy2[i]:
candy1[i] = candy1[i] - 1;
else:
candy2[i] = candy2[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
else:
v = candy2[i]-candy1[i];
if v>1:
candy2[i] = candy2[i]-(v/2);
candy1[i] = candy1[i]+(v/2);
v = candy2[i]-candy1[i];
if v==1:
if candy2[i]>candy1[i]:
candy2[i] = candy2[i]-1;
else:
candy1[i] = candy1[i] - 1;
candy1[i] = int(candy1[i]);
candy2[i] = int(candy2[i]);
for i in range(n):
print(candy1[i]);
``` | instruction | 0 | 10,018 | 9 | 20,036 |
No | output | 1 | 10,018 | 9 | 20,037 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
is_debug = False
q = int(input())
for i in range(0, q):
d = [int(x) for x in input().split()]
d = sorted(d)
print(f"d={d}", end="|") if is_debug else ''
a = d[0]
b = d[1]
c = d[2]
diff = b - a
print(f"a={a}, b={b}, c={c}, diff={diff}", end="|") if is_debug else ''
if diff > c:
print(f"{a+c}")
continue
else:
a = a + diff
c = c - diff
print(f"a={a}, c={c}", end="|") if is_debug else ''
e = int(c / 2)
a = a + e
print(f"{a}")
``` | instruction | 0 | 10,019 | 9 | 20,038 |
No | output | 1 | 10,019 | 9 | 20,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice and Bob have received three big piles of candies as a gift. Now they want to divide these candies as fair as possible. To do this, Alice takes one pile of candies, then Bob takes one of the other two piles. The last pile is split between Alice and Bob as they want: for example, it is possible that Alice takes the whole pile, and Bob gets nothing from it.
After taking the candies from the piles, if Alice has more candies than Bob, she discards some candies so that the number of candies she has is equal to the number of candies Bob has. Of course, Bob does the same if he has more candies.
Alice and Bob want to have as many candies as possible, and they plan the process of dividing candies accordingly. Please calculate the maximum number of candies Alice can have after this division process (of course, Bob will have the same number of candies).
You have to answer q independent queries.
Let's see the following example: [1, 3, 4]. Then Alice can choose the third pile, Bob can take the second pile, and then the only candy from the first pile goes to Bob β then Alice has 4 candies, and Bob has 4 candies.
Another example is [1, 10, 100]. Then Alice can choose the second pile, Bob can choose the first pile, and candies from the third pile can be divided in such a way that Bob takes 54 candies, and Alice takes 46 candies. Now Bob has 55 candies, and Alice has 56 candies, so she has to discard one candy β and after that, she has 55 candies too.
Input
The first line of the input contains one integer q (1 β€ q β€ 1000) β the number of queries. Then q queries follow.
The only line of the query contains three integers a, b and c (1 β€ a, b, c β€ 10^{16}) β the number of candies in the first, second and third piles correspondingly.
Output
Print q lines. The i-th line should contain the answer for the i-th query β the maximum number of candies Alice can have after the division, if both Alice and Bob act optimally (of course, Bob will have the same number of candies).
Example
Input
4
1 3 4
1 10 100
10000000000000000 10000000000000000 10000000000000000
23 34 45
Output
4
55
15000000000000000
51
Submitted Solution:
```
is_debug = False
q = int(input())
for _ in range(q):
print(f"{int(sum([int(x) for x in input().split()])/2)}")
``` | instruction | 0 | 10,020 | 9 | 20,040 |
No | output | 1 | 10,020 | 9 | 20,041 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy. | instruction | 0 | 10,051 | 9 | 20,102 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n = int(input())
a = list(map(int,input().split()))
if sum(a) == 1:
print(-1)
exit()
sm = sum(a)
nmax = 10**6+10
eratos = [0 for i in range(nmax+1)]
prime = []
cnt = 2
while True:
while cnt <= nmax and eratos[cnt]:
cnt += 1
if cnt > nmax:
break
eratos[cnt] = 1
prime.append(cnt)
for i in range(cnt**2,nmax+1,cnt):
eratos[i] = 1
dvls = set()
for i in prime:
while sm%i == 0:
dvls.add(i)
sm //= i
if sm != 1:
dvls.add(sm)
ansls = []
ls = []
for dv in dvls:
cnti = 0
ans = 0
if dv == 2:
for i in range(n):
if a[i]%2:
cnti += 1
if cnti%2:
pivot = i
else:
