text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Monocarp is playing a game "Assimilation IV". In this game he manages a great empire: builds cities and conquers new lands.
Monocarp's empire has n cities. In order to conquer new lands he plans to build one Monument in each city. The game is turn-based and, since Monocarp is still amateur, he builds exactly one Monument per turn.
Monocarp has m points on the map he'd like to control using the constructed Monuments. For each point he knows the distance between it and each city. Monuments work in the following way: when built in some city, a Monument controls all points at distance at most 1 to this city. Next turn, the Monument controls all points at distance at most 2, the turn after β at distance at most 3, and so on. Monocarp will build n Monuments in n turns and his empire will conquer all points that are controlled by at least one Monument.
Monocarp can't figure out any strategy, so during each turn he will choose a city for a Monument randomly among all remaining cities (cities without Monuments). Monocarp wants to know how many points (among m of them) he will conquer at the end of turn number n. Help him to calculate the expected number of conquered points!
Input
The first line contains two integers n and m (1 β€ n β€ 20; 1 β€ m β€ 5 β
10^4) β the number of cities and the number of points.
Next n lines contains m integers each: the j-th integer of the i-th line d_{i, j} (1 β€ d_{i, j} β€ n + 1) is the distance between the i-th city and the j-th point.
Output
It can be shown that the expected number of points Monocarp conquers at the end of the n-th turn can be represented as an irreducible fraction x/y. Print this fraction modulo 998 244 353, i. e. value x β
y^{-1} mod 998244353 where y^{-1} is such number that y β
y^{-1} mod 998244353 = 1.
Example
Input
3 5
1 4 4 3 4
1 4 1 4 2
1 4 4 4 3
Output
166374062
Note
Let's look at all possible orders of cities Monuments will be build in:
* [1, 2, 3]:
* the first city controls all points at distance at most 3, in other words, points 1 and 4;
* the second city controls all points at distance at most 2, or points 1, 3 and 5;
* the third city controls all points at distance at most 1, or point 1.
In total, 4 points are controlled.
* [1, 3, 2]: the first city controls points 1 and 4; the second city β points 1 and 3; the third city β point 1. In total, 3 points.
* [2, 1, 3]: the first city controls point 1; the second city β points 1, 3 and 5; the third city β point 1. In total, 3 points.
* [2, 3, 1]: the first city controls point 1; the second city β points 1, 3 and 5; the third city β point 1. In total, 3 points.
* [3, 1, 2]: the first city controls point 1; the second city β points 1 and 3; the third city β points 1 and 5. In total, 3 points.
* [3, 2, 1]: the first city controls point 1; the second city β points 1, 3 and 5; the third city β points 1 and 5. In total, 3 points.
The expected number of controlled points is (4 + 3 + 3 + 3 + 3 + 3)/(6) = 19/6 or 19 β
6^{-1} β‘ 19 β
166374059 β‘ 166374062 \pmod{998244353}
Submitted Solution:
```
a=input()
if a==('Aska'):
print(a)
```
No
| 104,700 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
num=int(input())
li=list(map(int,input().split()))
odd=0
even=0
li.sort()
#print(li)
for i in range(len(li)):
if i%2==0:
odd+=(3.1415926536)*(li[i]*li[i])
else:
even+=(3.1415926536)*(li[i]*li[i])
print(abs(odd-even))
```
| 104,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
from math import pi
n = int(input())
arr = [int(i) for i in input().strip().split()]
arr.sort(reverse = True)
if n % 2 != 0:
arr.append(0)
ans = 0
for i in range(0, len(arr), +2):
ans += arr[i]**2 - arr[i+1]**2
print(pi*ans)
```
| 104,702 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
from sys import stdin,stdout
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
from math import pi
n = ii1()
arr = sorted([0] + iia())
res = 0
if n < 2:
res = (arr[1])**2
else:
for i in range(n,0,-2):
if i > 0:
# print(i,i-1)
res += arr[i] ** 2 - arr[i - 1] ** 2
print(pi*res)
```
| 104,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
n = int(input())
r = [int(x) for x in input().split()]
r.sort(key = lambda x: -x)
res = 0
sign = True
for i in r:
res = res + i**2 if sign else res - i**2
sign = not sign
print(3.1415926536 * res)
```
| 104,704 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
a=int(input())
import math
b=list(map(int,input().split()))
b.sort()
if len(b)%2==1:
l=[b[i+1]*b[i+1]*math.pi-b[i]*b[i]*math.pi for i in range(len(b[1::])) if i%2==1]
c=round(b[0]*b[0]*math.pi+sum(l),10)
else:
l=[b[i]*b[i]*math.pi-b[i-1]*b[i-1]*math.pi for i in range(len(b)) if i%2==1]
c=round(sum(l),10)
print(c)
```
| 104,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
import math
x = int(input())
l = sorted(list(map(int,input().split())))
area = 0
if x % 2 == 0:
num = 1
while num < x:
area += abs((math.pi * (l[num - 1] ** 2)) - (math.pi*(l[num] ** 2)))
num += 2
print(area)
else:
num = 2
while num < x:
area += abs((math.pi * (l[num-1]**2)) - (math.pi*(l[num]**2)))
num += 2
print(area + (math.pi*(l[0]**2)))
```
| 104,706 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
# cook your dish here
import math
n=int(input())
a=list(map(int,input().split()))
for i in range(n):
a[i]=(math.pi)*(a[i]**2)
a=sorted(a)
c=0
if (n&1)==1:
for i in range(n):
if i&1==0:
if i==0:
c+=a[i]
else:
c+=a[i]-a[i-1]
else:
for i in range(n):
if i&1==1:
c+=a[i]-a[i-1]
print(c)
```
| 104,707 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Tags: geometry, sortings
Correct Solution:
```
from math import pi
n=input()
liste=input().split(" ")
liste=[ int(radius) for radius in liste]
liste.sort(reverse=True)
area=0
for i in range(len(liste)):
area+= -1*(i%2*2-1)*liste[i]**2
area*=pi
print(area)
```
| 104,708 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
import sys
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
# import numpy as np
starttime = time.time()
# import numpy as np
mod = int(pow(10, 9) + 7)
mod2 = 998244353
def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
# sys.setrecursionlimit(int(pow(10,6)))
sys.stdin = open("input.txt", "r")
# sys.stdout = open("../output.txt", "w")
except:
pass
n=L()[0]+1
A=[0]+sorted(L())
# for i in range(1,n):
# A[i]+=A[i-1]
ans=0
for i in range(n-1,0,-2):
ans+=(pi*(A[i]**2-A[i-1]**2))
print(ans)
endtime = time.time()
# print(f"Runtime of the program is {endtime - starttime}")
```
Yes
| 104,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
import math
n=int(input())
r=[int(i) for i in input().split()]
r.sort(reverse=True)
counter=0
red=0
while counter<n-1:
red+=(r[counter]**2-r[counter+1]**2)
counter+=2
if counter==n-1:
red+=r[counter]**2
print(red*math.pi)
```
Yes
| 104,710 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
#link: https://codeforces.com/problemset/problem/157/B
import math
n = int(input())
raios = list(map(int, input().split(" ")))
raios.sort(reverse=True)
r = math.pi
quadrados_pares = []
quadrados_impares = []
for x in range(0,len(raios)):
if x % 2 == 0:
quadrados_pares.append(raios[x]**2)
elif x % 2 == 1:
quadrados_impares.append(raios[x]**2)
resultado = sum(quadrados_pares) - sum(quadrados_impares)
area_resultante = resultado * r
print(area_resultante)
```
Yes
| 104,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
import math
n=int(input())
arr1=list(map(int,input().split()))
c=0
arr1.sort()
for i in range(len(arr1)):
if i%2==1:
c-=arr1[i]**2
else:
c+=arr1[i]**2
print(math.pi*abs(c))
```
Yes
| 104,712 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
def zip_sorted(a,b):
# sorted by a
a,b = zip(*sorted(zip(a,b)))
# sorted by b
sorted(zip(a, b), key=lambda x: x[1])
return a,b
import sys
input = sys.stdin.readline
I = lambda : list(map(int,input().split()))
S = lambda : list(map(str,input().split()))
import math
t,=I()
n = I()
n = sorted(n)
n1 = [0]+n
area = 0
i = 0
n2 = len(n1)
while i<n2:
area = area + ((math.pi)*(n1[n2-i-1])*(n1[n2-i-1]) - (math.pi)*(n1[n2-i-2])*(n1[n2-i-2]))
i = i+2
print(area)
'''
for i in range(n):
for j in range(n):
for k1 in range(len(a)):
for k2 in range(len(a)):
for k3 in range(len(a)):
'''
```
No
| 104,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
n=int(input())
ans=0
k=1
for r in sorted(map(int,input().split())):
ans+=r*r*k
print(abs(ans*3.1416))
```
No
| 104,714 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
from sys import stdin,stdout
ii1 = lambda: int(stdin.readline().strip())
is1 = lambda: stdin.readline().strip()
iia = lambda: list(map(int, stdin.readline().strip().split()))
isa = lambda: stdin.readline().strip().split()
mod = 1000000007
from math import pi
n = ii1()
arr = sorted([0] + iia())
res = 0
if 1 < 2:
res = arr[0]
else:
for i in range(n,0,-2):
if i > 0:
# print(i,i-1)
res += arr[i] ** 2 - arr[i - 1] ** 2
print(pi*res)
```
No
| 104,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One day, as Sherlock Holmes was tracking down one very important criminal, he found a wonderful painting on the wall. This wall could be represented as a plane. The painting had several concentric circles that divided the wall into several parts. Some parts were painted red and all the other were painted blue. Besides, any two neighboring parts were painted different colors, that is, the red and the blue color were alternating, i. e. followed one after the other. The outer area of the wall (the area that lied outside all circles) was painted blue. Help Sherlock Holmes determine the total area of red parts of the wall.
