message stringlengths 2 45.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 254 108k | cluster float64 3 3 | __index_level_0__ int64 508 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,169 | 3 | 120,338 |
Tags: geometry, math
Correct Solution:
```
import math
r,x,y,a,b=map(int,input().split())
#arr=list(map(int,input().split()))
print(math.ceil((((x-a)**2+(y-b)**2)**0.5)/(2*r)))
``` | output | 1 | 60,169 | 3 | 120,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,170 | 3 | 120,340 |
Tags: geometry, math
Correct Solution:
```
from math import sqrt, ceil
r, x, y, x1, y1 = map(int, input().split())
dist = sqrt((x1-x)**2 + (y1-y)**2)/2
# print(dist)
moves = ceil(dist/r)
print(moves)
``` | output | 1 | 60,170 | 3 | 120,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,171 | 3 | 120,342 |
Tags: geometry, math
Correct Solution:
```
r, x, y, xf, yf = list(map(int, input().split()))
from math import sqrt, ceil
distancia = sqrt((x-xf)**2 + (y-yf)**2)
# print(distancia)
print(int(ceil(distancia/(2*r))))
``` | output | 1 | 60,171 | 3 | 120,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,172 | 3 | 120,344 |
Tags: geometry, math
Correct Solution:
```
# 507B
from math import ceil,sqrt
t = input().split()
t = list(map(int, t))
r = t[0]
x = t[1]
y = t[2]
a = t[3]
b = t[4]
d = sqrt((x-a)**2+(y-b)**2)
if d == 0:
res = 0
else:
if r>d:
res = 1
else:
res = ceil(d/(2*r))
print(res)
``` | output | 1 | 60,172 | 3 | 120,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,173 | 3 | 120,346 |
Tags: geometry, math
Correct Solution:
```
import math
(r,xa,ya,xb,yb) = [int(x) for x in input().split()]
print(int(math.ceil(math.sqrt((xb-xa)**2+(yb-ya)**2)/(2*r))))
``` | output | 1 | 60,173 | 3 | 120,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image> | instruction | 0 | 60,174 | 3 | 120,348 |
Tags: geometry, math
Correct Solution:
```
import math
r , x , y , x1 , y1 = map(int , input().split())
d = math.sqrt((x - x1)**2 + (y-y1)**2)
print((math.ceil(d/(2*r))))
``` | output | 1 | 60,174 | 3 | 120,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
import math
from math import *
r,x,y,x1,y1 = [int(x) for x in input().split()]
dist=sqrt(((x-x1)**2)+((y-y1)**2))
ans=dist/(r*2)
print (math.ceil(ans))
``` | instruction | 0 | 60,175 | 3 | 120,350 |
Yes | output | 1 | 60,175 | 3 | 120,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
from math import ceil, sqrt
r, x, y, x_, y_ = map(int, input().split())
print(ceil(sqrt((x - x_)**2 + (y - y_)**2) / (2 * r)))
``` | instruction | 0 | 60,176 | 3 | 120,352 |
Yes | output | 1 | 60,176 | 3 | 120,353 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
import os
import math
import sys
debug = True
if debug and os.path.exists("input.in"):
input = open("input.in", "r").readline
else:
debug = False
input = sys.stdin.readline
def inp():
return (int(input()))
def inlt():
return (list(map(int, input().split())))
def insr():
s = input()
return s[:len(s) - 1] # Remove line char from end
def invr():
return (map(int, input().split()))
test_count = 1
if debug:
test_count = inp()
for t in range(test_count):
if debug:
print("Test Case #", t + 1)
r, x1, y1, x2, y2 = invr()
dist = math.sqrt((x1 - x2) ** 2 + (y1 - y2) ** 2)
print(math.ceil(dist / (2 * r)))
``` | instruction | 0 | 60,177 | 3 | 120,354 |
Yes | output | 1 | 60,177 | 3 | 120,355 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
r, x1, y1, x2, y2 = (float(x) for x in input().split())
dest = ((x1-x2)**2 + (y1-y2)**2)**0.5
if x1 == x2 and y1 == y2:
print(0)
else:
res = 0
while dest > 0:
dest -= 2*r
res += 1
print(res)
``` | instruction | 0 | 60,178 | 3 | 120,356 |
Yes | output | 1 | 60,178 | 3 | 120,357 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
r, x, y, x1, y1 = map(int, input().split())
ans = 0
ans += int(abs(x-x1)/(2*r))
ans += int(abs(y-y1)/(2*r))
if abs(x-x1)%(2*r)!=0 or abs(x-x1)%(2*r)!=0:
ans += 1
print(ans)
``` | instruction | 0 | 60,179 | 3 | 120,358 |
No | output | 1 | 60,179 | 3 | 120,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
#Codeforces507B
import math
liszt = list(input().split())
for i in range(len(liszt)):
liszt[i] = int(liszt[i])
d = math.pow((math.pow(liszt[3] - liszt[1], 2) + math.pow(liszt[4] - liszt[2], 2)), .5)
r = liszt[0]
if d%r == 0:
print(d/r)
else:
print(math.floor(d/r) + 2)
``` | instruction | 0 | 60,180 | 3 | 120,360 |
No | output | 1 | 60,180 | 3 | 120,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
l=lambda:map(int,input().split())
r,x,y,x1,y1=l()
d=int(((x-x1)**2+(y-y1)**2)**0.5)
from math import ceil
print(int(ceil(d/(2*r))))
``` | instruction | 0 | 60,181 | 3 | 120,362 |
No | output | 1 | 60,181 | 3 | 120,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr loves Geometry. One day he came up with a very interesting problem.
Amr has a circle of radius r and center in point (x, y). He wants the circle center to be in new position (x', y').
In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.
