message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:
1. For given numbers xi and vi assign value vi to element axi.
2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2.
3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
* if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105);
* if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n);
* if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105).
The input limits for scoring 30 points are (subproblem E1):
* It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.
The input limits for scoring 70 points are (subproblems E1+E2):
* It is guaranteed that there will be queries of the 1-st and 2-nd type only.
The input limits for scoring 100 points are (subproblems E1+E2+E3):
* No extra limitations.
Output
For each query print the calculated sum modulo 1000000000 (109).
Examples
Input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5
Output
12
32
8
50
Input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5
Output
12
45
Submitted Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
f = [1,1]
for i in range(2,n+1):
f.append(f[i-1] + f[i-2])
f[i] %= 10 ** 9
for q in range(m):
z,l,r = map(int,input().split())
if z == 1:
a[l-1] = r
else:
s = 0
for j in range(l-1,r):
s += (a[j] * f[j-l+1])
s %= 10 ** 9
print(s)
``` | instruction | 0 | 12,775 | 5 | 25,550 |
Yes | output | 1 | 12,775 | 5 | 25,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:
1. For given numbers xi and vi assign value vi to element axi.
2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2.
3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
* if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105);
* if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n);
* if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105).
The input limits for scoring 30 points are (subproblem E1):
* It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.
The input limits for scoring 70 points are (subproblems E1+E2):
* It is guaranteed that there will be queries of the 1-st and 2-nd type only.
The input limits for scoring 100 points are (subproblems E1+E2+E3):
* No extra limitations.
Output
For each query print the calculated sum modulo 1000000000 (109).
Examples
Input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5
Output
12
32
8
50
Input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5
Output
12
45
Submitted Solution:
```
n,m = map(int,input().split())
a = list(map(int,input().split()))
f = [1,1]
for i in range(2,n+1):
f.append(f[i-1] + f[i-2])
for q in range(m):
z,l,r = map(int,input().split())
if z == 1:
a[l-1] = r
else:
s = 0
for j in range(l-1,r):
s += (a[j] * f[j-l+1])
print(s)
``` | instruction | 0 | 12,776 | 5 | 25,552 |
No | output | 1 | 12,776 | 5 | 25,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:
1. For given numbers xi and vi assign value vi to element axi.
2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2.
3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
* if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105);
* if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n);
* if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105).
The input limits for scoring 30 points are (subproblem E1):
* It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.
The input limits for scoring 70 points are (subproblems E1+E2):
* It is guaranteed that there will be queries of the 1-st and 2-nd type only.
The input limits for scoring 100 points are (subproblems E1+E2+E3):
* No extra limitations.
Output
For each query print the calculated sum modulo 1000000000 (109).
Examples
Input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5
Output
12
32
8
50
Input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5
Output
12
45
Submitted Solution:
```
m,n = map(int, input().split())
a = list(map(int, input().split()))
f = [1]*n
for i in range(2,n):
f[i] = f[i-1]+f[i-2]
for i in range(n):
t,l,r = map(int, input().split())
if t==1:
a[l-1] = r
else:
sum_lr = 0
for x in range(r-l+1):
sum_lr += f[x] * a[l+x-1]
print(sum_lr)
``` | instruction | 0 | 12,777 | 5 | 25,554 |
No | output | 1 | 12,777 | 5 | 25,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:
1. For given numbers xi and vi assign value vi to element axi.
2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2.
3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
* if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105);
* if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n);
* if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105).
The input limits for scoring 30 points are (subproblem E1):
* It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.
The input limits for scoring 70 points are (subproblems E1+E2):
* It is guaranteed that there will be queries of the 1-st and 2-nd type only.
The input limits for scoring 100 points are (subproblems E1+E2+E3):
* No extra limitations.
Output
For each query print the calculated sum modulo 1000000000 (109).