Let us remind you that two circles are called concentric if their centers coincide. Several circles are called concentric if any two of them are concentric.
Input
The first line contains the single integer n (1 β€ n β€ 100). The second line contains n space-separated integers ri (1 β€ ri β€ 1000) β the circles' radii. It is guaranteed that all circles are different.
Output
Print the single real number β total area of the part of the wall that is painted red. The answer is accepted if absolute or relative error doesn't exceed 10 - 4.
Examples
Input
1
1
Output
3.1415926536
Input
3
1 4 2
Output
40.8407044967
Note
In the first sample the picture is just one circle of radius 1. Inner part of the circle is painted red. The area of the red part equals Ο Γ 12 = Ο.
In the second sample there are three circles of radii 1, 4 and 2. Outside part of the second circle is painted blue. Part between the second and the third circles is painted red. Part between the first and the third is painted blue. And, finally, the inner part of the first circle is painted red. Overall there are two red parts: the ring between the second and the third circles and the inner part of the first circle. Total area of the red parts is equal (Ο Γ 42 - Ο Γ 22) + Ο Γ 12 = Ο Γ 12 + Ο = 13Ο
Submitted Solution:
```
"""
~~ author : dokueki
~~ created : 11~05~2020
"""
import sys
from math import pi
INT_MAX = sys.maxsize
INT_MIN = -(sys.maxsize)-1
sys.setrecursionlimit(10**7)
mod = 1000000007
def IOE():
sys.stdin = open("input.txt", "r")
sys.stdout = open("output.txt", "w")
def main():
n = int(sys.stdin.readline())
a = list(map(int, sys.stdin.readline().split()))
a.sort()
s = 0
if n == 1:
print((a[0]**2) * pi)
else:
if len(a) & 1:
s += a[0] * pi
for i in range(2, n, 2):
s += abs(a[i] ** 2 - a[i - 1] ** 2) * pi
print(s)
else:
for i in range(1, n, 2):
s += abs(a[i] ** 2 - a[i - 1] ** 2) * pi
print(s)
if __name__ == "__main__":
try:
IOE()
except:
pass
main()
```
No
| 104,716 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
import math
tot = int(input())
arr = list(map(int,input().split(" ")))
p = tot - 1
p = int(math.log(p)/math.log(2))
k = range(1,2**p + 1)
ct = 1
prev = 0
for i in range(tot -1):
print(prev + arr[i])
prev = arr[i] + prev
if(arr[i]!=0):
p = int(math.log(tot - 1 - i)/math.log(2))
arr[i + 2 ** p] += arr[i]
# print(tp)
# for i in range(1,tot):
# print(tp)
# for i in k:
# print(sum(arr[:i]))
# for i in range(2**p + 1,tot):
# print(sum(arr[:i]) + sum(arr[:ct]))
# ct += 1
```
| 104,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
c = 0
for i in range(n-1):
j = i + 2**int(math.log(n-1-i, 2))
c += a[i]
a[j] += a[i]
a[i] = 0
print(c)
```
| 104,718 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
ans = []
cur = 0
for i in range(n - 1):
if a[i] == 0:
ans.append(cur)
else:
t = 0
while i + 2 ** t < n:
t += 1
t -= 1
cur += a[i]
a[i + 2 ** t] += a[i]
ans.append(cur)
for x in ans:
print(x)
```
| 104,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
from math import log,floor
n = int(input())
a = list(map(int,input().split()))
for k in range(1,n):
moves=0
b = a[:]
for i in range(1,k+1):
x = log(n-i,2)
j=floor(x)
while j>=0:
y = i+(2**j)
if y<=n:
moves += b[i-1]
b[y-1] += b[i-1]
b[i-1] = 0
break
else:
j -= 1
print(moves)
```
| 104,720 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
import sys
import math
#to read string
get_string = lambda: sys.stdin.readline().strip()
#to read list of integers
get_int_list = lambda: list( map(int,sys.stdin.readline().strip().split()) )
#to read integers
get_int = lambda: int(sys.stdin.readline())
#--------------------------------WhiteHat010--------------------------------------#
n = get_int()
lst = get_int_list()
prev = 0
for i in range(n-1):
t = int( math.log( n-i-1 , 2 ) )
prev = prev + lst[i]
lst[2**t + i] += lst[i]
print(prev)
```
| 104,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
ans = 0
for i in range(n-1):
if a[i] != 0:
for j in range(17,-1,-1):
if i+2**j < n:
a[i+2**j] += a[i]
ans += a[i]
a[i] = 0
print(ans)
```
| 104,722 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
from math import *
t = int(input())
array = []
i = t
array = list(map(int,input().split()))
ans = 0
for x in range(0,i-1):
ans += array[x]
diff = i - x -1
power = int(log(diff,2))
array[x+(2**power)]+=array[x]
print(ans)
```
| 104,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Tags: greedy
Correct Solution:
```
n=int(input())
L=[int(x) for x in input().split()]
count=0
for i in range(n-1):
t=0
while i+2**(t+1)<n:
t+=1
L[i+2**t]+=L[i]
count+=L[i]
L[i]=0
print(count)
```
| 104,724 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Submitted Solution:
```
n = int(input())
a = [int(t) for t in input().split()]
c = 0
for i in range(n - 1):
if a[i] > 0:
c += a[i]
print(c)
j = 0
while 2 ** j + i < n:
j += 1
a[2 ** (j - 1) + i] += a[i]
a[i] = 0
else:
print(c)
```
Yes
| 104,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
cnt = 0
for i in range(n-1):
t = 0;
while (i+(1<<t)) < n:
t+=1
j = i + (1<<(t-1))
a[j] += a[i]
cnt += a[i]
print(cnt)
```
Yes
| 104,726 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Submitted Solution:
```
import math
def calc_max_pow(i, n):
return int(math.log(n - i, 2) // 1)
n = int(input())
sequence = input().split(' ')
for i in range(n):
sequence[i] = int(sequence[i])
moves_before = 0
for i in range(n - 1):
moves_before += sequence[i]
print(moves_before)
max_pow_to_move = calc_max_pow(i + 1, n)
move_to = pow(2, max_pow_to_move) + i
sequence[move_to] += sequence[i]
```
Yes
| 104,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Submitted Solution:
```
import math
n = int(input())
a = [int(s) for s in input().split(' ')]
def calculate(mylist):
number_of_steps = []
length = len(mylist)
def getLastIndex(i,n):
t = int(math.log(n - i,2))
return i + 2 ** t
for i in range(1,n):
if(i == 1):
number_of_steps.append(mylist[0])
mylist[getLastIndex(1,length) - 1] += mylist[0]
else:
number_of_steps.append(number_of_steps[-1] + mylist[i-1])
mylist[getLastIndex(i,length) - 1] += mylist[i-1]
for i in number_of_steps:
print(i)
calculate(a)
```
Yes
| 104,728 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Smart Beaver from ABBYY began to develop a new educational game for children. The rules of the game are fairly simple and are described below.
The playing field is a sequence of n non-negative integers ai numbered from 1 to n. The goal of the game is to make numbers a1, a2, ..., ak (i.e. some prefix of the sequence) equal to zero for some fixed k (k < n), and this should be done in the smallest possible number of moves.
One move is choosing an integer i (1 β€ i β€ n) such that ai > 0 and an integer t (t β₯ 0) such that i + 2t β€ n. After the values of i and t have been selected, the value of ai is decreased by 1, and the value of ai + 2t is increased by 1. For example, let n = 4 and a = (1, 0, 1, 2), then it is possible to make move i = 3, t = 0 and get a = (1, 0, 0, 3) or to make move i = 1, t = 1 and get a = (0, 0, 2, 2) (the only possible other move is i = 1, t = 0).
You are given n and the initial sequence ai. The task is to calculate the minimum number of moves needed to make the first k elements of the original sequence equal to zero for each possible k (1 β€ k < n).
Input
The first input line contains a single integer n. The second line contains n integers ai (0 β€ ai β€ 104), separated by single spaces.