ans += i-pivot
else:
flg = 0
for i,ai in enumerate(a):
rai = ai%dv
if rai:
cnti += rai
if 1 <= cnti <= dv//2:
ls.append((i,rai))
elif not flg:
pivot = i
while ls:
j,num = ls.pop()
ans += (pivot-j)*num
if cnti < dv:
flg = 1
need = dv-cnti
else:
cnti -= dv
ls.append((pivot,cnti))
else:
if cnti >= dv:
ans += (need)*(i-pivot)
cnti -= dv
if cnti <= dv//2:
flg = 0
if rai-need:
ls.append((i,rai-need))
else:
pivot = i
need = dv-cnti
else:
ans += rai*(i-pivot)
need -= rai
ansls.append(ans)
print(min(ansls))
``` | output | 1 | 10,051 | 9 | 20,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy. | instruction | 0 | 10,052 | 9 | 20,104 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n = int(input())
a = list(map(int,input().split()))
su = sum(a)
if su == 1:
return -1
fac = []
if not su%2:
fac.append(2)
while not su%2:
su //= 2
for i in range(3,int(su**0.5)+1,2):
if not su%i:
fac.append(i)
while not su%i:
su //= i
if su != 1:
fac.append(su)
ans = 10**20
for i in fac:
ans1,car = 0,0
for j in range(n-1):
if a[j]+car < 0:
x = abs(a[j]+car)
ans1 += x
car = -x
continue
be = (a[j]+car)%i
ab = i-be
if ab < be:
car = -ab
ans1 += ab
else:
car = be
ans1 += be
if a[-1]+car < 0:
ans1 += abs(a[-1]+car)
ans = min(ans,ans1)
return ans
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
print(main())
``` | output | 1 | 10,052 | 9 | 20,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy. | instruction | 0 | 10,053 | 9 | 20,106 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
# η΄ ε ζ°ε解
def prime_decomposition(n):
i = 2
table = []
while i * i <= n:
while n % i == 0:
n //= i
table.append(i)
i += 1
if n > 1:
table.append(n)
return table
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
# γγγγ移εγγγ¦ε
±ιε ζ°γζγ€γγγ«γγ
su = sum(A)
if su == 1:
print(-1)
exit()
primes = sorted(set(prime_decomposition(su)))
ans = 10**18
for p in primes:
an = 0
half = p >> 1
cnt = 0
for a in A:
a %= p
cnt += a
if cnt <= half:
an += cnt
else:
if cnt < p:
an += p - cnt
else:
cnt -= p
if cnt <= half:
an += cnt
else:
an += p - cnt
if ans <= an:
break
else:
ans = min(ans, an)
print(ans)
``` | output | 1 | 10,053 | 9 | 20,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy. | instruction | 0 | 10,054 | 9 | 20,108 |
Tags: constructive algorithms, greedy, math, number theory, ternary search, two pointers
Correct Solution:
```
def simple_div(x):
if not x & 1:
yield 2
while not x & 1:
x >>= 1
i = 3
while i * i <= x:
if x % i == 0:
yield i
while x % i == 0:
x //= i
i += 2
if x != 1:
yield x
def __main__():
n = int(input())
a = list(map(int, input().split()))
sa = sum(a)
a.pop()
if sa == 1:
print(-1)
return
res = 1 << 64
for d in simple_div(sa):
tmp = 0
m = 0
half = d >> 1
for x in a:
m = (x + m) % d
tmp += m if m <= half else d - m
if res <= tmp:
break
else:
res = tmp
print(res)
__main__()
``` | output | 1 | 10,054 | 9 | 20,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy.
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n = int(input())
a = list(map(int,input().split()))
su = sum(a)
if su == 1:
return -1
fac = []
if not su%2:
fac.append(2)
while not su%2:
su //= 2
for i in range(3,int(su**0.5)+1,2):
if not su%i:
fac.append(i)
while not su%i:
su //= i
if su != 1:
fac.append(su)
ans = 10**15
for i in fac:
ans1,car = 0,0
for j in range(n-1):
if a[j]+car < 0:
x = abs(a[j]+car)
ans1 += x
car = x
continue
be = (a[j]+car)%i
ab = i-be
if ab < be:
car = ab
ans1 += ab
else:
car = be
ans1 += be
ans = min(ans,ans1)
return ans
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
print(main())
``` | instruction | 0 | 10,055 | 9 | 20,110 |
No | output | 1 | 10,055 | 9 | 20,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
a = list(map(int, input().split()))
S = sum(a)
primes = []
for p in range(2, S):
if p * p > S:
break
if S % p == 0:
primes.append(p)
while S % p == 0:
S //= p
if S > 1:
primes.append(S)
res = 10 ** 100
for p in primes:
cost = 0
ct = 0
for x in a:
ct = (ct + x) % p
cost += min(ct, p - ct);
if cost > res:
break
res = min(res, cost)
if S == 1:
res = -1
print(res)
``` | instruction | 0 | 10,056 | 9 | 20,112 |
No | output | 1 | 10,056 | 9 | 20,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy.