Examples
Input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5
Output
12
32
8
50
Input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5
Output
12
45
Submitted Solution:
```
import sys
f = [-1] * 200000
def fib(n):
if f[n] == -1:
k = n//2
while k > 0 and f[k] == -1:
k = k/2
for i in range(k, n):
f[i+1] = f[i] + f[i-1];
return f[n]
def solve(numbers, t, l, r, d):
if t == 1:
numbers[l-1] = r
elif t == 2:
total = 0
for j in range(l-1, r):
total = total + numbers[j] * fib(j)
print(total)
else:
for j in range(l-1, r):
numbers[j] = numbers[j] + d
n, m = sys.stdin.readline().rstrip().split()
n = int(n)
m = int(m)
numbers = list(map(lambda x: int (x), sys.stdin.readline().rstrip().split()))
f[0] = f[1] = 1
for i in range(m):
line = sys.stdin.readline().rstrip().split()
t = int(line[0])
d = 0
if t == 1:
l = int(line[1])
r = int(line[2])
elif t == 2:
l = int(line[1])
r = int(line[2])
else:
l = int(line[1])
r = int(line[2])
d = int(line[3])
solve(numbers, t, l, r, d)
``` | instruction | 0 | 12,778 | 5 | 25,556 |
No | output | 1 | 12,778 | 5 | 25,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
By the age of three Smart Beaver mastered all arithmetic operations and got this summer homework from the amazed teacher:
You are given a sequence of integers a1, a2, ..., an. Your task is to perform on it m consecutive operations of the following type:
1. For given numbers xi and vi assign value vi to element axi.
2. For given numbers li and ri you've got to calculate sum <image>, where f0 = f1 = 1 and at i ≥ 2: fi = fi - 1 + fi - 2.
3. For a group of three numbers li ri di you should increase value ax by di for all x (li ≤ x ≤ ri).
Smart Beaver planned a tour around great Canadian lakes, so he asked you to help him solve the given problem.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 2·105) — the number of integers in the sequence and the number of operations, correspondingly. The second line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 105). Then follow m lines, each describes an operation. Each line starts with an integer ti (1 ≤ ti ≤ 3) — the operation type:
* if ti = 1, then next follow two integers xi vi (1 ≤ xi ≤ n, 0 ≤ vi ≤ 105);
* if ti = 2, then next follow two integers li ri (1 ≤ li ≤ ri ≤ n);
* if ti = 3, then next follow three integers li ri di (1 ≤ li ≤ ri ≤ n, 0 ≤ di ≤ 105).
The input limits for scoring 30 points are (subproblem E1):
* It is guaranteed that n does not exceed 100, m does not exceed 10000 and there will be no queries of the 3-rd type.
The input limits for scoring 70 points are (subproblems E1+E2):
* It is guaranteed that there will be queries of the 1-st and 2-nd type only.
The input limits for scoring 100 points are (subproblems E1+E2+E3):
* No extra limitations.
Output
For each query print the calculated sum modulo 1000000000 (109).
Examples
Input
5 5
1 3 1 2 4
2 1 4
2 1 5
2 2 4
1 3 10
2 1 5
Output
12
32
8
50
Input
5 4
1 3 1 2 4
3 1 4 1
2 2 4
1 2 10
2 1 5
Output
12
45
Submitted Solution:
```
n,m = map(int, input().split())
a = list(map(int, input().split()))
f = [1]*n
for i in range(2,n):
f[i] = f[i-1]+f[i-2]
for i in range(n):
t,l,r = map(int, input().split())
if t==1:
a[l-1] = r
else:
sum_lr = 0
for x in range(r-l+1):
sum_lr += f[x] * a[l+x-1]
print(sum_lr%1000000000)
``` | instruction | 0 | 12,779 | 5 | 25,558 |
No | output | 1 | 12,779 | 5 | 25,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image> | instruction | 0 | 12,893 | 5 | 25,786 |
Tags: geometry, math
Correct Solution:
```
def main():
a, b = [int(i) for i in input().split(" ")]
eps = 0.0000000001
if (b > a):
print (-1)
elif b == a:
print(b)
elif b == 0:
print(0)
else:
x1, x2 = float('inf'), float('inf')
z1, z2 = (a-b)/2, (a+b)/2
k, kk = 1, 1
while kk:
while (z1/k) + eps >= b:
x1 = (z1/k)
k += kk
kk *= 2
k -= kk/2
kk //= 4
k, kk = 1, 1
while kk:
while (z2/k) + eps >= b:
x2 = (z2/k)
k += kk
kk *= 2
k -= kk/2
kk //= 4
if (min(x1, x2) != float('inf')):
print(min(x1, x2))
else:
print(-1)
if __name__ == "__main__":
main()
``` | output | 1 | 12,893 | 5 | 25,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
from math import floor
a, b = map(int, input().split())
if a < 0:
print(-1)
elif b < 0:
print(-1)
elif a < b:
print(-1)
elif a == 0 and b != 0:
print(-1)
elif a == b:
print(b)
elif b == 0:
print(0)
else:
n1 = floor((a - b) / (2 * b))
n2 = floor((a + b) / (2 * b))
if n1 == 0:
print((a + b) / (2 * n2))
else:
x1 = (a - b) / (2 * n1)
x2 = (a + b) / (2 * n2)
print(min(x1, x2))
``` | instruction | 0 | 12,899 | 5 | 25,798 |
Yes | output | 1 | 12,899 | 5 | 25,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
from math import floor
def main():
a, b = map(int, input().split())
if (b > a):
print(-1)
return
y = (a + b) / 2
k = y // b
print(y / k)
main()
``` | instruction | 0 | 12,900 | 5 | 25,800 |
Yes | output | 1 | 12,900 | 5 | 25,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
from math import floor
def main():
a, b = map(int, input().split())
if (b > a):
print(-1)
return
if (b == a):
print(a)
return
y = (a + b) / 2
k = y // b
print(y / k)
main()
``` | instruction | 0 | 12,901 | 5 | 25,802 |
Yes | output | 1 | 12,901 | 5 | 25,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
import decimal
a, b = map(int, input().split())
a = decimal.Decimal(a)
b = decimal.Decimal(b)
if a<b:
print('-1')
else:
k = a//b
ans = round((a+b)/k,12)
#print('1: '+str(ans))
#print('2: '+str(((a+b)/(k+1))))
if k%2==1:
ans = round((a+b)/(k+1),12)
print(ans)
``` | instruction | 0 | 12,902 | 5 | 25,804 |
Yes | output | 1 | 12,902 | 5 | 25,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
def main():
a, b = [int(i) for i in input().split(" ")]
if (b > a or a < 0):
print(-1)
else:
y1 = a+b
x = y1/2
x1 = float('inf')
k = 1
while k < 100000:
if (2*k*x >= a and a >= (2*k-1)*x):
x1 = x
k += 1
x = y1 / (2*k)
y2 = a-b
x = y2
x2 = float('inf')
k = 0
while k < 100000:
if (a >= 2*k*x and (2*k+1)*x >= a):
x2 = x
k += 1
x = y2 / (2*k)
j = min(x1, x2)
if (j != float('inf')):
print(j)
else:
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 12,903 | 5 | 25,806 |
No | output | 1 | 12,903 | 5 | 25,807 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
def main():
a, b = [int(i) for i in input().split(" ")]
eps = 0.0000000001
if (b > a):
print (-1)
elif b == a:
print(b)
elif b == 0:
print(0)
else:
x1, x2 = float('inf'), float('inf')
z1, z2 = (a-b)/2, (a+b)/2
k = 1
while (z1/k) + eps >= b:
x1 = (z1/k)
k *= 2
while (z1/k) + eps >= b:
x1 = (z1/k)
k += 1
k = 1
while (z2/k) + eps >= b:
x2 = (z2/k)
k *= 2
while (z2/k) + eps >= b:
x2 = (z2/k)
k += 1
if (min(x1, x2) != float('inf')):
print(min(x1, x2))
else:
print(-1)
if __name__ == "__main__":
main()
``` | instruction | 0 | 12,904 | 5 | 25,808 |
No | output | 1 | 12,904 | 5 | 25,809 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
from math import floor
a, b = map(int, input().split())
if a < 0:
print(-1)
elif b < 0:
print(-1)
elif a == 0 and b != 0:
print(-1)
elif a == b:
print(b)
elif b == 0:
print(0)
else:
n = floor((a - b) / (2 * b))
if n == 0:
n = floor((a + b) / (2 * b))
if n == 0:
print(-1)
x1 = 100000000000
else:
x1 = (a - b) / (2 * n)
n = floor((a + b) / (2 * b))
if n == 0:
x2 = 100000000000
else:
x2 = (a + b) / (2 * n)
print(min(x1, x2))
``` | instruction | 0 | 12,905 | 5 | 25,810 |
No | output | 1 | 12,905 | 5 | 25,811 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – ....
We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.
Input
Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).
Output
Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output - 1 as the answer.
Examples
Input
3 1
Output
1.000000000000
Input
1 3
Output
-1
Input
4 1
Output
1.250000000000
Note
You can see following graphs for sample 1 and sample 3.