Help Amr to achieve his goal in minimum number of steps.
Input
Input consists of 5 space-separated integers r, x, y, x' y' (1 ≤ r ≤ 105, - 105 ≤ x, y, x', y' ≤ 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.
Output
Output a single integer — minimum number of steps required to move the center of the circle to the destination point.
Examples
Input
2 0 0 0 4
Output
1
Input
1 1 1 4 4
Output
3
Input
4 5 6 5 6
Output
0
Note
In the first sample test the optimal way is to put a pin at point (0, 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter).
<image>
Submitted Solution:
```
import math
b=list(input().split())
b=[int(i) for i in b]
a=math.sqrt((b[1]-b[3])**2+(b[2]-b[4])**2)
if a%(2*b[0])==0:
print(int(a//(2*b[0])))
else:
print(int(a//(2*b[0])+2))
``` | instruction | 0 | 60,182 | 3 | 120,364 |
No | output | 1 | 60,182 | 3 | 120,365 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,415 | 3 | 120,830 |
"Correct Solution:
```
# modulo:mod
# inverse x^(-1)
def inv(x):
global mod
return pow(x,mod-2,mod)
# factorial x!
def fact(x):
global mod
res=1
for i in range(2,x+1):
res=res*i%mod
return res
# combination nCr
def combi(n,r):
if r<0 or r>n:
return 0
else:
return fact(n)*inv(fact(r))*inv(fact(n-r))%mod
mod=10**9+7
N=int(input())
x=[0]
x.extend(list(map(int,input().split())))
invs=[inv(d) for d in range(N)]
# print(invs)
ans=0;s=0
for d in range(1,(N-1)//2+1):
s=(s+x[N-d]-x[d])%mod
ans=(ans+s*invs[d]*invs[d+1])%mod
if d!=N-d-1:
ans=(ans+s*invs[N-d-1]*invs[N-d])%mod
for i in range(1,N):
ans=(ans+(x[N]-x[i])*invs[N-i])%mod
ans=ans*fact(N-1)%mod
print(ans)
``` | output | 1 | 60,415 | 3 | 120,831 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,416 | 3 | 120,832 |
"Correct Solution:
```
MOD=10**9+7
N=int(input())
x=list(map(int, input().split()))
y=[x[i+1]-x[i] for i in range(N-1)]
upp=[1]*(N+1)
for i in range(2, N+1):
upp[i]=upp[i-1]*i
upp[i]%=MOD
N_1=N-1
b=[0]*(N_1)
b[0]=upp[N_1]
tmp=upp[N_1]
for i in range(1, N_1):
a=tmp*pow(i+1, MOD-2, MOD)%MOD
b[i]=(b[i-1]+a)%MOD
print(sum([b[i]*y[i]%MOD for i in range(N_1)])%MOD)
``` | output | 1 | 60,416 | 3 | 120,833 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,417 | 3 | 120,834 |
"Correct Solution:
```
n = int(input())
x = list( map(int, input().split()))
MOD = 10**9 + 7
def make_mod_inv(l,p): # lまでの逆元を作る
mod_inv = [0, 1] + [0] * (l+3) # inv[n] = n^(-1) mod p, 0! = 1 だけど便宜上inv[0]=0にしてる
for i in range(2, l+5):
mod_inv[i] = -mod_inv[p % i] * (p // i) % p
return mod_inv
mod_inv = make_mod_inv(n-1,MOD)
ans = 0
factorial = 1
for i in range(2,n):
factorial = (factorial * i%MOD) %MOD
p = [0]*(n-1)
p[0] = 1 * factorial%MOD
for i in range(1,n-1):
p[i] = p[i-1] + factorial*mod_inv[i+1]
p[i] %= MOD
for i in range(n-1):
ans += (x[i+1]-x[i])%MOD*p[i]
ans %= MOD
print(int(ans))
``` | output | 1 | 60,417 | 3 | 120,835 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,418 | 3 | 120,836 |
"Correct Solution:
```
mod = 10**9+7
# N = 15
N = int(input())
xx = list(map(int, input().split()))
a = 1
for i in range(2,N):
a *= i
a %= mod
aa = [a]
for i in range(N-2):
aa.append((aa[-1]+a*pow(i+2,mod-2,mod))%mod)
dx = [xx[i]-xx[i-1] for i in range(1,N)]
ans = 0
for a,d in zip(aa,dx):
ans += a*d
ans %= mod
print(ans%mod)
``` | output | 1 | 60,418 | 3 | 120,837 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,419 | 3 | 120,838 |
"Correct Solution:
```
n = int(input())
x = list(map(int, input().split()))
ans = 0
mod = 1000000007
p = 1
for i in range(1, n):
p *= i
p %= mod
for i in range(n-1):
ans += (x[n-1]-x[i])*pow(i+1, mod-2, mod)
ans *= p
ans %= 1000000007
print(ans)
``` | output | 1 | 60,419 | 3 | 120,839 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,420 | 3 | 120,840 |
"Correct Solution:
```
N, *X = map(int, open(0).read().split())
MOD = 10 ** 9 + 7
fact = 1
for i in range(2, N):
fact = fact * i % MOD
fact_list = [fact]
for i in range(2, N):
fact_list.append(fact_list[-1] + fact * pow(i, MOD - 2, MOD) % MOD)
ans = 0
for i in range(N - 1):
ans = (ans + (X[i + 1] - X[i]) * fact_list[i]) % MOD
print(ans % MOD)
``` | output | 1 | 60,420 | 3 | 120,841 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,421 | 3 | 120,842 |
"Correct Solution:
```
import math
n = int(input())
x = list(map(int,input().split()))
mod = 10**9 + 7
f = math.factorial(n-1)
f %= mod
def modinv(a, mod=10**9+7):
return pow(a, mod-2, mod)
l = []
for i in range(n-1):
d = x[i+1] - x[i]
l.append(d)
# print(l)
count = 0
k = f
plus = 0
for i in range(1,n):
# print(k)
k = f + plus
count += l[i-1] * k
plus += f * modinv(i+1,mod)
print(int(count)%mod)
``` | output | 1 | 60,421 | 3 | 120,843 |