Submitted Solution:
```
def simple_div(x):
if x % 2 == 0:
yield 2
i = 3
while i * i <= x:
if x % i == 0:
while x % i == 0:
x //= i
yield i
i += 2
if x != 1:
yield x
def __main__():
n = int(input())
a = list(map(int, input().split()))
sa = sum(a)
if sa == 1:
print(-1)
return
res = 10**100
for d in simple_div(sa):
b = a[0:]
tmp = 0
for i in range(n - 1):
shift = b[i] % d
if shift <= d // 2:
b[i] -= shift
b[i + 1] += shift
tmp += shift
else:
b[i] += d - shift
b[i + 1] -= d - shift
tmp += d - shift
if tmp < res:
res = tmp
print(res)
__main__()
``` | instruction | 0 | 10,057 | 9 | 20,114 |
No | output | 1 | 10,057 | 9 | 20,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the harder version of the problem. In this version, 1 β€ n β€ 10^6 and 0 β€ a_i β€ 10^6. You can hack this problem if you locked it. But you can hack the previous problem only if you locked both problems
Christmas is coming, and our protagonist, Bob, is preparing a spectacular present for his long-time best friend Alice. This year, he decides to prepare n boxes of chocolate, numbered from 1 to n. Initially, the i-th box contains a_i chocolate pieces.
Since Bob is a typical nice guy, he will not send Alice n empty boxes. In other words, at least one of a_1, a_2, β¦, a_n is positive. Since Alice dislikes coprime sets, she will be happy only if there exists some integer k > 1 such that the number of pieces in each box is divisible by k. Note that Alice won't mind if there exists some empty boxes.
Charlie, Alice's boyfriend, also is Bob's second best friend, so he decides to help Bob by rearranging the chocolate pieces. In one second, Charlie can pick up a piece in box i and put it into either box i-1 or box i+1 (if such boxes exist). Of course, he wants to help his friend as quickly as possible. Therefore, he asks you to calculate the minimum number of seconds he would need to make Alice happy.
Input
The first line contains a single integer n (1 β€ n β€ 10^6) β the number of chocolate boxes.
The second line contains n integers a_1, a_2, β¦, a_n (0 β€ a_i β€ 10^6) β the number of chocolate pieces in the i-th box.
It is guaranteed that at least one of a_1, a_2, β¦, a_n is positive.
Output
If there is no way for Charlie to make Alice happy, print -1.
Otherwise, print a single integer x β the minimum number of seconds for Charlie to help Bob make Alice happy.
Examples
Input
3
4 8 5
Output
9
Input
5
3 10 2 1 5
Output
2
Input
4
0 5 15 10
Output
0
Input
1
1
Output
-1
Note
In the first example, Charlie can move all chocolate pieces to the second box. Each box will be divisible by 17.
In the second example, Charlie can move a piece from box 2 to box 3 and a piece from box 4 to box 5. Each box will be divisible by 3.
In the third example, each box is already divisible by 5.
In the fourth example, since Charlie has no available move, he cannot help Bob make Alice happy.