<image> <image>
Submitted Solution:
```
def gex_min_x(x0, ty):
while x0 / 2 >= ty:
x0 /= 2
return x0
a, b = list(map(int, input().split()))
x1 = a - b
x2 = a + b
if x1 < 0:
print(-1)
elif x1 == 0:
print('%.10f' % (x2 / 2))
elif x1 / 2 <= b:
res2 = gex_min_x(x2 / 2, b)
print('%.10f' % res2)
else:
res1 = gex_min_x(x1 / 2, b)
res2 = gex_min_x(x2 / 2, b)
print('%.10f' % min(res1, res2))
``` | instruction | 0 | 12,906 | 5 | 25,812 |
No | output | 1 | 12,906 | 5 | 25,813 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5
Submitted Solution:
```
def main():
input()
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = [(a[i], "a") for i in range(len(a))]
c.extend([(b[i], "b") for i in range(len(b))])
c.sort(key=lambda x: x[0])
more_then = {}
count = 0
for i in range(len(c)):
if c[i][1] == 'b':
more_then[c[i][0]] = i - count
count += 1
ans = []
for i in range(len(b)):
ans.append(more_then[b[i]])
print(" ".join(map(str, ans)))
if __name__ == "__main__":
main()
``` | instruction | 0 | 12,916 | 5 | 25,832 |
Yes | output | 1 | 12,916 | 5 | 25,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5
Submitted Solution:
```
l = list(map(int,input().split()))
a1 = list(map(int,input().split()))
a1.sort()
a2 = list(map(int,input().split()))
print(a1)
print(a2)
ans=[]
for i in a2:
x=0;
while (a1[x]<=i):
x+=1;
if (x==l[0]):break;
ans.append(str(x));
joiner=" ";
print (joiner.join(ans));
``` | instruction | 0 | 12,919 | 5 | 25,838 |
No | output | 1 | 12,919 | 5 | 25,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5
Submitted Solution:
```
n=2
result=[]
sth=[int(n) for n in input().split()]
a_l=sth[0]
b_l=sth[1]
a=[int(a_l) for a_l in input().split()]
b=[int(b_l) for b_l in input().split()]
a.sort()
for i in range(b_l):
number = b[i]
low = 0
high = a_l - 1
while low <= high:
mid = (low + high) // 2
if number < a[mid]:
high = mid - 1
elif number > a[mid]:
low = mid + 1
else:
break
if mid != 0:
if a[mid] == a[mid-1]:
result.append(mid + a.count(a[mid]))
elif a[mid] == number:
result.append(max(high,low)+1)
else:
result.append(max(high,low))
elif mid != a_l-1:
if a[mid] == a[mid+1]:
result.append(mid + a.count(a[mid]))
elif a[mid] == number:
result.append(max(high,low)+1)
else:
result.append(max(high,low))
else:
if a[0]<b[0]:
result.append(1)
else:
result.append(0)
print(*result)
``` | instruction | 0 | 12,920 | 5 | 25,840 |
No | output | 1 | 12,920 | 5 | 25,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two arrays of integers a and b. For each element of the second array bj you should find the number of elements in array a that are less than or equal to the value bj.
Input
The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the sizes of arrays a and b.
The second line contains n integers — the elements of array a ( - 109 ≤ ai ≤ 109).
The third line contains m integers — the elements of array b ( - 109 ≤ bj ≤ 109).
Output
Print m integers, separated by spaces: the j-th of which is equal to the number of such elements in array a that are less than or equal to the value bj.