Provide a correct Python 3 solution for this coding contest problem.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044 | instruction | 0 | 60,422 | 3 | 120,844 |
"Correct Solution:
```
MOD = 10**9+7
n = int(input())
F = [1]*(n+1)
for i in range(n):
F[i+1] = F[i]*(i+1)
F[i+1] %= MOD
X = list(map(int, input().split()))
D = [X[i+1]-X[i] for i in range(n-1)]
T = [1]*n
for i in range(2, n):
T[i] = T[i-1]*i+F[i-1]
T[i] %= MOD
t = 1
for i in range(1, n-1):
t *= (n-i)
t %= MOD
T[n-1-i] *= t
T[n-1-i] %= MOD
ans = 0
for i in range(1, n):
ans += D[i-1]*T[i]
ans %= MOD
print(ans)
``` | output | 1 | 60,422 | 3 | 120,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
def egcd(a, b):
(x, lastx) = (0, 1)
(y, lasty) = (1, 0)
while b != 0:
q = a // b
(a, b) = (b, a % b)
(x, lastx) = (lastx - q * x, x)
(y, lasty) = (lasty - q * y, y)
return (lastx, lasty, a)
# ax ≡ 1 (mod m)なるxを返す
def modinv(a, m):
(inv, q, gcd_val) = egcd(a, m)
return inv % m
MOD=10**9+7
n=int(input())
x=list(map(int,input().split()))
P=1
for i in range(1,n):
P=(P*i)%MOD
s=0
ans=0
for i in range(1,n):
l=x[i]-x[i-1]
s=s+P*modinv(i,MOD)
s%MOD
ans+=l*s
ans%=MOD
print(ans%MOD)
``` | instruction | 0 | 60,423 | 3 | 120,846 |
Yes | output | 1 | 60,423 | 3 | 120,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
# -*- coding: utf-8 -*-
# dwacon6th-prelims/dwacon6th_prelims_b
import sys
s2nn = lambda s: [int(c) for c in s.split(' ')]
i2s = lambda: sys.stdin.readline().rstrip()
i2n = lambda: int(i2s())
i2nn = lambda: s2nn(i2s())
MAX = 10**5 + 100
MOD = 10**9 + 7
fac = [1, 1] + [0] * MAX
finv = [1, 1] + [0] * MAX
inv = [1, 1] + [0] * MAX
coef = [1, 1] + [0] * MAX
for i in range(2, MAX):
fac[i] = fac[i - 1] * i % MOD
inv[i] = MOD - inv[MOD % i] * (MOD // i) % MOD
finv[i] = finv[i - 1] * inv[i] % MOD
coef[i] = (coef[i - 1] * i + fac[i - 1]) % MOD
def main():
N = i2n()
n = N - 1
x = i2nn()
total = 0
for i in range(1, N):
d = x[i] - x[i - 1]
total = (total + coef[i] * fac[n] % MOD * finv[i] % MOD * d % MOD) % MOD
print(total)
return
main()
``` | instruction | 0 | 60,424 | 3 | 120,848 |
Yes | output | 1 | 60,424 | 3 | 120,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
N = int(input())
A = list(map(int, input().split()))
mod = int(1e+9 + 7)
p = mod - 2
S = []
while p != 0:
S = [p%2] + S[:]
p //= 2
frac = 1
for i in range(N - 1):
frac *= i+1
frac %= mod
T = 0
for i in range(N - 1):
k = 1
for j in range(len(S)):
if S[j] == 1:
k *= i+1
k %= mod
if j != len(S) - 1:
k *= k
k %= mod
T += (frac * k * (A[N - 1] - A[i])) % mod
T %= mod
print(T%mod)
``` | instruction | 0 | 60,425 | 3 | 120,850 |
Yes | output | 1 | 60,425 | 3 | 120,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
mod=10**9+7
# mod inverse(逆元)
# 今回はfermatの小定理で求める
def mod_inv(x):
return pow(x,mod-2,mod)
n=int(input())
A=list(map(int,input().split()))
# (N-1)!を求める
fac=1
for i in range(1,n):
fac = fac*i%mod
# 各区間の個数テーブルを先に求める
cnt=[0]*n
for i in range(1,n):
cnt[i] = fac * mod_inv(i) % mod
cnt[i] = (cnt[i]+cnt[i-1]) % mod
# 計算
ans=0
for i in range(n-1):
ans += (A[i+1]-A[i]) * cnt[i+1]
ans %= mod
print(ans)
``` | instruction | 0 | 60,426 | 3 | 120,852 |
Yes | output | 1 | 60,426 | 3 | 120,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
#B
mod = 10**9 + 7
N = int(input())
perm = 1
for i in range(N-1):
perm*=(i+1)
X = list(map(int,input().split()))
D = []
for i in range(N-1):
D.append(X[i+1]-X[i])
kai = 1
ans = 0
for k in range(N-1):
kai*=(k+1)
ans += int(perm*D[k]*(2-(1/kai)))
ans%=mod
print(kai)
ans%=mod
print(ans)
``` | instruction | 0 | 60,427 | 3 | 120,854 |
No | output | 1 | 60,427 | 3 | 120,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
import math
mod=10**9+7
n=int(input())
x=list(map(int,input().split()))
kai=math.factorial(n-1)
a=[]
for i in range(n-1):
a.append(x[i+1]-x[i])
ans=0
count=0
a1=sum(a)
for i in range(n-1):
count+=kai/(i+1)
ans+=a[i]*count
print(int(ans%mod))
``` | instruction | 0 | 60,428 | 3 | 120,856 |
No | output | 1 | 60,428 | 3 | 120,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
import sys
import math
#import numpy as np
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split())) # applies to numbers only
rs = lambda: stdin.readline().rstrip() # ignore trailing spaces
N = ri()
X = rl()
MOD = 10 ** 9 + 7
limit = math.factorial(N-1)
last = X[-1]
X = [last - x for x in X] #最後までの距離
answer = 0
for i in range(1, N):
answer += (X[i-1] * limit / i)%MOD
answer %= MOD
print(answer)
``` | instruction | 0 | 60,429 | 3 | 120,858 |
No | output | 1 | 60,429 | 3 | 120,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N slimes standing on a number line. The i-th slime from the left is at position x_i.