The input limitations for getting 20 points are:
* 1 β€ n β€ 300
The input limitations for getting 50 points are:
* 1 β€ n β€ 2000
The input limitations for getting 100 points are:
* 1 β€ n β€ 105
Output
Print exactly n - 1 lines: the k-th output line must contain the minimum number of moves needed to make the first k elements of the original sequence ai equal to zero.
Please do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams, or the %I64d specifier.
Examples
Input
4
1 0 1 2
Output
1
1
3
Input
8
1 2 3 4 5 6 7 8
Output
1
3
6
10
16
24
40
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
rs = 0
for i in range(1,n):
if a[i - 1]!=0:
rs+=a[i - 1]
print(rs)
t = 0
while(i+2**t <= n): t+=1
t-=1
a[int(i + 2**t) - 1] += a[i - 1]
```
No
| 104,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 β€ n β€ 105) β length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive β numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Tags: greedy, hashing, string suffix structures
Correct Solution:
```
def pow(x, exp, mod):
res = 1
while exp:
if exp & 1:
res = (res * x) % mod
x = (x * x) % mod
exp >>= 1
return res
MOD = 2 ** 121 - 1
M = int(1e9) + 1
n = int(input())
vals = list(map(int, input().split()))
groups = dict()
for i in range(n):
groups.setdefault(vals[i], []).append(i)
powsA = [1]
for i in range(n):
powsA.append((powsA[-1] * M) % MOD)
hashes = [0] * (n + 1)
for i in range(n):
hashes[i + 1] = (hashes[i] * M + vals[i]) % MOD
def get_hash(p, l):
res = hashes[p + l] - (hashes[p] * powsA[l]) % MOD
if res < 0:
res += MOD
elif res > MOD:
res -= MOD
return res
best = 0
i = 0
while i < n:
val = vals[i]
for j in groups[val]:
if j <= i:
continue
l = j - i
if j + l <= n and get_hash(i, l) == get_hash(j, l):
best = max(best, j)
i = j - 1
break
i += 1
res = vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
| 104,730 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 β€ n β€ 105) β length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive β numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
n = int(input())
number = [int(x) for x in input().split()]
x = list(set(number))
print(len(x))
for i in x:
print(i,end=" ")
```
No
| 104,731 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 β€ n β€ 105) β length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive β numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))[:n]
b = []
for i in range(len(a)):
if a[i] not in b:
b.append(a[i])
print(len(b))
print(*b)
```
No
| 104,732 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 β€ n β€ 105) β length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive β numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
def get_pow_list(x, exp, mod):
res = [1]
for i in range(exp):
res.append((res[-1] * x) % mod)
return res
MOD = 2 ** 61 - 1
n = int(input())
orig_vals = list(map(int, input().split()))
sv = list(set(orig_vals))
sv_inv = {sv[i]: i + 1 for i in range(len(sv))}
vals = [sv_inv[x] for x in orig_vals]
groups = dict()
for i in range(n):
groups.setdefault(vals[i], [])
groups[vals[i]].append(i)
pows = [get_pow_list(x, n, MOD) for x in [2, len(sv) + 1]]
logs = [-1]
while len(logs) - 1 < n:
logs.extend([logs[-1] + 1] * len(logs))
hashes = [[0] * (4 * n) for i in range(logs[n] + 1)]
for i in range(n):
hashes[0][i] = vals[i]
for l in range(1, len(hashes)):
e = pows[0][l - 1]
for i in range(n):
hashes[l][i] = hashes[l - 1][i] * pows[1][e] + hashes[l - 1][i + e]
def get_hash(h, p, l, mod):
if l == 0:
return 0
elif ((l - 1) & l) == 0:
return h[logs[l]][p]
else:
nl = pows[0][logs[l]]
return get_hash(h, p, nl, mod) * pows[1][l - nl] + get_hash(
h, p + nl, l - nl, mod
)
best = 0
for i in range(n - 1, -1, -1):
val = vals[i]
for jj in range(len(groups[val])):
j = groups[val][jj]
if i == j:
break
l = i - j
if get_hash(hashes, i, l, MOD) == get_hash(hashes, j, l, MOD):
best = max(best, j)
if best != 0:
break
res = orig_vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
No
| 104,733 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Bob saw a string. It contained so many different letters, that the letters were marked by numbers, but at the same time each letter could be met in the string at most 10 times. Bob didn't like that string, because it contained repeats: a repeat of length x is such a substring of length 2x, that its first half coincides character by character with its second half. Bob started deleting all the repeats from the string. He does it as follows: while it's possible, Bob takes the shortest repeat, if it is not unique, he takes the leftmost one, and deletes its left half and everything that is to the left of this repeat.
You're given the string seen by Bob. Find out, what it will look like after Bob deletes all the repeats in the way described above.
Input
The first input line contains integer n (1 β€ n β€ 105) β length of the string. The following line contains n space-separated integer numbers from 0 to 109 inclusive β numbers that stand for the letters of the string. It's guaranteed that each letter can be met in the string at most 10 times.
Output
In the first line output the length of the string's part, left after Bob's deletions. In the second line output all the letters (separated by a space) of the string, left after Bob deleted all the repeats in the described way.
Examples
Input
6
1 2 3 1 2 3
Output
3
1 2 3
Input
7
4 5 6 5 6 7 7
Output
1
7
Submitted Solution:
```
def pow(x, exp, mod):
res = 1
while exp:
if exp & 1:
res = (res * x) % mod
x = (x * x) % mod
exp >>= 1
return res
MOD = 2 ** 121 - 1
M = int(1e9) + 1
n = int(input())
vals = list(map(int, input().split()))
groups = dict()
for i in range(n):
groups.setdefault(vals[i], []).append(i)
powsA = [1]
for i in range(n):
powsA.append((powsA[-1] * M) % MOD)
hashes = [0] * (n + 1)
for i in range(n):
hashes[i + 1] = (hashes[i] * M + vals[i]) % MOD
def get_hash(p, l):
res = hashes[p + l] - (hashes[p] * powsA[l]) % MOD
if res < 0:
res += MOD
elif res > MOD:
res -= MOD
return res
best = 0
for i in range(n - 1, -1, -1):
val = vals[i]
for j in groups[val]:
if i == j:
break
l = i - j
if i + l <= n and get_hash(i, l) == get_hash(j, l):
best = max(best, i)
if best != 0:
break
res = vals[best:]
print(len(res))
print(" ".join(map(str, res)))
```
No
| 104,734 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
def main():
n = int(input())
if (n < 3):
print(-1)
else:
n -= 1
p = n
b = 10
res = 1
while p:
if p & 0x1: res *= b
b *= b
p >>= 1
for i in range(0, 1000):
if (res + i) % 210 == 0:
print(10 ** n + i)
exit()
main()
```
| 104,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n=int(input())
if n==1 or n==2:
print(-1)
elif n==3:
print(210)
else:
x=10**(n-1)
while x%210!=0:
x=x+1
print(x)
```
| 104,736 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
import math
n = int(input())
n -= 1
if n >= 2:
print(10 ** n + 210 - (10 ** n) % 210)
else:
print(-1)
```
| 104,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n = int(input())
if n <= 2:
print(-1)
elif n==3:
print(210)
else:
x = 2*pow(10,n-1,21)%21
a,b = x//10, x%10
print("1"+"0"*(n-4)+str(a)+str(b)+"0")
```
| 104,738 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n = int(input())
if n < 3 :
print(-1)
else :
ans = pow(10, (n - 1)) + (210 - (pow(10, (n - 1)) % 210))
print(ans)
```
| 104,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n = int(input())
if (n < 3):
print(-1)
exit()
cur = 10 ** (n - 1)
ans = 2 * 3 * 5 * 7
nex = cur//ans
print(ans * (nex + 1))
```
| 104,740 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n = int(input())
z = 10**(n-1)
if n<3:
print(-1)
else:
print(z+(210-z%210))
```
| 104,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Tags: math, number theory
Correct Solution:
```
n = int(input())
x = 10 ** (n-1)
res = x + 210 - x % 210
if res % 210 == 0 and res >= x and res <= x * 10:
print(res)
else:
print(-1)
```
| 104,742 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
n = int(input()) - 1
print(10 ** n + 210 - 10 * pow(10, n - 1, 21) if n > 1 else -1)
```
Yes
| 104,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
from fractions import gcd
from random import randint
if __name__ == "__main__":
n = int(input())
if n < 3:
print(-1)
exit(0)
res = (210 -(10**(n-1) % 210)) + 10**(n-1);
print(res)
```
Yes
| 104,744 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
n=int(input())
rem=10**(n-1)%210
var=210-rem
if n<3: #248B
print(-1)
elif n==3:
print(210)
else:
print(10**(n-1)+var)
```
Yes
| 104,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
n = int(input())
s = 10 ** (n - 1)
while s % 21 != 0:
s += 10
if n > 2:
print(s)
else:
print(-1)
```
Yes
| 104,746 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
n = int(input())
n = 10 ** (n - 1)
while n % 21 != 0:
n += 10
if n > 2:
print(n)
else:
print(-1)
```
No
| 104,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
import math
n=int(input())
if n == 1 or n==2:
print(-1)
else:
a = 210
r = int(math.pow(10, n - 1))
for i in range(r, r + 211):
if (i % a == 0):
print(i)
break;
```
No
| 104,748 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
from collections import defaultdict
dic = defaultdict(int)
l = int(input())
n = 2*3*5*7
for i in range(1000):
n *= 2
dic[len(str(n))] = n
print(dic[l]) if dic[l]!=0 else print(-1)
```
No
| 104,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.
Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.
A number's length is the number of digits in its decimal representation without leading zeros.
Input
A single input line contains a single integer n (1 β€ n β€ 105).
Output
Print a single integer β the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.
Examples
Input
1
Output
-1
Input
5
Output
10080
Submitted Solution:
```
n = int(input())
if n < 3:
print(-1)
else:
print(10 ** (n - 1) // 210 + 1)
```
No
| 104,750 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
_ = input()
a = list(map(int, input().split(' ')))
m = input()
res = list()
for _ in range(int(m)):
w, h = map(int, input().split(' '))
#print(w, h)
cmax = max(a[0], a[w-1])
res.append(cmax)
# for x in range(w):
# a[x] = cmax + h
a[0] = cmax + h
a[w-1] = cmax + h
#print(a)
print('\n'.join(map(str, res)))
```
| 104,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
n=int(input())
A=[0]+[int(x) for x in input().split()]
m=int(input())
curHeight=0
# print(A)
Answer=[]
for i in range(m):
a,b=map(int, input().split())
curHeight=max(curHeight,A[a])
Answer.append(str(curHeight))
curHeight+=b
print("\n".join(Answer))
```
| 104,752 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
m = int(input())
ans = 0
res = [0 for i in range(m)]
x = 0
for i in range(m):
w, h = map(int, input().split())
ans = max(a[w - 1], ans + x)
x = h
res[i] = ans
print(*res)
```
| 104,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
def check(n , a , m , dimensions):
result = list()
prev = 0
for ele in dimensions:
w = ele[0]
h = ele[1]
if w <= prev:
maxHeight = a[prev]
else:
maxHeight = max(a[prev:w])
prev = w - 1
result.append(maxHeight)
a[prev] = maxHeight + h
return result
if __name__ == "__main__":
n = int(input().rstrip())
a = list(map(int , input().rstrip().split()))
m = int(input().rstrip())
dimensions = list()
for i in range(m):
temp = list(map(int , input().rstrip().split()))
w = temp[0]
h = temp[1]
dimensions.append((w , h))
result = check(n , a , m , dimensions)
for i in result:
print (i)
```
| 104,754 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
g=l[0]
m=int(input())
S=''
for i in range(m) :
a,b=map(int,input().split())
t=max(l[0],l[a-1])
S+=str(t)+'\n'
l[0]=t+b
l[a-1]=t+b
print(S)
```
| 104,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
m = int(input())
for i in range(m):
w,h = map(int,input().split())
ans = max(l[0],l[w-1])
print(ans)
l[0]=ans+h
```
| 104,756 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
from math import log2, log, ceil
# swap_array function
def swaparr(arr, a, b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp
## gcd function
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if (k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return res
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0):
hi = len(a)
while lo < hi:
mid = (lo + hi) // 2
if a[mid] < x:
lo = mid + 1
else:
hi = mid
return lo
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while (n % 2 == 0):
primes[2] = primes.get(2, 0) + 1
n = n // 2
for i in range(3, int(n ** 0.5) + 2, 2):
while (n % i == 0):
primes[i] = primes.get(i, 0) + 1
n = n // i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0):
return 0
while (y > 0):
if ((y & 1) == 1):
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a, b):
temp = a
a = b
b = temp
return a, b
# find function
def find(x, link):
while (x != link[x]):
x = link[x]
return x
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x, y = swap(x, y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n + 1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n + 1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e6 + 5)
'''
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i * i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve()
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x // spf[x]
return ret
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)'''
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def find_parent(parent, i,cnt):
cnt+=1
if parent[i-1] ==i:
return cnt
if parent[i-1] !=i:
find_parent(parent, parent[i-1],cnt)
##########################################################
#for _ in range(int(input())):
# from collections import deque
from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#for _ in range(int(input())):
#n,k= map(int, input().split())
from collections import Counter
import math
#for i in range(int(input())):
#n,k=map(int, input().split())
n=int(input())
arr=list(map(int,input().split()))
m=int(input())
box=[]
for i in range(m):
u,v=map(int,input().split())
box.append((u,v))
ans=0
val=0
for i in range(m):
ans=max(val+ans,arr[box[i][0]-1])
val=box[i][1]
print(ans)
```
| 104,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Tags: data structures, implementation
Correct Solution:
```
from collections import Counter, defaultdict, OrderedDict, deque
from bisect import bisect_left, bisect_right
from functools import reduce, lru_cache
from typing import List
import itertools
import sys
import math
import heapq
import string
import random
MIN, MAX, MOD = -0x3f3f3f3f, 0x3f3f3f3f, 1000000007
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
n = N()
a = RLL()
m = N()
hit = 0
for _ in range(m):
w, h = RL()
res = max(a[w - 1], hit)
print(res)
hit = max(hit, res + h)
# mx = max(a[:w])
# print(mx)
# for i in range(w):
# a[i] = mx + h
```
| 104,758 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
stair_number = int(input())
stair_heights = list(map(int,input().split()))
box_number = int(input())
x = stair_heights[0]
for _ in range(box_number) :
w , h = input().split()
w ,h = int(w) , int(h)
if x >= stair_heights[w-1] :
k = x
else :
k = stair_heights[w-1]
print(k)
x = k + h
```
Yes
| 104,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, staircase = int(input()), [int(i) for i in input().split()]
boxes, best = [[int(i) for i in input().split()] for _ in range(int(input()))], 0
for box in boxes:
best = max(staircase[box[0] - 1], best)
print(best)
best += box[1] # only track leftmost stair
```
Yes
| 104,760 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
m = int(input())
l = [list(map(int,input().split())) for _ in range(m)]
ans = [a[l[0][0]-1]]
lasth = l[0][1]
for _,[w,h] in enumerate(l[1:]):
ans.append(max(ans[-1] + lasth,a[w-1]))
lasth = h
print(*ans,sep='\n')
```
Yes
| 104,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
__author__ = "runekri3"
stairs_amount = int(input())
stair_heights = list(map(int, input().split()))
boxes_amount = int(input())
boxes = []
for _ in range(boxes_amount):
boxes.append(list(map(int, input().split())))
for width, height in boxes:
box_bottom = max(stair_heights[0], stair_heights[width - 1])
print(box_bottom)
stair_heights[0] = box_bottom + height
```
Yes
| 104,762 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
from sys import stdin, stdout
number = int(stdin.readline())
stair_case = list(map(int, stdin.readline().split()))
boxes = int(stdin.readline())
current = []
levels = {}
mx_width = 0
cnt = 0
for index, x in enumerate(stair_case):
levels[index + 1] = x
for x in range(boxes):
w, h = map(int, input().split())
if w > mx_width and cnt < stair_case[-1]:
cnt = levels[w]
stdout.write(str(cnt) + '\n')
cnt += h
mx_width = w
else:
stdout.write(str(cnt) + '\n')
cnt += h
```
No
| 104,763 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
m = int(input())
boxes = []
for i in range(m):
w, h = list(map(int, input().split()))
boxes.append([w,h])
for i in range(m):
w, h = boxes[i]
t = a[w-1]
print(t)
for i in range(w):
a[i] = t+h
```
No
| 104,764 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def seive(n):
primes = [True]*(n+1)
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
return primes
def factors(n):
factors = []
x = 2
while x*x <= n:
while n % x == 0:
factors.append(x)
n //= x
if n > 1:
factors.append(x)
return factors
# Functions: list of factors, seive of primes, gcd of two numbers
def main():
try:
n = inp()
a = inlt()
m = inp()
mn = a[0]
for i in range(m):
w, h = invr()
print(max(mn,a[w-1]))
mn = a[w-1] + h
a[w-1] = max(mn, a[w-1])
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 104,765 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima's got a staircase that consists of n stairs. The first stair is at height a1, the second one is at a2, the last one is at an (1 β€ a1 β€ a2 β€ ... β€ an).
Dima decided to play with the staircase, so he is throwing rectangular boxes at the staircase from above. The i-th box has width wi and height hi. Dima throws each box vertically down on the first wi stairs of the staircase, that is, the box covers stairs with numbers 1, 2, ..., wi. Each thrown box flies vertically down until at least one of the two following events happen:
* the bottom of the box touches the top of a stair;
* the bottom of the box touches the top of a box, thrown earlier.