Submitted Solution:
```
# η΄ ε ζ°ε解
def prime_decomposition(n):
i = 2
table = []
while i * i <= n:
while n % i == 0:
n //= i
table.append(i)
i += 1
if n > 1:
table.append(n)
return table
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
# γγγγ移εγγγ¦ε
±ιε ζ°γζγ€γγγ«γγ
su = sum(A)
if su == 1:
print(-1)
exit()
primes = list(set(prime_decomposition(su)))
ans = float("inf")
for p in primes:
an = 0
half = p // 2
cnt = 0
B = [a%p for a in A]
for a in B:
cnt += a
if cnt <= half:
an += cnt
else:
an += max(p - cnt, 0)
if cnt >= p:
cnt -= p
ans = min(ans, an)
print(ans)
``` | instruction | 0 | 10,058 | 9 | 20,116 |
No | output | 1 | 10,058 | 9 | 20,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,143 | 9 | 20,286 |
Tags: constructive algorithms, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')
for _ in range(int(input())):
n = int(input())
print(n)
print(*range(1, n + 1))
``` | output | 1 | 10,143 | 9 | 20,287 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,144 | 9 | 20,288 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=[]
for i in range(1,n+1):
a.append(i)
print(n)
print(*a)
``` | output | 1 | 10,144 | 9 | 20,289 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,145 | 9 | 20,290 |
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input().strip())
for _ in range(t):
n = int(input().strip())
ans = [i for i in range(2,n + 1)]
print(len(ans))
print(' '.join(str(i) for i in ans))
``` | output | 1 | 10,145 | 9 | 20,291 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,146 | 9 | 20,292 |
Tags: constructive algorithms, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
print(n-1)
j=n
for i in range(2,n+1):
print(i,end=" ")
``` | output | 1 | 10,146 | 9 | 20,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,147 | 9 | 20,294 |
Tags: constructive algorithms, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
l=[x+1 for x in range(n)]
print(n)
print(*l)
``` | output | 1 | 10,147 | 9 | 20,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,148 | 9 | 20,296 |
Tags: constructive algorithms, math
Correct Solution:
```
#bsdwalon aasan question diya karo
for _ in range(int(input())):
n = int(input())
print(n-1)
temp = []
for i in range(2, n+1):
temp.append(i)
print(*temp)
``` | output | 1 | 10,148 | 9 | 20,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,149 | 9 | 20,298 |
Tags: constructive algorithms, math
Correct Solution:
```
a=int(input())
for i in range(a):
b=int(input())
arr=[str(i) for i in range(b,1,-1)]
arr=arr[::-1]
print(len(arr))
print(' '.join(arr))
``` | output | 1 | 10,149 | 9 | 20,299 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12]. | instruction | 0 | 10,150 | 9 | 20,300 |
Tags: constructive algorithms, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
print(n)
a = [i+1 for i in range(n)]
print(*a)
``` | output | 1 | 10,150 | 9 | 20,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
from __future__ import division, print_function
import os
import sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
#########
def solve():
a = int(input())
print(a)
lst = []
for i in range(1,a+1):
lst.append(i)
print(*lst)
#########
def main():
testcases = 1
testcases = int(input())
for _ in range(testcases):
solve()
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 10,151 | 9 | 20,302 |
Yes | output | 1 | 10,151 | 9 | 20,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
T = int(input())
for _ in range(T):
N = int(input())
print(N)
print(*range(1, N + 1))
``` | instruction | 0 | 10,152 | 9 | 20,304 |
Yes | output | 1 | 10,152 | 9 | 20,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
print(n)
for i in range(1, n + 1):
print(i, end=' ')
``` | instruction | 0 | 10,153 | 9 | 20,306 |
Yes | output | 1 | 10,153 | 9 | 20,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
print(n)
for i in range(n):
print(i+1,end=" ")
print()
``` | instruction | 0 | 10,154 | 9 | 20,308 |
Yes | output | 1 | 10,154 | 9 | 20,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
import math
def lcm(a, b):
m = a * b
while a != 0 and b != 0:
if a > b:
a %= b
else:
b %= a
return m // (a + b)
def dtod(a):
s=''
while a!=0:
s+=str(a%2)
a=int(a/2)
return s[::-1]
n = int(input())
for i in range(n):
a = int(input())
print(a - 1)
b = list(range(a, 1, -1))
for item in b:
print((str(item) + ' ') * (item - 2) + str(item), end = ' ')
``` | instruction | 0 | 10,155 | 9 | 20,310 |
No | output | 1 | 10,155 | 9 | 20,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
print(n)
for i in range(n, 0, -1):
print(i, end = " ")
print()
``` | instruction | 0 | 10,156 | 9 | 20,312 |
No | output | 1 | 10,156 | 9 | 20,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
import sys
import math
from collections import Counter
from collections import OrderedDict
from collections import defaultdict
from functools import reduce
#from itertools import groupby
sys.setrecursionlimit(10**6)
def inputt():
return sys.stdin.readline().strip()
def printt(n):
sys.stdout.write(str(n)+'\n')
def listt():
return [int(i) for i in inputt().split()]
def gcd(a,b):
return math.gcd(a,b)
def lcm(a,b):
return (a*b) // gcd(a,b)
def factors(n):
step = 2 if n%2 else 1
return set(reduce(list.__add__,([i, n//i] for i in range(1, int(math.sqrt(n))+1, step) if n % i == 0)))
def comb(n,k):
factn=math.factorial(n)
factk=math.factorial(k)
fact=math.factorial(n-k)
ans=factn//(factk*fact)
return ans
def is_prime(n):
if n <= 1:
return False
if n == 2:
return True
if n > 2 and n % 2 == 0:
return False
max_div = math.floor(math.sqrt(n))
for i in range(3, 1 + max_div, 2):
if n % i == 0:
return False
return True
def maxpower(n,x):
B_max = int(math.log(n, x)) + 1 #tells upto what power of x n is less than it like 1024->5^4
return B_max
t=int(input())
for _ in range(t):
n=int(inputt())
print(n)
for i in range(n,0,-1):
print(i,end=" ")
print()
``` | instruction | 0 | 10,157 | 9 | 20,314 |
No | output | 1 | 10,157 | 9 | 20,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n bags with candies, initially the i-th bag contains i candies. You want all the bags to contain an equal amount of candies in the end.