It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.
Niwango will perform N-1 operations. The i-th operation consists of the following procedures:
* Choose an integer k between 1 and N-i (inclusive) with equal probability.
* Move the k-th slime from the left, to the position of the neighboring slime to the right.
* Fuse the two slimes at the same position into one slime.
Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.
Constraints
* 2 \leq N \leq 10^{5}
* 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}
* x_i is an integer.
Input
Input is given from Standard Input in the following format:
N
x_1 x_2 \ldots x_N
Output
Print the answer.
Examples
Input
3
1 2 3
Output
5
Input
12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932
Output
750927044
Submitted Solution:
```
import itertools
N = int(input())
x = list(map(int,input().split()))
mod = 10 ** 9 + 7
dist = [[0]*N for _ in range(N)]
for i in range(N):
for j in range(N):
dist[i][j] = abs(x[j] - x[i]) % mod
#print(dist)
perm = list(itertools.permutations(list(range(N - 1))))
#print(perm)
ans = 0
for p in perm:
tmp = list(p)
tmp.append(N-1)
for i in range(N - 1):
ans += dist[tmp[i]][tmp[i+1]]% mod
print(ans % mod)
``` | instruction | 0 | 60,430 | 3 | 120,860 |
No | output | 1 | 60,430 | 3 | 120,861 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Asimov, a robotics researcher, released cleaning robots he developed (see Problem B). His robots soon became very popular and he got much income. Now he is pretty rich. Wonderful.
First, he renovated his house. Once his house had 9 rooms that were arranged in a square, but now his house has N × N rooms arranged in a square likewise. Then he laid either black or white carpet on each room.
Since still enough money remained, he decided to spend them for development of a new robot. And finally he completed.
The new robot operates as follows:
* The robot is set on any of N × N rooms, with directing any of north, east, west and south.
* The robot detects color of carpets of lefthand, righthand, and forehand adjacent rooms if exists. If there is exactly one room that its carpet has the same color as carpet of room where it is, the robot changes direction to and moves to and then cleans the room. Otherwise, it halts. Note that halted robot doesn't clean any longer. Following is some examples of robot's movement.
<image>
Figure 1. An example of the room
In Figure 1,
* robot that is on room (1,1) and directing north directs east and goes to (1,2).
* robot that is on room (0,2) and directing north directs west and goes to (0,1).
* robot that is on room (0,0) and directing west halts.
* Since the robot powered by contactless battery chargers that are installed in every rooms, unlike the previous robot, it never stops because of running down of its battery. It keeps working until it halts.
Doctor's house has become larger by the renovation. Therefore, it is not efficient to let only one robot clean. Fortunately, he still has enough budget. So he decided to make a number of same robots and let them clean simultaneously.
The robots interacts as follows:
* No two robots can be set on same room.
* It is still possible for a robot to detect a color of carpet of a room even if the room is occupied by another robot.
* All robots go ahead simultaneously.
* When robots collide (namely, two or more robots are in a single room, or two robots exchange their position after movement), they all halt. Working robots can take such halted robot away.
On every room dust stacks slowly but constantly. To keep his house pure, he wants his robots to work so that dust that stacked on any room at any time will eventually be cleaned.
After thinking deeply, he realized that there exists a carpet layout such that no matter how initial placements of robots are, this condition never can be satisfied. Your task is to output carpet layout that there exists at least one initial placements of robots that meets above condition. Since there may be two or more such layouts, please output the K-th one lexicographically.
Constraints
* Judge data consists of at most 100 data sets.
* 1 ≤ N < 64
* 1 ≤ K < 263
Input
Input file contains several data sets. One data set is given in following format:
N K
Here, N and K are integers that are explained in the problem description.
The end of input is described by a case where N = K = 0. You should output nothing for this case.
Output
Print the K-th carpet layout if exists, "No" (without quotes) otherwise.
The carpet layout is denoted by N lines of string that each has exactly N letters. A room with black carpet and a room with white carpet is denoted by a letter 'E' and '.' respectively. Lexicographically order of carpet layout is defined as that of a string that is obtained by concatenating the first row, the second row, ..., and the N-th row in this order.
Output a blank line after each data set.
Example
Input
2 1
2 3
6 4
0 0
Output
..
..
No
..EEEE
..E..E
EEE..E
E..EEE
E..E..