We only consider touching of the horizontal sides of stairs and boxes, at that touching with the corners isn't taken into consideration. Specifically, that implies that a box with width wi cannot touch the stair number wi + 1.
You are given the description of the staircase and the sequence in which Dima threw the boxes at it. For each box, determine how high the bottom of the box after landing will be. Consider a box to fall after the previous one lands.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of stairs in the staircase. The second line contains a non-decreasing sequence, consisting of n integers, a1, a2, ..., an (1 β€ ai β€ 109; ai β€ ai + 1).
The next line contains integer m (1 β€ m β€ 105) β the number of boxes. Each of the following m lines contains a pair of integers wi, hi (1 β€ wi β€ n; 1 β€ hi β€ 109) β the size of the i-th thrown box.
The numbers in the lines are separated by spaces.
Output
Print m integers β for each box the height, where the bottom of the box will be after landing. Print the answers for the boxes in the order, in which the boxes are given in the input.
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 6 6
4
1 1
3 1
1 1
4 3
Output
1
3
4
6
Input
3
1 2 3
2
1 1
3 1
Output
1
3
Input
1
1
5
1 2
1 10
1 10
1 10
1 10
Output
1
3
13
23
33
Note
The first sample are shown on the picture.
<image>
Submitted Solution:
```
#pyrival orz
import os
import sys
from io import BytesIO, IOBase
input = sys.stdin.readline
############ ---- Input Functions ---- ############
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
def gcd(a, b):
if b == 0:
return a
else:
return gcd(b, a%b)
def seive(n):
primes = [True]*(n+1)
for i in range(2, n):
if not primes[i]:
continue
j = 2*i
while j <= n:
primes[j] = False
j += i
return primes
def factors(n):
factors = []
x = 2
while x*x <= n:
while n % x == 0:
factors.append(x)
n //= x
if n > 1:
factors.append(x)
return factors
# Functions: list of factors, seive of primes, gcd of two numbers
def main():
try:
n = inp()
a = inlt()
m = inp()
mn = a[0]
for i in range(m):
w, h = invr()
print(max(mn,a[w-1]))
mn = a[w-1] + h
a[w-1] = mn
except Exception as e:
print(e)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 104,766 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
N = 100100
def main():
n = int(input())
d=[0]*N
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m,c=[0]*N,[0]*N
m[0] = b[0]
c[0] = d[0]
z = 1
p = 0
for i in range(1,n):
p = min(p,z)
while(p+1<z and m[p+1]*a[i]+c[p+1]<=m[p]*a[i]+c[p]):
p=p+1
d[i] = m[p]*a[i]+c[p]
while(z>=2 and (c[z-2]-c[z-1])*(b[i]-m[z-1])>=(c[z-1]-d[i])*(m[z-1]-m[z-2]) ):
z=z-1
m[z],c[z]=b[i],d[i]
z=z+1
#for j in range(n):print(d[j])
print(d[n-1])
main()
# Made By Mostafa_Khaled
```
| 104,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
read = lambda: map(int, input().split())
n = int(input())
a = list(read())
b = list(read())
dp = [0] * n
ch = [0]
def get(i, x):
return b[i] * x + dp[i]
def f1():
if len(ch) < 2:
return 0
return get(ch[0], a[i]) >= get(ch[1], a[i])
def f2():
if len(ch) < 2:
return 0
i1 = ch[-1]
x = (dp[i1] - dp[i]) / (b[i] - b[i1])
return get(ch[-2], x) <= get(i, x)
for i in range(1, n):
while f1():
ch.pop(0)
dp[i] = get(ch[0], a[i])
while f2():
ch.pop()
ch.append(i)
print(dp[n - 1])
```
| 104,768 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
c = [0]*n
st = [0]*n
first=0
second=0
for i in range(1,n):
#Delete from front until the value of lines is increasing w.r.to value a[i]
while(second-first>0 and a[i] * b[st[first]] + c[st[first]] >= a[i] * b[st[first+1]] + c[st[first+1]]):
first = first + 1;
c[i] = a[i] * b[st[first]] + c[st[first]];
#Delete from back until the intersection of lines last two lines in stack is to the left of the new line and the last line
while(second-first>0 and (b[st[second]] - b[i])*(c[st[second]] - c[st[second-1]]) > (c[i] - c[st[second]])*(b[st[second-1]] - b[st[second]])):
second = second-1;
second = second+1
st[second] = i;
print(c[n-1]);
```
| 104,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
def tree_cutting(n,A,B):
C=[0 for _ in range(n)]
hullx=[0 for _ in range(n)]
hully=[0 for _ in range(n)]
sz=0
p=0
C[0]=0
hullx[0]=B[0]
hully[0]=C[0]
for i in range(1,n):
p=min(sz,p)
while sz>0 and p<sz and (hully[p+1]-hully[p])/(hullx[p]-hullx[p+1])<=A[i]:
p+=1
C[i]=hully[p]+hullx[p]*A[i]
sz+=1
hullx[sz]=B[i]
hully[sz]=C[i]
while sz>1 and (hully[sz-2]-hully[sz-1])/(hullx[sz-1]-hullx[sz-2])>=(hully[sz-2]-hully[sz])/(hullx[sz]-hullx[sz-2]):
hullx[sz-1]=hullx[sz]
hully[sz-1]=hully[sz]
sz-=1
return C[n-1]
print(tree_cutting(n,A,B))
```
| 104,770 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = [0 for i in range(0, n + 1)]
b = [0 for i in range(0, n + 1)]
a = list(map(int, input().split()))
b = list(map(int, input().split()))
dp = [0 for i in range(0, n + 1)]
c = [[0 for i in range(0, 3)] for j in range(0, n + 1)]
stack = []
stack.append(0)
stack.append(1)
dp[0] = 0
dp[1] = a[1] * b[0]
def intersection(x, y):
return int((dp[y] - dp[x]) / (b[x] - b[y]))
last = 0
c[last] = [0, intersection(0, 1), 0]
last+=1
c[last] = [intersection(0, 1), 1000000001, 1]
last1=0
for i in range(2, n):
while (last1 >= 0):
if (c[last1][0] <= a[i] and c[last1][1] >= a[i]):
dp[i] = dp[c[last1][2]] + b[c[last1][2]] * a[i]
#print(i,dp[i])
break
elif c[last1][0] > a[i]:
last1 -= 1
else:
last1 += 1
while stack:
top = stack[-1]
if len(stack) >= 2:
second_top = stack[-2]
if intersection(second_top, i) < intersection(top,second_top):
stack.pop()
last -= 1
else:
break
else:
break
stack.append(i)
last += 1
c[last] = [intersection(top, i),1000000001,i]
c[last-1]= [c[last-1][0],intersection(top,i),c[last-1][2]]
print(dp[n-1])
```
| 104,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
n = int(input())
a = [0] + [int(x) for x in input().split()]
b = [0] + [int(x) for x in input().split()]
q = [0 for i in range(n + 1)]
f = [0 for _ in range(n + 1)]
l, r, q[1] = 1, 1, 1
for i in range(2, n + 1):
while l < r and f[q[l + 1]] - f[q[l]] < a[i] * (b[q[l]] - b[q[l + 1]]):
l += 1
f[i] = f[q[l]] + b[q[l]] * a[i]
while l < r and (f[q[r]] - f[q[r - 1]]) * (b[q[r]] - b[i]) >= (f[i] - f[q[r]]) * (b[q[r - 1]] - b[q[r]]):
r -= 1
r += 1
q[r] = i
print(f[n])
```
| 104,772 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 1 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
```
| 104,773 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Tags: dp, geometry
Correct Solution:
```
N = 100100
def main():
n = int(input())
d=[0]*N
a = list(map(int, input().split()))
b = list(map(int, input().split()))
m,c=[0]*N,[0]*N
m[0] = b[0]
c[0] = d[0]