To achieve this, you will:
* Choose m such that 1 β€ m β€ 1000
* Perform m operations. In the j-th operation, you will pick one bag and add j candies to all bags apart from the chosen one.
Your goal is to find a valid sequence of operations after which all the bags will contain an equal amount of candies.
* It can be proved that for the given constraints such a sequence always exists.
* You don't have to minimize m.
* If there are several valid sequences, you can output any.
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 β€ t β€ 100). Description of the test cases follows.
The first and only line of each test case contains one integer n (2 β€ nβ€ 100).
Output
For each testcase, print two lines with your answer.
In the first line print m (1β€ m β€ 1000) β the number of operations you want to take.
In the second line print m positive integers a_1, a_2, ..., a_m (1 β€ a_i β€ n), where a_j is the number of bag you chose on the j-th operation.
Example
Input
2
2
3
Output
1
2
5
3 3 3 1 2
Note
In the first case, adding 1 candy to all bags except of the second one leads to the arrangement with [2, 2] candies.
In the second case, firstly you use first three operations to add 1+2+3=6 candies in total to each bag except of the third one, which gives you [7, 8, 3]. Later, you add 4 candies to second and third bag, so you have [7, 12, 7], and 5 candies to first and third bag β and the result is [12, 12, 12].
Submitted Solution:
```
"""
Code of Ayush Tiwari
Codeforces: servermonk
Codechef: ayush572000
"""
import sys
input = sys.stdin.buffer.readline
def solution():
# This is the main code
n=int(input())
print(n)
l=[]
for i in range(n,0,-1):
l.append(i)
print(*l)
t=int(input())
for _ in range(t):
solution()
``` | instruction | 0 | 10,158 | 9 | 20,316 |
No | output | 1 | 10,158 | 9 | 20,317 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,386 | 9 | 20,772 |
Tags: binary search, implementation
Correct Solution:
```
#------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def value():return tuple(map(int,input().split()))
def array():return [int(i) for i in input().split()]
def Int():return int(input())
def Str():return input()
def arrayS():return [i for i in input().split()]
#-------------------------code---------------------------#
#vsInput()
def canMake(x):
needed=[x*i for i in quantity]
#print(x,needed)
powder=magic
for i in range(n):
powder-=max(0,needed[i]-have[i])
if(powder<0):
return False
return True
n,magic=value()
quantity=array()
have=array()
low=0
high=10**15
ans=0
while(low<=high):
mid=(low+high)//2
if(canMake(mid)):
ans=mid
low=mid+1
else:
high=mid-1
print(ans)
``` | output | 1 | 10,386 | 9 | 20,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,387 | 9 | 20,774 |
Tags: binary search, implementation
Correct Solution:
```
n,k=map(int,input().split())
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
def bs(m):
c=0
for i in range(n):
if a[i]*m>b[i]:
c+=abs(b[i]-(a[i]*m))
if c<=k:
return True
else:
return False
l,h=0,10**10
ans=0
while(l<=h):
m=l+(h-l)//2
if bs(m):
ans=m
l=m+1
else:
h=m-1
print(ans)
``` | output | 1 | 10,387 | 9 | 20,775 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,388 | 9 | 20,776 |
Tags: binary search, implementation
Correct Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
lo = 0
hi = 2*1e9
def p(cookies):
powder = k
for i in range(len(b)):
have = b[i]
one = a[i]
remainder = have - (one * cookies)
if remainder < 0:
powder += remainder
if powder < 0:
return False
return True
while lo < hi:
m = (lo + hi) // 2
if p(m):
lo = m + 1
else:
hi = m
print(int(lo if p(lo) else lo - 1))
"""
"""
``` | output | 1 | 10,388 | 9 | 20,777 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,389 | 9 | 20,778 |
Tags: binary search, implementation
Correct Solution:
```
"""
#If FastIO not needed, used this and don't forget to strip
#import sys, math
#input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict as dd, deque as dq
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
MOD = 10**9+7
"""
Without the powder we can make min(AN//BN)
With the powder, we need sum(AN % BN) to make one more
Then we need sum(AN) for every other