EEEE.. | instruction | 0 | 60,583 | 3 | 121,166 |
"Correct Solution:
```
# AOJ 1024 Cleaning Robot 2.0
# Python3 2018.7.5 bal4u
mv = ((-1,0), (0,1), (1,0), (0,-1))
d2c = {0:'.', 1:'E'}
while True:
n, k = map(int, input().split())
if n == 0: break
k, n1 = k-1, n-1
if (n & 1) or k >= (1<<(n>>1)):
print("No\n")
continue
arr = [[-1 for c in range(n)] for r in range(n)]
for c in range(n): arr[0][c] = (k>>((n1-c)>>1)) & 1
for r in range(n1):
for c in range(n):
f, t = 0, arr[r][c]
for i in range(4):
nr, nc = r + mv[i][0], c + mv[i][1]
if nr >= 0 and nr < n and nc >= 0 and nc < n and arr[nr][nc] == t: f += 1
arr[r+1][c] = 1-t if f==2 else t
for r in range(n): print(''.join([d2c[arr[r][c]] for c in range(n)]))
print()
``` | output | 1 | 60,583 | 3 | 121,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,689 | 3 | 121,378 |
Tags: greedy
Correct Solution:
```
def main():
n,b,a = map(int,input().split())
array = list(map(int,input().split()))
max_a = a
for i in range(n-1):
#print(b,a)
if array[i] == 1:
if b > 0:
if a < max_a:
b -= 1
a += 1
else:
if a > 0:
a -= 1
elif b > 0:
b -= 1
else:
print(i)
return
elif a > 0:
a -= 1
else:
print(i)
return
else:
if a > 0:
a -= 1
elif b > 0:
b -= 1
else:
print(i)
return
if a > 0 or b > 0:
print(n)
return
print(n-1)
main()
``` | output | 1 | 60,689 | 3 | 121,379 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,690 | 3 | 121,380 |
Tags: greedy
Correct Solution:
```
n, b, a = map(int, input().split())
fa = a
l = list(map(int, input().split()))
mx = 0
for i in l:
if i == 0 and fa > 0:
fa -= 1
mx += 1
elif i == 0 and fa == 0:
b -= 1
mx += 1
elif i == 1 and fa == a:
fa -= 1
mx += 1
elif i == 1 and b > 0:
mx += 1
if fa + 1 > a:
b -= 1
else:
b -= 1
fa += 1
elif i == 1 and b == 0:
fa -= 1
mx += 1
if fa == 0 and b == 0:
break
print(mx)
``` | output | 1 | 60,690 | 3 | 121,381 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,691 | 3 | 121,382 |
Tags: greedy
Correct Solution:
```
ll=lambda:map(int,input().split())
t=lambda:int(input())
ss=lambda:input()
#from math import log10 ,log2,ceil,factorial as f,gcd
#from itertools import combinations_with_replacement as cs
#from functools import reduce
from math import gcd
'''
for _ in range(t()):
n=t()
'''
n,b,ac=ll()
s=list(ll())
t=ac
final=0
for i in range(n):
if s[i]==1 and b:
if ac!=t:
ac+=1
b-=1
else:
ac-=1
elif s[i]==0 and ac:
ac-=1
elif ac and not b:
final=min(ac+final,n)
break
elif b and not ac:
b-=1
else:
break
final+=1
print(final)
``` | output | 1 | 60,691 | 3 | 121,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,692 | 3 | 121,384 |
Tags: greedy
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left
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")
import time
start_time = time.time()
import collections as col
import math, string
from functools import reduce
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
"""
If we can charge the accumulator (path[i] = 0 and charge < A and battery > 0), use battery
Else use accumulator if possible, otherwise battery, otherwise break
"""
def solve():
N, B, A = getInts()
A_cap = A
path = getInts()
i = 0
while i < N:
if path[i] == 1 and A < A_cap and B > 0:
B -= 1
A += 1
i += 1
elif A > 0:
A -= 1
i += 1
elif B > 0:
B -= 1
i += 1
else:
break
return i
#for _ in range(getInt()):
print(solve())
``` | output | 1 | 60,692 | 3 | 121,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,693 | 3 | 121,386 |
Tags: greedy
Correct Solution:
```
n, b, a = list(map(int, input().split()))
l = list(map(int, input().split()))
aa = a
for i in range(n):
if a == 0 and b == 0:
print(i)
break
if l[i] == 0:
if a > 0:
a -= 1
else:
b -= 1
if l[i] == 1:
if a == aa:
a -= 1
else:
if b > 0:
b -= 1
a = min(aa, a + 1)
else:
a -= 1
if i == n - 1:
print(n)
``` | output | 1 | 60,693 | 3 | 121,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,694 | 3 | 121,388 |
Tags: greedy
Correct Solution:
```
def main():
n, a, b = (int(i) for i in input().split())
ma = a
mb = b
S = [int(i) for i in input().split()]
ans = 0
for s in S:
if s == 0 and b > 0:
b -= 1
elif s == 0 and b == 0 and a > 0:
a -= 1
elif s == 1 and a > 0 and b < mb:
a -= 1
b += 1
elif s == 1 and b > 0:
b -= 1
elif s == 1 and b == 0 and a > 0:
a -= 1
elif a == 0 and b == 0:
break
ans += 1
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 60,694 | 3 | 121,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,695 | 3 | 121,390 |
Tags: greedy
Correct Solution:
```
n, b, a = map(int, input().split())
max_a = a
dist = 0
def acc():
global a
global b
global dist
if a>0: a-=1; dist+=1
elif b>0: b-=1; dist+=1
else: print(dist); exit()
def bat():
global a
global b
global dist
global max_a
if a==max_a: a-=1; dist+=1
elif b>0: b-=1; dist+=1; a+=1
elif a>0: a-=1; dist+=1
else: print(dist); exit()
s = list(map(int, input().split()))
acc()
for i in s[1:]:
if i==0: acc()
else: bat()
print(n)
``` | output | 1 | 60,695 | 3 | 121,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order. | instruction | 0 | 60,696 | 3 | 121,392 |
Tags: greedy
Correct Solution:
```
(n, b, a) = [int(x) for x in input().split(' ')]
arr = [int(x) for x in input().split(' ')]
# print(n, b, a, arr)
ln = 0
max_a = a
for i in arr:
if a == 0 and b == 0:
break
elif a == 0:
if i == 1:
a = min(a + 1, max_a)
b -= 1
elif b == 0:
a -= 1
elif i == 1 and a < max_a:
a += 1
b -= 1
else:
a -= 1
ln += 1
# print(i)
print(ln)
``` | output | 1 | 60,696 | 3 | 121,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n, b, a = map(int, input().split())
s = list(map(int, input().split()))
rs = 0
tmp_a = a
nap = True
for i in s:
nap = True
if i == 0:
if a > 0:
a -=1
elif b > 0:
b -= 1
else:
break
rs += 1
else:
if (b == 0 and a > 0) or (a == tmp_a):
a -= 1
nap = False
elif b > 0:
b -= 1
else:
break
if nap:
a += 1
a = min(tmp_a, a)
rs += 1
print(rs)
``` | instruction | 0 | 60,697 | 3 | 121,394 |
Yes | output | 1 | 60,697 | 3 | 121,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n,b,a=map(int,input().split())
lst=list(map(int,input().split()))
a1=a
b1=b
cnt=0
for i in range(n):
if lst[i] == 1:
if b1 == 0:
if a1 > 0:
a1-=1
else:
if a1==a:
a1-=1
else:
b1-=1
a1+=1
else:
if a1 > 0:
a1-=1
else:
b1-=1
cnt+=1
if b1==0 and a1==0:
break
print(cnt)
``` | instruction | 0 | 60,698 | 3 | 121,396 |
Yes | output | 1 | 60,698 | 3 | 121,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
xbn = [int(elem) for elem in input().split()]
s = [int(ggg) for ggg in input().split()]
bat = xbn[1]
akk = xbn[2]
BAT = xbn[1]
AKK = xbn[2]
road = 0
fin = 0
for i in range(xbn[0]):
if s[i]==1:
if akk<AKK and bat>0:
bat-=1
akk+=1
road+=1
else:
if akk>0:
akk-=1
road+=1
else:
fin = road
break
else:
if akk>0:
akk-=1
road+=1
elif bat>0:
road+=1
bat-=1
else:
break
print (road)
``` | instruction | 0 | 60,699 | 3 | 121,398 |
Yes | output | 1 | 60,699 | 3 | 121,399 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
def solve():
n, b, a = map(int, input().split())
S = [int(k) for k in input().split()]
ans = 0
curr_a = a
curr_b = b
for i in range(n):
if curr_a == 0 and curr_b == 0:
break
if S[i] == 1:
if curr_a < a and curr_b > 0:
ans += 1
curr_b -= 1
curr_a += 1
continue
if curr_a > 0:
curr_a -= 1
ans += 1
continue
if curr_b > 0:
curr_b -= 1
ans += 1
else:
if curr_a > 0:
ans += 1
curr_a -= 1
continue
if curr_b > 0:
ans += 1
curr_b -= 1
print (ans)
solve()
``` | instruction | 0 | 60,700 | 3 | 121,400 |
Yes | output | 1 | 60,700 | 3 | 121,401 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n,b,a=map(int,input().split())
aa=a
it=list(map(int,input().split()))
done=0
for i in range(len(it)):
#print(done,a,b)
if b==0 and a==0:
break
if it[i]==1:
if b>0:
b-=1
if a<aa:
a+=1
done+=1
continue
break
done+=1
if a>0:
a-=1
elif b>0:
b-=1
else:
done-=1
break
print(done)
``` | instruction | 0 | 60,701 | 3 | 121,402 |
No | output | 1 | 60,701 | 3 | 121,403 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n,b,a=map(int,input().split())
m=a
s=list(map(int,input().split()))
for i in range(n):
if s[i]==0:
if a>0:
a-=1
elif b>0:
b-=1
else:
if b>0 and a+1<=m:
b-=1
a+=1
elif a>0:
a-=1
else:
b-=1
if a==0 and b==0:
print(i+1)
break
``` | instruction | 0 | 60,702 | 3 | 121,404 |
No | output | 1 | 60,702 | 3 | 121,405 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n=input()
n=n.split(' ')
m=input()
m=m.split(' ')
n[0]=int(n[0])
n[1]=int(n[1])
n[2]=int(n[2])
c=n[2]
t=0
for i in m:
if i=='0':
if n[2]>0:
n[2]=n[2]-1
t=t+1
elif n[1]>0:
n[1]=n[1]-1
t=t+1
elif i=='1':
if n[2]==c:
n[2]=n[2]-1
t=t+1
elif n[1]>0:
n[1]=n[1]-1
t=t+1
if n[2]<c:
n[2]=n[2]+1
elif n[2]>0:
n[2]=n[2]-1
t=t+1
print(n[2],n[1])
print(t)
``` | instruction | 0 | 60,703 | 3 | 121,406 |
No | output | 1 | 60,703 | 3 | 121,407 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a robot staying at X=0 on the Ox axis. He has to walk to X=n. You are controlling this robot and controlling how he goes. The robot has a battery and an accumulator with a solar panel.
The i-th segment of the path (from X=i-1 to X=i) can be exposed to sunlight or not. The array s denotes which segments are exposed to sunlight: if segment i is exposed, then s_i = 1, otherwise s_i = 0.
The robot has one battery of capacity b and one accumulator of capacity a. For each segment, you should choose which type of energy storage robot will use to go to the next point (it can be either battery or accumulator). If the robot goes using the battery, the current charge of the battery is decreased by one (the robot can't use the battery if its charge is zero). And if the robot goes using the accumulator, the current charge of the accumulator is decreased by one (and the robot also can't use the accumulator if its charge is zero).