z = 1
p = 0
for i in range(1,n):
p = min(p,z)
while(p+1<z and m[p+1]*a[i]+c[p+1]<=m[p]*a[i]+c[p]):
p=p+1
d[i] = m[p]*a[i]+c[p]
while(z>=2 and (c[z-2]-c[z-1])*(b[i]-m[z-1])>=(c[z-1]-d[i])*(m[z-1]-m[z-2]) ):
z=z-1
m[z],c[z]=b[i],d[i]
z=z+1
#for j in range(n):print(d[j])
print(d[n-1])
main()
```
| 104,774 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
c = [0] * n
stk = [0]
for i in range(1, n):
while len(stk) > 1 and c[stk[1]] - c[stk[0]] <= a[i] * (b[stk[0]] -
b[stk[1]]):
del stk[0]
c[i] = c[stk[0]] + a[i] * b[stk[0]]
while len(stk) > 1 and ((c[stk[-1]] - c[stk[-2]]) * (b[stk[-1]] - b[i]) >
(b[stk[-2]] - b[stk[-1]]) * (c[i] - c[stk[-1]])):
del stk[-1]
stk.append(i)
print(c[n - 1])
```
Yes
| 104,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def dot(a, b, c):
return a + c*b
def main():
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
dp = [ 0 for _ in range(n) ]
st = [0 for _ in range(n) ]
i, j = 0, 0
for k in range(1,n):
while(j - i > 0 and dot(dp[st[i]],b[st[i]],a[k]) >= dot(dp[st[i+1]],b[st[i+1]],a[k])):
i+=1
dp[k] = a[k]*b[st[i]] + dp[st[i]]
while(j - i > 0 and (b[st[j]] - b[k])*(dp[st[j]] - dp[st[j-1]]) > (dp[k] - dp[st[j]])*(b[st[j-1]] - b[st[j]])):
j-=1
j+=1
st[j] = k
print(dp[-1])
if __name__ == "__main__":
main()
```
Yes
| 104,776 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
#d_n= [C_m+A_n*B_m] for m in range()
#A=[ ->> ]
#B=[ <<- ]
def intersection(p,q):
return (q[1]-p[1])/(p[0]-q[0])
def tree_cutting(n,A,B):
I=[[0,0] for _ in range(n)]
C=[0 for _ in range(n)]
C[0]=0
I[0][0]=-float("inf")
I[0][1]=float("inf")
C[1]=C[0]+A[1]*B[0]
hull=[[B[0],C[0]],[B[1],C[1]]]
I[0][1]=intersection(hull[-1],hull[-2])
I[1][0]=I[0][1]
I[1][1]=float("inf")
curr=1
k=0
for i in range(2,n):
k=min(k,curr)-1
while True:
k+=1
if I[k][0]<=A[i] and A[i]<=I[k][1]:
j=k
break
C[i]=hull[k][1]+A[i]*hull[k][0]
p=[B[i],C[i]]
while intersection(p,hull[-2])<=intersection(hull[-1],hull[-2]):
hull.pop()
curr-=1
if len(hull)<2: break
if B[i]!=hull[-1][0]:
hull.append(p)
I[curr][1]=intersection(hull[-1],hull[-2])
curr+=1
I[curr][0]=intersection(hull[-1],hull[-2])
I[curr][1]=+float("inf")
else:
I[curr][1]=+float("inf")
return C[n-1]
if __name__=="__main__":
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
print(tree_cutting(n,A,B))
```
Yes
| 104,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 1 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
if left >= right: left = right - 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
```
Yes
| 104,778 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
# k, i, j
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 0 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 < right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
right -= 1
if left >= right: left = right - 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
```
No
| 104,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
def cross(i, j, k, b, cost):
return (cost[i] - cost[j]) * (b[i] - b[k]) - (b[i] - b[j]) * (cost[i] - cost[k])
def dot(i, j, a, b, cost):
return cost[j] + a[i] * b[j]
def CF319C():
N = int(input())
a = tuple(map(int, input().split()))
b = tuple(map(int, input().split()))
cost = [0] * N
hull = [0] * N # Make hull considering total minimum cost and charge of individual trees
left = 0
right = 0 # Index of items in hull
for i in range(1, N):
# Find the last cut tree which incurs minimum cost for ith tree
while left + 1 <= right and dot(i, hull[left], a, b, cost) >= dot(i, hull[left + 1], a, b, cost):
# Only increase left when it decreases cutting cost
left += 1
cost[i] = dot(i, hull[left], a, b, cost)
# print("Before ", hull[:right+1])
while right >= 2 and cross(hull[right], hull[right - 1], i, b, cost) >= 0:
# print("removed ", i, hull[right])
right -= 1
right += 1
hull[right] = i
return cost[-1]
if __name__ == '__main__':
res = CF319C()
print(res)
```
No
| 104,780 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
MD = 0x7fffffff
aa, bb, dp = [], [], []
def cross2(x1,y1,x2,y2):
h = x1*(y1/MD) - x2*(y1/MD)
l = x1*(y2%MD) - x2*(y1%MD)
h += l/MD
l %= MD
if h != 0:
if h < 0:
return -1
else:
return 1
else:
if l == 0:
return 0
else:
if l < 0:
return -1
else:
return 1
def cross(i,j,k):
return cross2(bb[j] - bb[i], dp[j] - dp[i], bb[k] - bb[i], dp[k] - dp[i])
def dot(i, a):
return dp[i] + a * bb[i]
def main():
n = int(input())
a = list(map(int, input().split(' ')))
b = list(map(int, input().split(' ')))
qq = []
cnt, h = 1, 0
for i in range(len(a)):
aa.append(a[i])
for i in range(len(b)):
bb.append(b[i])
for _ in range(n):
dp.append(0)
qq.append(0)
del a,b
for i in range(1,n):
a = aa[i]
while h+1 < cnt and dot(qq[h+1], a) <= dot(qq[h], a):
h+=1
dp[i] = dot(qq[h],a)
while cnt >= 2 and cross(qq[cnt-2], qq[cnt-1], i) >= 0:
cnt-=1
if h >= cnt:
h = cnt-1
qq[cnt] = i
cnt+=1
# print(dp[-1])
print(aa)
print(bb)
print(dp)
if __name__ == "__main__":
main()
```
No
| 104,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kalila and Dimna are two jackals living in a huge jungle. One day they decided to join a logging factory in order to make money.
The manager of logging factory wants them to go to the jungle and cut n trees with heights a1, a2, ..., an. They bought a chain saw from a shop. Each time they use the chain saw on the tree number i, they can decrease the height of this tree by one unit. Each time that Kalila and Dimna use the chain saw, they need to recharge it. Cost of charging depends on the id of the trees which have been cut completely (a tree is cut completely if its height equal to 0). If the maximum id of a tree which has been cut completely is i (the tree that have height ai in the beginning), then the cost of charging the chain saw would be bi. If no tree is cut completely, Kalila and Dimna cannot charge the chain saw. The chainsaw is charged in the beginning. We know that for each i < j, ai < aj and bi > bj and also bn = 0 and a1 = 1. Kalila and Dimna want to cut all the trees completely, with minimum cost.
They want you to help them! Will you?
Input
The first line of input contains an integer n (1 β€ n β€ 105). The second line of input contains n integers a1, a2, ..., an (1 β€ ai β€ 109). The third line of input contains n integers b1, b2, ..., bn (0 β€ bi β€ 109).
It's guaranteed that a1 = 1, bn = 0, a1 < a2 < ... < an and b1 > b2 > ... > bn.
Output
The only line of output must contain the minimum cost of cutting all the trees completely.
Please, do not write the %lld specifier to read or write 64-bit integers in Π‘++. It is preferred to use the cin, cout streams or the %I64d specifier.