"""
def solve():
N, K = getInts()
A = getInts()
B = getInts()
curr_min = 10**10
for i in range(N):
curr_min = min(curr_min, B[i]//A[i])
ans = curr_min
rem = []
for i in range(N):
rem.append(B[i]-curr_min*A[i])
def works(D):
req = 0
for i in range(N):
req += max(0,D*A[i]-rem[i])
return req <= K
assert works(0)
assert not works(10**18)
left = 0
right = 10**18
while right-left > 1:
mid = (right+left)//2
if works(mid):
left = mid
else:
right = mid
return ans+left
#for _ in range(getInt()):
print(solve())
``` | output | 1 | 10,389 | 9 | 20,779 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,390 | 9 | 20,780 |
Tags: binary search, implementation
Correct Solution:
```
n, k = map(int, input().split())
b = list(map(int, input().split()))
a = list(map(int, input().split()))
c = []
x = 0
for i in range(n):
c.append([a[i]//b[i], a[i], b[i]])
c.append([10**12, 0, 10**10])
c = sorted(c)
ans = c[0][0]
suma = 0
sumb = 0
i = 0
n += 1
while i < n:
j = i
cura = 0
curb = 0
while i < n and c[j][0] == c[i][0]:
cura += c[i][1]
curb += c[i][2]
i += 1
if sumb * c[j][0] - suma <= k:
ans = max(ans, c[j][0])
else:
ans = max(ans, min((suma + k)//sumb, c[j][0]-1))
break
suma += cura
sumb += curb
print(ans)
``` | output | 1 | 10,390 | 9 | 20,781 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,391 | 9 | 20,782 |
Tags: binary search, implementation
Correct Solution:
```
read = lambda: map(int, input().split())
n, k = read()
a = list(read())
b = list(read())
c = [0] * n
r = [0] * n
for i in range(n):
c[i] = b[i] // a[i]
r[i] = a[i] - b[i] % a[i]
def f(x):
k1 = k
for i in range(n):
if c[i] >= x:
continue
cnt = (x - c[i] - 1) * a[i] + r[i]
k1 -= cnt
if k1 < 0:
return False
return True
L, R = 0, 10 ** 10
while R - L > 1:
M = (L + R) // 2
if f(M): L = M
else: R = M
ans = L
print(ans)
``` | output | 1 | 10,391 | 9 | 20,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,392 | 9 | 20,784 |
Tags: binary search, implementation
Correct Solution:
```
import sys,math,bisect
sys.setrecursionlimit(10**5)
from random import randint
inf = float('inf')
mod = 10**9+7
"========================================"
def lcm(a,b):
return int((a/math.gcd(a,b))*b)
def gcd(a,b):
return int(math.gcd(a,b))
def tobinary(n):
return bin(n)[2:]
def binarySearch(a,x):
i = bisect.bisect_left(a,x)
if i!=len(a) and a[i]==x:
return i
else:
return -1
def lowerBound(a, x):
i = bisect.bisect_left(a, x)
if i:
return (i-1)
else:
return -1
def upperBound(a,x):
i = bisect.bisect_right(a,x)
if i!= len(a)+1 and a[i-1]==x:
return (i-1)
else:
return -1
def primesInRange(n):
ans = []
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
for p in range(2, n+1):
if prime[p]:
ans.append(p)
return ans
def primeFactors(n):
factors = []
while n % 2 == 0:
factors.append(2)
n = n // 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
factors.append(i)
n = n // i
if n > 2:
factors.append(n)
return factors
def isPrime(n,k=5):
if (n <2):
return True
for i in range(0,k):
a = randint(1,n-1)
if(pow(a,n-1,n)!=1):
return False
return True
"========================================="
"""
n = int(input())
n,k = map(int,input().split())
arr = list(map(int,input().split()))
"""
from collections import deque,defaultdict,Counter
from heapq import heappush, heappop,heapify
import string
for _ in range(1):
n,k=map(int,input().split())
need=list(map(int,input().split()))
have=list(map(int,input().split()))
def canMake(need,have,toMake,k):
req = 0
for i in range(len(need)):
if need[i]*toMake <= have[i]:
pass
else:
req+= (need[i]*toMake)-have[i]
if req<=k:
return True
return False
ans = 0
l=1
r=1e20
while l<=r:
mid=(l+r)//2
if canMake(need,have,mid,k):
l=mid+1
ans=max(ans,mid)
else:
r=mid-1
print(int(ans))
``` | output | 1 | 10,392 | 9 | 20,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3 | instruction | 0 | 10,393 | 9 | 20,786 |
Tags: binary search, implementation
Correct Solution:
```
def process(key, a,b, k,n):
#print(key, nc,end=" ")
for i in range(n):
y=b[i]-(a[i]*key)
if y<0:
k+=y
if k<0:
return False
#print(k, res)
if k>=0:
return True
else:
return False
def bin_src(a,b,n,k,lo,hi):
while (hi-lo>1):
mid = ((hi +lo) // 2)
val = process(mid,a,b,k,n)