If the current segment is exposed to sunlight and the robot goes through it using the battery, the charge of the accumulator increases by one (of course, its charge can't become higher than it's maximum capacity).
If accumulator is used to pass some segment, its charge decreases by 1 no matter if the segment is exposed or not.
You understand that it is not always possible to walk to X=n. You want your robot to go as far as possible. Find the maximum number of segments of distance the robot can pass if you control him optimally.
Input
The first line of the input contains three integers n, b, a (1 ≤ n, b, a ≤ 2 ⋅ 10^5) — the robot's destination point, the battery capacity and the accumulator capacity, respectively.
The second line of the input contains n integers s_1, s_2, ..., s_n (0 ≤ s_i ≤ 1), where s_i is 1 if the i-th segment of distance is exposed to sunlight, and 0 otherwise.
Output
Print one integer — the maximum number of segments the robot can pass if you control him optimally.
Examples
Input
5 2 1
0 1 0 1 0
Output
5
Input
6 2 1
1 0 0 1 0 1
Output
3
Note
In the first example the robot can go through the first segment using the accumulator, and charge levels become b=2 and a=0. The second segment can be passed using the battery, and charge levels become b=1 and a=1. The third segment can be passed using the accumulator, and charge levels become b=1 and a=0. The fourth segment can be passed using the battery, and charge levels become b=0 and a=1. And the fifth segment can be passed using the accumulator.
In the second example the robot can go through the maximum number of segments using battery two times and accumulator one time in any order.
Submitted Solution:
```
n, b, a = map(int, input().split())
s = map(int, input().split())
cb = b
ca = a
c = 0
for x in s:
if x == 1 and cb > 0 and ca < a:
c += 1
ca += 1
b -= 1
elif x == 0 and ca > 0:
ca -= 1
c += 1
elif cb > 0:
cb -= 1
c += 1
elif ca > 0:
ca -= 1
c += 1
if ca == 0 and cb == 0:
break
print(c)
``` | instruction | 0 | 60,704 | 3 | 121,408 |
No | output | 1 | 60,704 | 3 | 121,409 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a system of pipes. It consists of two rows, each row consists of n pipes. The top left pipe has the coordinates (1, 1) and the bottom right — (2, n).
There are six types of pipes: two types of straight pipes and four types of curved pipes. Here are the examples of all six types:
<image> Types of pipes
You can turn each of the given pipes 90 degrees clockwise or counterclockwise arbitrary (possibly, zero) number of times (so the types 1 and 2 can become each other and types 3, 4, 5, 6 can become each other).
You want to turn some pipes in a way that the water flow can start at (1, 0) (to the left of the top left pipe), move to the pipe at (1, 1), flow somehow by connected pipes to the pipe at (2, n) and flow right to (2, n + 1).
Pipes are connected if they are adjacent in the system and their ends are connected. Here are examples of connected pipes:
<image> Examples of connected pipes
Let's describe the problem using some example:
<image> The first example input
And its solution is below:
<image> The first example answer
As you can see, the water flow is the poorly drawn blue line. To obtain the answer, we need to turn the pipe at (1, 2) 90 degrees clockwise, the pipe at (2, 3) 90 degrees, the pipe at (1, 6) 90 degrees, the pipe at (1, 7) 180 degrees and the pipe at (2, 7) 180 degrees. Then the flow of water can reach (2, n + 1) from (1, 0).
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries. Then q queries follow.
Each query consists of exactly three lines. The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of pipes in each row. The next two lines contain a description of the first and the second rows correspondingly. Each row description consists of n digits from 1 to 6 without any whitespaces between them, each digit corresponds to the type of pipe in the corresponding cell. See the problem statement to understand which digits correspond to which types of pipes.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For the i-th query print the answer for it — "YES" (without quotes) if it is possible to turn some pipes in a way that the water flow can reach (2, n + 1) from (1, 0), and "NO" otherwise.
Example
Input
6
7
2323216
1615124
1
3
4
2
13
24
2
12
34
3
536
345
2
46
54
Output
YES
YES
YES
NO
YES
NO
Note
The first query from the example is described in the problem statement. | instruction | 0 | 60,737 | 3 | 121,474 |
Tags: dp, implementation
Correct Solution:
```
from typing import List
TOP = 0
BOTTOM = 1
STRAIGHT = 0
CURVED = 1
def pipes(n: int, pipes_system: List[List[int]]) -> bool:
switch = []
for i in range(n):
if determine_pipe(pipes_system[TOP][i]) == CURVED and determine_pipe(pipes_system[BOTTOM][i]) == CURVED:
switch.append(i)
if len(switch) % 2 != 1:
return False
cur_row = TOP
switch = set(switch)
for i in range(n):
if i in switch:
cur_row = BOTTOM if cur_row == TOP else TOP
elif determine_pipe(pipes_system[cur_row][i]) != STRAIGHT:
return False
return True
def determine_pipe(type: int) -> int:
if 1 <= type <= 2:
return STRAIGHT
return CURVED
def test():
assert pipes(7, [[2, 3, 2, 3, 2, 1, 6], [1, 6, 1, 5, 1, 2, 4]])
assert pipes(1, [[3], [4]])
assert pipes(2, [[1, 3], [2, 4]])
assert not pipes(2, [[1, 2], [3, 4]])
assert pipes(3, [[5, 3, 6], [3, 4, 5]])
assert not pipes(2, [[4, 6], [5, 4]])
if __name__ == '__main__':
q = int(input())
for _ in range(q):
n = int(input())
first_row = [int(c) for c in input()]
second_row = [int(c) for c in input()]
result = pipes(n, [first_row, second_row])
if result:
print("YES")
else:
print("NO")
``` | output | 1 | 60,737 | 3 | 121,475 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a system of pipes. It consists of two rows, each row consists of n pipes. The top left pipe has the coordinates (1, 1) and the bottom right — (2, n).