Examples
Input
5
1 2 3 4 5
5 4 3 2 0
Output
25
Input
6
1 2 3 10 20 30
6 5 4 3 2 0
Output
138
Submitted Solution:
```
#d_n= [C_m+A_n*B_m] for m in range()
#A=[ ->> ]
#B=[ <<- ]
def intersection(p,q):
return (q[1]-p[1])/(p[0]-q[0])
def tree_cutting(n,A,B):
I=[[0,0] for _ in range(n)]
C=[0]*(n+1)
C[0]=0
I[0][0]=-float("inf")
I[0][1]=float("inf")
C[1]=C[0]+A[0]*B[0]
hull=[[B[0],C[0]],[B[1],C[1]]]
I[0][1]=intersection(hull[-1],hull[-2])
I[1][0]=I[0][1]
I[1][1]=float("inf")
curr=1
k=0
for i in range(2,n):
k=min(k,curr)-1
while True:
k+=1
if I[k][0]<=A[i] and A[i]<=I[k][1]:
j=k
break
C[i]=hull[k][1]+A[i]*hull[k][0]
p=[B[i],C[i]]
while intersection(p,hull[-2])<=intersection(hull[-1],hull[-2]):
hull.pop()
curr-=1
if len(hull)<2: break
if B[i]!=hull[-1][0]:
hull.append(p)
I[curr][1]=intersection(hull[-1],hull[-2])
curr+=1
I[curr][0]=intersection(hull[-1],hull[-2])
I[curr][1]=+float("inf")
else:
I[curr][1]=+float("inf")
return C[n-1]
if __name__=="__main__":
n=int(input())
A=list(map(int,input().strip().split()))
B=list(map(int,input().strip().split()))
print(tree_cutting(n,A,B))
```
No
| 104,782 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
for T in range(1):
n, m = map(int, input().split())
p = list(map(int, input().split()))
q = list(map(int, input().split()))
ok, ng = 10**11, -1
while ok-ng > 1:
x = (ok+ng)//2
flg = True
j = 0
for i in range(n):
l = max(p[i] - q[j], 0)
if l > x:
flg = False
break
while j < m and q[j] < p[i]:
j += 1
if j == m:
break
while j < m:
r = q[j] - p[i]
if min(l, r)*2 + max(l, r) <= x:
j += 1
continue
else:
break
if j == m:
break
if flg == False or j < m:
ng = x
else:
ok = x
print(ok)
```
| 104,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**12)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
| 104,784 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.2e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright: pidx += 1
hidx += 1
if pidx != len(p): works = False
if works: ll = avg
else: ss = avg + 1
print(ss)
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
| 104,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
x=len(a)-1
y=len(b)-1
while x>=0 and y>=0:
if b[y]<=mi:
if abs(b[y]-ma)<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
elif b[y]>=ma:
if abs(b[y]-mi)<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
else:
if abs(ma-mi)+min(abs(b[y]-mi),abs(b[y]-ma))<=d1:
x-=1
if x==-1:
break
ma=a[x]
else:
y-=1
mi=a[x]
ma=a[x]
return x==-1
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**10)*3
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1,s):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
| 104,786 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**11)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
| 104,787 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.1e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright: pidx += 1
hidx += 1
if pidx != len(p): works = False
if works: ll = avg
else: ss = avg + 1
print(ss)
if sys.hexversion == 50594544 : sys.stdin = open("test.txt")
solve()
```
| 104,788 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Tags: binary search, greedy, two pointers
Correct Solution:
```
import sys
from itertools import *
from math import *
def solve():
n, m = map(int, input().split())
h = list(map(int, input().split()))
p = list(map(int, input().split()))
ss, ll = 0, int(2.1e10)
while ss < ll:
avg = (ss + ll) // 2
works = True
hidx = 0
pidx = 0
while hidx < len(h) and pidx < len(p):
leftget = p[pidx]
curpos = h[hidx]
if curpos - leftget > avg:
works = False
break
getbacktime = max(0, 2*(curpos - leftget))
alsotoright = max(0, avg - getbacktime)
leftime = max(0, curpos - leftget)
remtime = max(0, (avg - leftime) // 2)
furthestright = curpos + max(alsotoright, remtime)
while pidx < len(p) and p[pidx] <= furthestright:
pidx += 1
hidx += 1
if pidx != len(p):
works = False
if works:
ll = avg
else:
ss = avg + 1
print(ss)
solve()
```
| 104,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
import sys
def input():
return sys.stdin.buffer.readline()[:-1]
for T in range(1):
n, m = map(int, input().split())
p = list(map(int, input().split()))
q = list(map(int, input().split()))
ok, ng = 10**11, 0
while ok-ng > 1:
x = (ok+ng)//2
flg = True
j = 0
for i in range(n):
l = max(p[i] - q[j], 0)
if l > x:
flg = False
break
while j < m and q[j] < p[i]:
j += 1
if j == m:
break
while j < m:
r = q[j] - p[i]
if min(l, r)*2 + max(l, r) <= x:
j += 1
continue
else:
break
if j == m:
break
if flg == False or j < m:
ng = x
else:
ok = x
print(ok)
```
No
| 104,790 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
def can(d,a,b):
d1=d
mi=a[-1]
ma=a[-1]
while len(a)>0 and len(b)>0:
if b[-1]<=mi:
if abs(b[-1]-ma)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
elif b[-1]>=ma:
if abs(b[-1]-mi)<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
else:
if abs(ma-mi)+min(abs(b[-1]-mi),abs(b[-1]-ma))<=d1:
a.pop()
if len(a)==0:
break
ma=a[-1]
else:
b.pop()
mi=a[-1]
ma=a[-1]
return len(a)==0
n,m=map(int,input().split())
s=list(map(int,input().split()))[::-1]
s1=list(map(int,input().split()))[::-1]
high=(10**10)+1
low=0
while high-low>1:
mid=(high+low)//2
if can(mid,s1.copy(),s.copy()):
high=mid
else:
low=mid
if can(low,s1,s):
print(low)
else:
print(high)
```
No
| 104,791 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
#!/usr/bin/env python3
import bisect
def pred(hs, ps, t):
p = ps[0]
for h in hs:
if p < h - t:
return False
elif p < h:
q = max(h,
h + t - 2 * (h - p),
h + (t - (h - p)) // 2)
else:
q = h + t
if ps[-1] <= q:
return True
p = ps[bisect.bisect_right(ps, q)]
return False
n, m = map(int,input().split())
h = list(map(int,input().split()))
p = list(map(int,input().split()))
low, high = -1, max(h[-1], p[-1])
while low + 1 < high:
mid = (low + high) // 2
if pred(h,p,mid):
high = mid
else:
low = mid
print(high)
```
No
| 104,792 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
<image>
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
Input
The first line of the input contains two space-separated integers n, m (1 β€ n, m β€ 105) β the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 β€ hi β€ 1010, hi < hi + 1) β the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 β€ pi β€ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in Π‘++. It is recommended to use the cin, cout streams or the %I64d specifier.
Output
Print a single number β the minimum time required, in seconds, to read all the needed tracks.
Examples
Input
3 4
2 5 6
1 3 6 8
Output
2
Input
3 3
1 2 3
1 2 3
Output
0
Input
1 2
165
142 200
Output
81
Note
The first test coincides with the figure. In this case the given tracks can be read in 2 seconds in the following way:
1. during the first second move the 1-st head to the left and let it stay there;
2. move the second head to the left twice;
3. move the third head to the right twice (note that the 6-th track has already been read at the beginning).
One cannot read the tracks in 1 second as the 3-rd head is at distance 2 from the 8-th track.
Submitted Solution:
```
def can(k):
s2=s.copy()
x=0
y=0
while len(s2)>0 and x<len(s1):
#print(s2,s1,y)
if abs(s2[-1]-s1[x])+y<=k:
y+=abs(s2[-1]-s1[x])
s2[-1]=s1[x]
x+=1
else:
y=0
s2.pop()
if x==len(s1):
return True
else:
return False
import sys
n,m=map(int,input().split())
s=list(map(int,input().split()))
s=s[::-1]
s1=list(map(int,input().split()))
if n==1:
if s[0]>=s1[-1]:
print(s[0]-s1[0])
elif s[0]<=s1[0]:
print(s1[-1]-s[0])
else:
print(s[0]-s1[0]+(s1[-1]-s1[0]))
sys.exit()
ans=0
low=0
high=10000000002
aux1=10**12
aux2=10**12
if s[0]>s1[0]:
aux1=s[0]-s1[0]
s[0]=s1[0]
if s[-1]<s1[-1]:
aux2=s1[-1]-s[-1]
s[-1]=s1[-1]
while high-low>1:
mid=(low+high)//2
if can(mid):
high=mid
else:
low=mid
if can(low):
print(min(low,aux1,aux2))
else:
print(min(low+1,aux1,aux2))
```
No
| 104,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima loves Inna very much. He decided to write a song for her. Dima has a magic guitar with n strings and m frets. Dima makes the guitar produce sounds like that: to play a note, he needs to hold one of the strings on one of the frets and then pull the string. When Dima pulls the i-th string holding it on the j-th fret the guitar produces a note, let's denote it as aij. We know that Dima's guitar can produce k distinct notes. It is possible that some notes can be produced in multiple ways. In other words, it is possible that aij = apq at (i, j) β (p, q).
Dima has already written a song β a sequence of s notes. In order to play the song, you need to consecutively produce the notes from the song on the guitar. You can produce each note in any available way. Dima understood that there are many ways to play a song and he wants to play it so as to make the song look as complicated as possible (try to act like Cobein).
We'll represent a way to play a song as a sequence of pairs (xi, yi) (1 β€ i β€ s), such that the xi-th string on the yi-th fret produces the i-th note from the song. The complexity of moving between pairs (x1, y1) and (x2, y2) equals <image> + <image>. The complexity of a way to play a song is the maximum of complexities of moving between adjacent pairs.
Help Dima determine the maximum complexity of the way to play his song! The guy's gotta look cool!
Input
The first line of the input contains four integers n, m, k and s (1 β€ n, m β€ 2000, 1 β€ k β€ 9, 2 β€ s β€ 105).
Then follow n lines, each containing m integers aij (1 β€ aij β€ k). The number in the i-th row and the j-th column (aij) means a note that the guitar produces on the i-th string and the j-th fret.
The last line of the input contains s integers qi (1 β€ qi β€ k) β the sequence of notes of the song.
Output
In a single line print a single number β the maximum possible complexity of the song.