if val==True:
lo=mid
else:
hi=mid
if(process(lo,a,b,k,n)==True):
return lo
else:
return hi
def main():
nk = list(map(int, input().split(" ")))
n,k=nk[0],nk[1]
a=list(map(int,input().split(" ")))
b=list(map(int,input().split(" ")))
print(bin_src(a,b,n,k, 0, (2*(10**9))+1))
if __name__=="__main__":
main()
``` | output | 1 | 10,393 | 9 | 20,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
def ok(x):
need = 0
for i in range(n):
tmp = a[i] * x - b[i]
if tmp > 0:
need += tmp
return need <= k
n, k = (int(_) for _ in input().split())
a = [int(_) for _ in input().split()]
b = [int(_) for _ in input().split()]
lo, hi = 0, 2 * 10 ** 9
while lo <= hi:
mid = (lo + hi) >> 1
if ok(mid):
lo = mid + 1
else:
hi = mid - 1
print(hi)
``` | instruction | 0 | 10,394 | 9 | 20,788 |
Yes | output | 1 | 10,394 | 9 | 20,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
I=lambda:list(map(int,input().split()))
n,k=I()
a=I()
b=I()
l=0
r=2*10**9
while l<r:
m=(l+r)//2+1;s=sum(max(a[i]*m-b[i],0)for i in range(n))
if s>k:r=m-1
else:l=m
print(l)
``` | instruction | 0 | 10,395 | 9 | 20,790 |
Yes | output | 1 | 10,395 | 9 | 20,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
"""
keywords : python input intput python how to take input in python
Python Input :-
1. Just one value : a = int(input())
2. Two or three consecutive values : m,n=map(int,input().split())
3. Array as a space separated line : ar=list(map(int,input().split()))
"""
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
def check(mid):
left = 0
for i in range(n):
left += max(0, mid * a[i] - b[i])
return left <= k
def binsearch(lo, hi):
ans = int(-1e18)
while lo <= hi:
mid = (lo + hi + 1) // 2
#print("min = ", mid)
if check(mid):
ans = max(ans, mid)
lo = mid + 1
else:
hi = mid - 1
return ans
ans = binsearch(0, int(1e18))
print(ans)
``` | instruction | 0 | 10,396 | 9 | 20,792 |
Yes | output | 1 | 10,396 | 9 | 20,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
def check(a,b,midd,k):
ans1=0
for i in range(n):
if a[i]*midd<=b[i]:
continue
ans1+=(midd*a[i]-b[i])
if ans1>k:return False
return True
n,k=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
l=0
r=3*(10**9)
ans=0
while l<=r:
mid=(l+r)//2
if check(a,b,mid,k):
l=mid+1
ans=max(ans,mid)
else:r=mid-1
print(ans)
``` | instruction | 0 | 10,397 | 9 | 20,794 |
Yes | output | 1 | 10,397 | 9 | 20,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
from math import ceil,sqrt,gcd,log,floor
from collections import deque
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().strip().split(" "))
def li(): return list(mi())
def msi(): return map(str,input().strip().split(" "))
def lsi(): return list(msi())
#for _ in range(ii()):
n,k=mi()
a=li()
b=li()
l=0
r=1000000005
while(l<r):
m=(l+r)//2
s=0
for i in range(n):
if(b[i]<(m*a[i])):
s+=(m*a[i]-b[i])
if(s>k):
r=m-1
else:
m1=m
s1=s
l=m+1
k-=s1
m1+=(k//sum(a))
print(m1)
``` | instruction | 0 | 10,398 | 9 | 20,796 |
No | output | 1 | 10,398 | 9 | 20,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
n,k=input().split()
n=int(n)
k=int(k)
a=[]
a=input().split()
b=input().split()
for i in range(n):
a[i],b[i]=int(a[i]),int(b[i])
min=b[0]//a[0]
for i in range(1,n):
if(min>b[i]//a[i]):
min=b[i]//a[i]
qty=min
for i in range(n):
b[i]-=a[i]*min
diff=0
for i in range(n):
if(a[i]-b[i]>0):
diff+=a[i]-b[i]
else:
k+=b[i]-a[i]
if(diff>k):
print(qty)
elif(diff==k):
print(qty+1)
else:
k-=diff
total=sum(a)
print(qty+k//total+1)
``` | instruction | 0 | 10,399 | 9 | 20,798 |
No | output | 1 | 10,399 | 9 | 20,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
def main():
from heapq import heappushpop, heapify
n, k = map(int, input().split())
l = list((b // a, a - (b % a), a) for a, b in zip(map(int, input().split()), map(int, input().split())))
heapify(l)
b = r = a = 10 ** 9
while True:
b, r, a = heappushpop(l, (b, r, a))
if k < r:
break
else:
k -= r
b += 1
r = a
print(b)
if __name__ == '__main__':
main()
``` | instruction | 0 | 10,400 | 9 | 20,800 |
No | output | 1 | 10,400 | 9 | 20,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The term of this problem is the same as the previous one, the only exception β increased restrictions.