Examples
Input
5 4
1 3 5 7 9
6 4 2 8
Output
3 2 1 4
Input
5 5
1 2 1 2 5
3 1 4 1 5
Output
4 2 4 2 5
Submitted Solution:
```
"""
Author : co_devil Chirag Garg
Institute : JIIT
"""
from math import *
from sys import stdin, stdout
import itertools
import os
import sys
import threading
from collections import deque, Counter, OrderedDict, defaultdict
from heapq import *
#from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
# from bisect import bisect_left,bisect_right
# from decimal import *,threading
from fractions import Fraction
mod = 100000009
def ii(): return int(input())
def si(): return str(input())
def mi(): return map(int, input().split())
def li(): return list(mi())
def fii(): return int(stdin.readline())
def fsi(): return str(stdin.readline())
def fmi(): return map(int, stdin.readline().split())
def fli(): return list(fmi())
abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
'z': 25}
def getKey(item): return item[0]
def sort2(l): return sorted(l, key=getKey)
def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
def decimalToBinary(n): return bin(n).replace("0b", "")
def ntl(n): return [int(i) for i in str(n)]
def powerMod(x, y, p):
res = 1
x %= p
while y > 0:
if y & 1:
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
graph = defaultdict(list)
visited = [0] * 1000000
col = [-1] * 1000000
def bfs(d, v):
q = []
q.append(v)
visited[v] = 1
while len(q) != 0:
x = q[0]
q.pop(0)
for i in d[x]:
if visited[i] != 1:
visited[i] = 1
q.append(i)
print(x)
def make_graph(e):
d = {}
for i in range(e):
x, y = mi()
if x not in d:
d[x] = [y]
else:
d[x].append(y)
if y not in d:
d[y] = [x]
else:
d[y].append(x)
return d
def gr2(n):
d = defaultdict(list)
for i in range(n):
x, y = mi()
d[x].append(y)
return d
def connected_components(graph):
seen = set()
def dfs(v):
vs = set([v])
component = []
while vs:
v = vs.pop()
seen.add(v)
vs |= set(graph[v]) - seen
component.append(v)
return component
ans = []
for v in graph:
if v not in seen:
d = dfs(v)
ans.append(d)
return ans
def primeFactors(n):
s = set()
while n % 2 == 0:
s.add(2)
n = n // 2
for i in range(3, int(sqrt(n)) + 1, 2):
while n % i == 0:
s.add(i)
n = n // i
if n > 2:
s.add(n)
return s
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1:
return
yield start
start += len(sub)
def SieveOfEratosthenes(n, isPrime):
isPrime[0] = isPrime[1] = False
for i in range(2, n):
isPrime[i] = True
p = 2
while (p * p <= n):
if (isPrime[p] == True):
i = p * p
while (i <= n):
isPrime[i] = False
i += p
p += 1
return isPrime
def dijkstra(edges, f, t):
g = defaultdict(list)
for l, r, c in edges:
g[l].append((c, r))
q, seen, mins = [(0, f, ())], set(), {f: 0}
while q:
(cost, v1, path) = heappop(q)
if v1 not in seen:
seen.add(v1)
path = (v1, path)
if v1 == t:
return (cost, path)
for c, v2 in g.get(v1, ()):
if v2 in seen:
continue
prev = mins.get(v2, None)
next = cost + c
if prev is None or next < prev:
mins[v2] = next
heappush(q, (next, v2, path))
return float("inf")
n, m = mi()
a = li()
b = li()
d = {}
l = [0]*m
for i in range(m):
d[b[i]] = 0
c = b
a = sorted(a)
b = sorted(b)
j = 0
i = 0
while i < n and j < m:
if a[i] <= b[j]:
i += 1
l[j] += 1
else:
j += 1
if j < m:
l[j] = l[j-1]
for i in range(m):
d[b[i]] = l[i]
z = []
for i in range(m):
z.append(d[c[i]])
print(*z)
``` | instruction | 0 | 12,921 | 5 | 25,842 |
No | output | 1 | 12,921 | 5 | 25,843 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,108 | 5 | 26,216 |
"Correct Solution:
```
input()
a=[int(i) for i in input().split()]
ans=1
if 0 in a:
print(0)
exit()
for i in a:
ans*=i
if ans>10**18:
print(-1)
exit()
print(ans)
``` | output | 1 | 13,108 | 5 | 26,217 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,109 | 5 | 26,218 |
"Correct Solution:
```
N=int(input())
ans=1
A=list(map(int, input().split()))
for i in range(N):
ans=ans*A[i]
if ans>10**18 or ans<0:
ans=-1
print(ans)