There are six types of pipes: two types of straight pipes and four types of curved pipes. Here are the examples of all six types:
<image> Types of pipes
You can turn each of the given pipes 90 degrees clockwise or counterclockwise arbitrary (possibly, zero) number of times (so the types 1 and 2 can become each other and types 3, 4, 5, 6 can become each other).
You want to turn some pipes in a way that the water flow can start at (1, 0) (to the left of the top left pipe), move to the pipe at (1, 1), flow somehow by connected pipes to the pipe at (2, n) and flow right to (2, n + 1).
Pipes are connected if they are adjacent in the system and their ends are connected. Here are examples of connected pipes:
<image> Examples of connected pipes
Let's describe the problem using some example:
<image> The first example input
And its solution is below:
<image> The first example answer
As you can see, the water flow is the poorly drawn blue line. To obtain the answer, we need to turn the pipe at (1, 2) 90 degrees clockwise, the pipe at (2, 3) 90 degrees, the pipe at (1, 6) 90 degrees, the pipe at (1, 7) 180 degrees and the pipe at (2, 7) 180 degrees. Then the flow of water can reach (2, n + 1) from (1, 0).
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries. Then q queries follow.
Each query consists of exactly three lines. The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of pipes in each row. The next two lines contain a description of the first and the second rows correspondingly. Each row description consists of n digits from 1 to 6 without any whitespaces between them, each digit corresponds to the type of pipe in the corresponding cell. See the problem statement to understand which digits correspond to which types of pipes.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For the i-th query print the answer for it — "YES" (without quotes) if it is possible to turn some pipes in a way that the water flow can reach (2, n + 1) from (1, 0), and "NO" otherwise.
Example
Input
6
7
2323216
1615124
1
3
4
2
13
24
2
12
34
3
536
345
2
46
54
Output
YES
YES
YES
NO
YES
NO
Note
The first query from the example is described in the problem statement. | instruction | 0 | 60,738 | 3 | 121,476 |
Tags: dp, implementation
Correct Solution:
```
q = int(input())
for r in range(q):
n = int(input())
rows = [[0 if int(c) <= 2 else 1 for c in input()]]
rows.append([0 if int(c) <= 2 else 1 for c in input()])
switch = 0
row = 0
i = 0
while i < n - 1:
if rows[row][i]:
if switch:
switch = 0
i += 1
else:
switch = 1
row = 1 if row == 0 else 0
else:
if switch:
print("NO")
break
else:
i += 1
else:
if (row == 1 and not rows[row][i]) or (row == 0 and rows[row][i] and rows[1][i]):
print("YES")
else:
print("NO")
``` | output | 1 | 60,738 | 3 | 121,477 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a system of pipes. It consists of two rows, each row consists of n pipes. The top left pipe has the coordinates (1, 1) and the bottom right — (2, n).
There are six types of pipes: two types of straight pipes and four types of curved pipes. Here are the examples of all six types:
<image> Types of pipes
You can turn each of the given pipes 90 degrees clockwise or counterclockwise arbitrary (possibly, zero) number of times (so the types 1 and 2 can become each other and types 3, 4, 5, 6 can become each other).
You want to turn some pipes in a way that the water flow can start at (1, 0) (to the left of the top left pipe), move to the pipe at (1, 1), flow somehow by connected pipes to the pipe at (2, n) and flow right to (2, n + 1).
Pipes are connected if they are adjacent in the system and their ends are connected. Here are examples of connected pipes:
<image> Examples of connected pipes
Let's describe the problem using some example:
<image> The first example input
And its solution is below:
<image> The first example answer
As you can see, the water flow is the poorly drawn blue line. To obtain the answer, we need to turn the pipe at (1, 2) 90 degrees clockwise, the pipe at (2, 3) 90 degrees, the pipe at (1, 6) 90 degrees, the pipe at (1, 7) 180 degrees and the pipe at (2, 7) 180 degrees. Then the flow of water can reach (2, n + 1) from (1, 0).
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 10^4) — the number of queries. Then q queries follow.
Each query consists of exactly three lines. The first line of the query contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of pipes in each row. The next two lines contain a description of the first and the second rows correspondingly. Each row description consists of n digits from 1 to 6 without any whitespaces between them, each digit corresponds to the type of pipe in the corresponding cell. See the problem statement to understand which digits correspond to which types of pipes.
It is guaranteed that the sum of n over all queries does not exceed 2 ⋅ 10^5.
Output
For the i-th query print the answer for it — "YES" (without quotes) if it is possible to turn some pipes in a way that the water flow can reach (2, n + 1) from (1, 0), and "NO" otherwise.
Example
Input
6
7
2323216
1615124
1
3
4
2
13
24
2
12
34
3
536
345
2
46
54
Output
YES
YES
YES
NO
YES
NO
Note
The first query from the example is described in the problem statement. | instruction | 0 | 60,739 | 3 | 121,478 |
Tags: dp, implementation
Correct Solution:
```
q = int(input())
for iter in range(q):
arr = []
n = int(input())
for read in range(2):
arr.append([int(x) for x in list(input())])
curr_i = 0
curr_j = 0
flag = 0
while curr_j<n:
ele = arr[curr_i][curr_j]
if ele == 1 or ele == 2:
curr_j += 1
else:
curr_i = 1-curr_i
new_ele = arr[curr_i][curr_j]
if new_ele == 1 or new_ele == 2:
flag=1
break
curr_j += 1
if curr_i == 0 or flag==1:
print("NO")
else:
print("YES")
``` | output | 1 | 60,739 | 3 | 121,479 |
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