Input
The first line contains two positive integers n and k (1 β€ n β€ 100 000, 1 β€ k β€ 109) β the number of ingredients and the number of grams of the magic powder.
The second line contains the sequence a1, a2, ..., an (1 β€ ai β€ 109), 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 β€ bi β€ 109), 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
1 1000000000
1
1000000000
Output
2000000000
Input
10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000
1 1 1 1 1 1 1 1 1 1
Output
0
Input
3 1
2 1 4
11 3 16
Output
4
Input
4 3
4 3 5 6
11 12 14 20
Output
3
Submitted Solution:
```
n,k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
r = min([B[i]//A[i] for i in range(n)])
for i in range(n):
B[i] = B[i] - r*A[i]
s = sum(A)
r2 = k // s
k = k - r2*s
r += r2
ok = 1
while ok:
L = [0 for _ in range(n)]
for i in range(n):
B[i] = B[i] - A[i]
if B[i] < 0:
L[i] = -B[i]
B[i] = 0
if sum(L) <= k:
r += 1
k = k - sum(L)
else:
ok = 0
print(r)
``` | instruction | 0 | 10,401 | 9 | 20,802 |
No | output | 1 | 10,401 | 9 | 20,803 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is organizing a party.
At the party, each guest will receive one or more snack pieces.
Takahashi predicts that the number of guests at this party will be A or B.
Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.
We assume that a piece cannot be divided and distributed to multiple guests.
Constraints
* 1 \leq A, B \leq 10^5
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.
Examples
Input
2 3
Output
6
Input
123 456
Output
18696
Input
100000 99999
Output
9999900000 | instruction | 0 | 11,424 | 9 | 22,848 |
"Correct Solution:
```
import fractions
A,B=map(int,input().split())
f=A*B//fractions.gcd(A,B)
print(f)
``` | output | 1 | 11,424 | 9 | 22,849 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is organizing a party.
At the party, each guest will receive one or more snack pieces.
Takahashi predicts that the number of guests at this party will be A or B.
Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.
We assume that a piece cannot be divided and distributed to multiple guests.
Constraints
* 1 \leq A, B \leq 10^5
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.
Examples
Input
2 3
Output
6
Input
123 456
Output
18696
Input
100000 99999
Output
9999900000 | instruction | 0 | 11,425 | 9 | 22,850 |
"Correct Solution:
```
a,b=map(int,input().split())
import fractions as fr
print(int(a*b/fr.gcd(a,b)))
``` | output | 1 | 11,425 | 9 | 22,851 |
Provide a correct Python 3 solution for this coding contest problem.
Takahashi is organizing a party.
At the party, each guest will receive one or more snack pieces.
Takahashi predicts that the number of guests at this party will be A or B.
Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.
We assume that a piece cannot be divided and distributed to multiple guests.
Constraints
* 1 \leq A, B \leq 10^5
* A \neq B
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
A B
Output
Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.
Examples
Input
2 3
Output
6
Input
123 456
Output
18696
Input
100000 99999
Output
9999900000 | instruction | 0 | 11,426 | 9 | 22,852 |
"Correct Solution:
```
import fractions
A,B=map(int,input().split())
a=fractions.gcd(A,B)
print((A * B) // a)
``` | output | 1 | 11,426 | 9 | 22,853 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.