``` | output | 1 | 13,109 | 5 | 26,219 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,110 | 5 | 26,220 |
"Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
a.sort()
t=1
for i in a:
t*=i
if t>10**18:
t=-1
break
print(t)
``` | output | 1 | 13,110 | 5 | 26,221 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,111 | 5 | 26,222 |
"Correct Solution:
```
n,*a=map(int,open(0).read().split())
a.sort()
ans=1
for ai in a:
ans*=ai
if ans>10**18:
print(-1)
exit()
print(ans)
``` | output | 1 | 13,111 | 5 | 26,223 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,112 | 5 | 26,224 |
"Correct Solution:
```
a,b=open(0);c=1;
for i in sorted(b.split()):
c*=int(i)
if c>10**18:c=-1;break
print(c)
``` | output | 1 | 13,112 | 5 | 26,225 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,113 | 5 | 26,226 |
"Correct Solution:
```
n = int(input())
a = map(int, input().split())
a = sorted(a)
s = 1
for x in a:
s *= x;
if s > 1e18:
print(-1)
exit()
print(s)
``` | output | 1 | 13,113 | 5 | 26,227 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,114 | 5 | 26,228 |
"Correct Solution:
```
N=int(input())
A=[*map(int,input().split())]
a=1
if 0 in A:
a=0
else:
while A:
a*=A.pop()
if a>10**18:
a=-1
break
print(a)
``` | output | 1 | 13,114 | 5 | 26,229 |
Provide a correct Python 3 solution for this coding contest problem.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0 | instruction | 0 | 13,115 | 5 | 26,230 |
"Correct Solution:
```
n,*a=map(int,open(0).read().split())
a.sort()
ans=1
for i in a:
ans*=i
if ans>1e18:
print(-1)
exit()
print(ans)
``` | output | 1 | 13,115 | 5 | 26,231 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
n=int(input())
A=list(map(int,input().split()))
A.sort()
ans = 1
for i in A:
ans *= i
if ans>10**18:
print(-1)
exit()
print(ans)
``` | instruction | 0 | 13,116 | 5 | 26,232 |
Yes | output | 1 | 13,116 | 5 | 26,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
N = int(input())
ans = 1
for a in sorted(input().split()):
ans *= int(a)
if ans > 10**18:
print(-1)
exit()
print(ans)
``` | instruction | 0 | 13,117 | 5 | 26,234 |
Yes | output | 1 | 13,117 | 5 | 26,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
x=int(input())
s=list(map(int,input().split()))
s.sort()
res=1
for n in s:
res*=n
if res>1e18:
res=-1
print(-1)
exit(0)
print(res)
``` | instruction | 0 | 13,118 | 5 | 26,236 |
Yes | output | 1 | 13,118 | 5 | 26,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
n=int(input())
A=list(map(int,input().split()))
s=1
if 0 in A:
s=0
else:
for a in A:
s*=a
if s>10**18:
s=-1
break
print(s)
``` | instruction | 0 | 13,119 | 5 | 26,238 |
Yes | output | 1 | 13,119 | 5 | 26,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
N = int(input())
As = list(map(lambda x: int(x), input().split(" ")))
LIMIT = 10**18
result = 1
for a in As:
result *= a
if result > LIMIT:
result = -1
break
print(result)
``` | instruction | 0 | 13,120 | 5 | 26,240 |
No | output | 1 | 13,120 | 5 | 26,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
N=int(input())
L=list(map(int,input().split()))
A=1
if 0 in L:
print(0)
for i in range(N):
A=A*L[i]
if 10**18<A:
print(-1)
exit()
print(A)
``` | instruction | 0 | 13,121 | 5 | 26,242 |
No | output | 1 | 13,121 | 5 | 26,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
n=int(input())
A=list(map(int,input().split()))
if 0 in a:
print(0)
else:
a=1
for i in A:
a=a*i
if a>1000000000000000000:
print(-1)
else:
print(a)
``` | instruction | 0 | 13,122 | 5 | 26,244 |
No | output | 1 | 13,122 | 5 | 26,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given N integers A_1, ..., A_N, compute A_1 \times ... \times A_N.
However, if the result exceeds 10^{18}, print `-1` instead.
Constraints
* 2 \leq N \leq 10^5
* 0 \leq A_i \leq 10^{18}
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
A_1 ... A_N
Output
Print the value A_1 \times ... \times A_N as an integer, or `-1` if the value exceeds 10^{18}.
Examples
Input
2
1000000000 1000000000
Output
1000000000000000000
Input
3
101 9901 999999000001
Output
-1
Input
31
4 1 5 9 2 6 5 3 5 8 9 7 9 3 2 3 8 4 6 2 6 4 3 3 8 3 2 7 9 5 0
Output
0
Submitted Solution:
```
N=int(input())
A=input().split()
answer=1
for i in range(N):
answer=answer*int(A[i])
if answer>1*10**18:
print(-1)
else:
print(answer)
``` | instruction | 0 | 13,123 | 5 | 26,246 |
No | output | 1 | 13,123 | 5 | 26,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Find the number of the possible tuples of sequences (A_0,A_1,...,A_N) that satisfy all of the following conditions, modulo M:
* For every i (0\leq i\leq N), A_i is a sequence of length i consisting of integers between 1 and K (inclusive);
* For every i (1\leq i\leq N), A_{i-1} is a subsequence of A_i, that is, there exists 1\leq x_i\leq i such that the removal of the x_i-th element of A_i would result in a sequence equal to A_{i-1};
* For every i (1\leq i\leq N), A_i is lexicographically larger than A_{i-1}.
Constraints
* 1 \leq N,K \leq 300
* 2 \leq M \leq 10^9
* N, K and M are integers.
Input
Input is given from Standard Input in the following format:
N K M
Output
Print the number of the possible tuples of sequences (A_0,A_1,...,A_N), modulo M.
Examples
Input
2 2 100
Output
5
Input
4 3 999999999
Output
358
Input
150 150 998244353
Output
186248260
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
vector<vector<int64_t> > get_combination(int N, int M) {
vector<vector<int64_t> > mat(N + 1, vector<int64_t>(N + 1));
for (int i = 0; i <= N; i++) {
for (int j = 0; j <= i; j++) {
if (j == 0 || j == i)
mat[i][j] = 1;
else
mat[i][j] = mat[i - 1][j - 1] + mat[i - 1][j];
mat[i][j] %= M;
}
}
return (mat);
}
int main() {
int N, K, M;
cin >> N >> K >> M;
auto C = get_combination(N, M);
int dp[302][302] = {{}};
for (int i = 0; i <= K; i++) {
dp[1][i] = 1;
}
for (int i = 2; i <= N + 1; i++) {
int sub[302] = {};
for (int j = K - 1; j >= 0; j--) {
for (int k = 0; k < i; k++) {
sub[k] += dp[k][j + 1];
sub[k] %= M;
}
for (int k = 1; k <= i; k++) {
dp[i][j] += 1LL * dp[i - k][j] * C[i - 2][k - 1] % M * sub[k] % M;
dp[i][j] %= M;
}
}
}
cout << dp[N + 1][0] << endl;
}
``` | instruction | 0 | 13,176 | 5 | 26,352 |
No | output | 1 | 13,176 | 5 | 26,353 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,203 | 5 | 26,406 |
"Correct Solution:
```
while True:
q = int(input())
if q < 1: break
x = q / 2
err = 0.00001*q
while True:
x2 = x * x
t = x2 * x - q
if -err < t < err: break
x = x - t / (3 * x2)
print(x)
``` | output | 1 | 13,203 | 5 | 26,407 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,204 | 5 | 26,408 |
"Correct Solution:
```
while True:
q = int(input())
if q == -1:
break
x = q / 2
end = q * 10 ** -5
while abs(x ** 3 - q) >= end:
x = (2 * x ** 3 + q) / (3 * x ** 2)
print(x)
``` | output | 1 | 13,204 | 5 | 26,409 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,205 | 5 | 26,410 |
"Correct Solution:
```
while True:
q=int(input())
if q<1: break
x=q/2
while abs(x**3-q)>=0.00001*q:
x-=(x**3-q)/(3*x*x)
print(x)
``` | output | 1 | 13,205 | 5 | 26,411 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,206 | 5 | 26,412 |
"Correct Solution:
```
while True:
try:
n=int(input())
if n==-1:
break
x=n/2
y=n*10**-5
while abs(x**3-n)>=y:
x = (2 * x ** 3 + n) / (3 * x ** 2)
print(f'{x:.6f}')
except EOFError:
break
``` | output | 1 | 13,206 | 5 | 26,413 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,207 | 5 | 26,414 |
"Correct Solution:
```
def calc_third_root(q):
x = q / 2
while not (abs(x ** 3 - q) < 0.00001 * q):
x = x - (x ** 3 - q) / (3.0 * x ** 2)
return x
while 1:
q = int(input())
if q == -1:
break
result = calc_third_root(q)
print("{:.6f}".format(result))
``` | output | 1 | 13,207 | 5 | 26,415 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,208 | 5 | 26,416 |
"Correct Solution:
```
while 1:
try:
q = float(input())
if q == -1:
break
x = q/2
while abs(x ** 3 - q) >= 0.00001 * q:
x = x - (x ** 3 - q) / (3 * x ** 2)
print(x)
except:
break
``` | output | 1 | 13,208 | 5 | 26,417 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,209 | 5 | 26,418 |
"Correct Solution:
```
# coding: utf-8
# Your code here!
def san(x):
y=q/2
while True:
y=y-(y**3-x)/(3*y**2)
if abs(y**3-x)<0.00001*x:
break
print(f'{y:.6f}')
while True:
q=int(input())
if q==-1:
break
san(q)
``` | output | 1 | 13,209 | 5 | 26,419 |
Provide a correct Python 3 solution for this coding contest problem.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212 | instruction | 0 | 13,210 | 5 | 26,420 |
"Correct Solution:
```
while 1:
q=int(input())
if q==-1:break
ans=q/2
while abs(ans**3-q)>=q*10**-5:
ans-=(ans**3-q)/(3*ans**2)
print(ans)
``` | output | 1 | 13,210 | 5 | 26,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
while 1:
q=float(input())
if q<1:
break
xn=q/2
while 1:
xn=xn-(((xn*xn*xn)-q)/(3*(xn**2)))
if abs((xn*xn*xn)-q)<0.00001*q:
print(format(xn,'.6f'))
break
``` | instruction | 0 | 13,211 | 5 | 26,422 |
Yes | output | 1 | 13,211 | 5 | 26,423 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
import sys
f = sys.stdin
while True:
q = int(f.readline())
if q == -1:
break
x = q / 2
while abs(x ** 3 - q) >= 0.00001 * q:
x = x - (x ** 3 - q) / (3 * x ** 2)
print('{:.6f}'.format(x))
``` | instruction | 0 | 13,212 | 5 | 26,424 |
Yes | output | 1 | 13,212 | 5 | 26,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
def third_root(q):
x=q/2
while abs(x**3-q)>=q*10**(-5):
x=x-(x**3-q)/(3*x**2)
return x
while 1:
q=int(input())
if q==-1:break
ans=third_root(q)
print(ans)
``` | instruction | 0 | 13,213 | 5 | 26,426 |
Yes | output | 1 | 13,213 | 5 | 26,427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
while True:
q = int(input())
if q == -1:
break
n = 1
while True:
if n == 1:
x = q / 2
n = n + 1
if n >= 2:
y = x
x = y - (y**3 - q) / (3 * (y**2))
n = n + 1
if abs(x**3 - q) < 0.00001 * q:
break
print(x)
``` | instruction | 0 | 13,214 | 5 | 26,428 |
Yes | output | 1 | 13,214 | 5 | 26,429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
# Aizu Problem 0080: Third Root
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
def third_root(q):
x = q / 2
while abs(x**3 - q) > 1e-8:
x = x - (x**3 - q) / (3 * x**2)
return x
for line in sys.stdin:
n = int(line)
if n == -1:
break
print("%.6f" % third_root(n))
``` | instruction | 0 | 13,215 | 5 | 26,430 |
No | output | 1 | 13,215 | 5 | 26,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
def rec(x1,q):
return x1-((x1**3-q)/(3*(x1**2)))
q = []
while True:
try:
tmp = int(input())
if tmp == -1:
break
q.append(tmp)
except EOFError:
break
for i in range(len(q)):
x1 = float(q[i]/2)
for j in range(500000):
x1 = rec(x1,q[i])
print(x1)
``` | instruction | 0 | 13,216 | 5 | 26,432 |
No | output | 1 | 13,216 | 5 | 26,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.
Put a positive number $ \ frac {q} {2} $ in $ x_1 $
$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….
While doing this calculation
When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.
Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".
input
Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.
The number of datasets does not exceed 50.
output
Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.
Example
Input
15
15
-1
Output
2.466212
2.466212
Submitted Solution:
```
# Aizu Problem 0080: Third Root
#
import sys, math, os
# read input:
PYDEV = os.environ.get('PYDEV')
if PYDEV=="True":
sys.stdin = open("sample-input.txt", "rt")
for line in sys.stdin:
n = int(line)
if n == -1:
break
print("%.6f" % (pow(n, 1. / 3)))
``` | instruction | 0 | 13,217 | 5 | 26,434 |
No | output | 1 | 13,217 | 5 | 26,435 |
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