Examples
Input
4 6 5 7
3 1 2 2 3 1
3 2 2 2 5 5
4 2 2 2 5 3
3 2 2 1 4 3
2 3 1 4 1 5 1
Output
8
Input
4 4 9 5
4 7 9 5
1 2 1 7
8 3 4 9
5 7 7 2
7 1 9 2 5
Output
4
Tags: brute force, implementation, math
Correct Solution:
```
def solution() : # ζ倧ηθ·η¦»ζ₯θͺδΊθ§θ½ιθΏηηΉ
n,m,k,s = map(int, input().split())
dis = lambda a,b : abs(a[0] - b[0]) + abs(a[1] - b[1])
corner = [(0,0), (0,m-1), (n-1,0), (n-1,m-1)]
vertex = [[(n,m), (n,-1), (-1,m), (-1,-1)] for _ in range(k+1)]
for i in range(n) :
for j,note in enumerate(map(int, input().split())) :
vertex[note] = [
(i,j) if dis((i,j), c) < dis(v, c) else v
for v,c in zip(vertex[note], corner)]
maxdis = [[-1] * (k+1) for _ in range(k+1)]
pairs = [(0,3),(3,0),(1,2),(2,1)]
for i in range(1, k+1) :
for j in range(i, k+1) :
vi,vj = vertex[i],vertex[j]
maxdis[i][j] = max(dis(vi[a], vj[b]) for a,b in pairs)
maxdis[j][i] = maxdis[i][j]
s = list(map(int, input().split()))
print(max(maxdis[s[i]][s[i+1]] for i in range(len(s) - 1)))
solution()
```
| 104,794 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima loves Inna very much. He decided to write a song for her. Dima has a magic guitar with n strings and m frets. Dima makes the guitar produce sounds like that: to play a note, he needs to hold one of the strings on one of the frets and then pull the string. When Dima pulls the i-th string holding it on the j-th fret the guitar produces a note, let's denote it as aij. We know that Dima's guitar can produce k distinct notes. It is possible that some notes can be produced in multiple ways. In other words, it is possible that aij = apq at (i, j) β (p, q).
Dima has already written a song β a sequence of s notes. In order to play the song, you need to consecutively produce the notes from the song on the guitar. You can produce each note in any available way. Dima understood that there are many ways to play a song and he wants to play it so as to make the song look as complicated as possible (try to act like Cobein).
We'll represent a way to play a song as a sequence of pairs (xi, yi) (1 β€ i β€ s), such that the xi-th string on the yi-th fret produces the i-th note from the song. The complexity of moving between pairs (x1, y1) and (x2, y2) equals <image> + <image>. The complexity of a way to play a song is the maximum of complexities of moving between adjacent pairs.
Help Dima determine the maximum complexity of the way to play his song! The guy's gotta look cool!
Input
The first line of the input contains four integers n, m, k and s (1 β€ n, m β€ 2000, 1 β€ k β€ 9, 2 β€ s β€ 105).
Then follow n lines, each containing m integers aij (1 β€ aij β€ k). The number in the i-th row and the j-th column (aij) means a note that the guitar produces on the i-th string and the j-th fret.
The last line of the input contains s integers qi (1 β€ qi β€ k) β the sequence of notes of the song.
Output
In a single line print a single number β the maximum possible complexity of the song.
Examples
Input
4 6 5 7
3 1 2 2 3 1
3 2 2 2 5 5
4 2 2 2 5 3
3 2 2 1 4 3
2 3 1 4 1 5 1
Output
8
Input
4 4 9 5
4 7 9 5
1 2 1 7
8 3 4 9
5 7 7 2
7 1 9 2 5
Output
4
Submitted Solution:
```
def solution() : # ζ倧ηθ·η¦»ζ₯θͺδΊθ§θ½ιθΏηηΉ
n,m,k,s = map(int, input().split())
dis = lambda a,b : abs(a[0] - b[0]) + abs(a[1] - b[1])
corner = [(0,0), (0,m-1), (n-1,0), (n-1,m-1)]
vertex = [[(n,m), (n,-1), (-1,m), (-1,-1)] for _ in range(k+1)]
for i in range(n) :
for j,note in enumerate(map(int, input().split())) :
vertex[note] = [
(i,j) if dis((i,j), c) < dis(v, c) else v
for v,c in zip(vertex[note], corner)]
maxdis = [[0] * (k+1) for _ in range(k+1)]
pairs = [(0,3),(3,0),(1,2),(2,1)]
for i in range(1, k+1) :
for j in range(i+1, k+1) :
vi,vj = vertex[i],vertex[j]
maxdis[i][j] = max(dis(vi[a], vj[b]) for a,b in pairs)
maxdis[j][i] = maxdis[i][j]
s = list(map(int, input().split()))
print(max(maxdis[s[i]][s[i+1]] for i in range(len(s) - 1)))
solution()
```
No
| 104,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Tags: implementation
Correct Solution:
```
n = int(input())
l = []
m = []
for i in range(n):
X1, Y1, X2, Y2 = map(int, input().split())
m1 = (abs(X2 - X1)) ** 3
l.append([X1, Y1, X2, Y2])
m.append(m1)
X1, Y1, X2, Y2 = l[0][0], l[0][1], l[0][2], l[0][3]
if X1 > X2: X2, X1 = X1, X2
if Y1 > Y2: Y1, Y2 = Y2, Y1
ans = n
for i in range(1, n):
for j in range(i, 0, -1):
x1, y1, x2, y2 = l[j - 1][0], l[j - 1][1], l[j - 1][2], l[j - 1][3]
if x1 > x2: x2, x1 = x1, x2
if y1 > y2: y1, y2 = y2, y1
numx, numy, den = 0, 0 , 0
for k in range(i, j - 1, -1):
a = (l[k][0] + l[k][2]) / 2
b = (l[k][1] + l[k][3]) / 2
m = (abs(l[k][2] - l[k][0])) ** 3
numx += m * a
numy += m * b
den += m
xcm = numx / den
ycm = numy / den
if x1 <= xcm <= x2 and y1 <= ycm <= y2: pass
else:
ans = i
break
if ans != n: break
print(ans)
```
| 104,796 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Tags: implementation
Correct Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][3]),max(a[j][0],a[j][2]),max(a[j][1],a[j][3])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*((abs(a[j][0]-a[j][2])**3)/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][3])/2-c[1])*((abs(a[j][0]-a[j][2])**3)/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
```
| 104,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
k = 1
ob = r = 0
for i in range(n-1):
b = list(map(int,input().split()))
if b[1]==a[1]:
if a[0]<b[0] and a[2]<b[2]:
if a[2]-b[0]>=b[2]-a[2]:
k+=1
else: break
else:
if b[2]-a[0]>=a[0]-b[0]:
k+=1
else:
break
else:
if a[1]<b[1] and a[3]<b[3]:
if a[3]-b[1]>=b[3]-a[3]:
k+=1
else: break
else:
if b[3]-a[1]>=a[1]-b[1]:
k+=1
else:
break
a = b
print(k)
```
No
| 104,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once Vasya played bricks. All the bricks in the set had regular cubical shape. Vasya vas a talented architect, however the tower he built kept falling apart.
Let us consider the building process. Vasya takes a brick and puts it on top of the already built tower so that the sides of the brick are parallel to the sides of the bricks he has already used. Let's introduce a Cartesian coordinate system on the horizontal plane, where Vasya puts the first brick. Then the projection of brick number i on the plane is a square with sides parallel to the axes of coordinates with opposite corners in points (xi, 1, yi, 1) and (xi, 2, yi, 2). The bricks are cast from homogeneous plastic and the weight of a brick a Γ a Γ a is a3 grams.
It is guaranteed that Vasya puts any brick except the first one on the previous one, that is the area of intersection of the upper side of the previous brick and the lower side of the next brick is always positive.
We (Vasya included) live in a normal world where the laws of physical statics work. And that is why, perhaps, if we put yet another brick, the tower will collapse under its own weight. Vasya puts the cubes consecutively one on top of the other until at least one cube loses the balance and falls down. If it happens, Vasya gets upset and stops the construction. Print the number of bricks in the maximal stable tower, that is the maximal number m satisfying the condition that all the towers consisting of bricks 1, 2, ..., k for every integer k from 1 to m remain stable.
Input
The first input file contains an integer n (1 β€ n β€ 100) which is the number of bricks. Each of the next n lines contains four numbers xi, 1, yi, 1, xi, 2, yi, 2 (xi, 1 β xi, 2, |xi, 1 - xi, 2| = |yi, 1 - yi, 2|) which are the coordinates of the opposite angles of the base of the brick number i. The coordinates are integers and their absolute value does not exceed 50.
The cubes are given in the order Vasya puts them. It is guaranteed that the area of intersection of the upper side of the brick number i - 1 and the lower side of the brick number i is strictly strictly greater than zero for all i β₯ 2.
Output
Print the number of bricks in the maximal stable tower.
Examples
Input
2
0 0 3 3
1 0 4 3
Output
2
Input
2
0 0 3 3
2 0 5 3
Output
1
Input
3
0 0 3 3
1 0 4 3
2 0 5 3
Output
3
Submitted Solution:
```
a=[list(map(int,input().split())) for i in range(int(input()))]
n=len(a)
z=n
for i in range(n):
if i!=0:
c=[(a[i][0]+a[i][2])/2,(a[i][1]+a[i][3])/2]
m=abs(a[i][0]-a[i][2])**3
for j in range(i-1,-1,-1):
x1,y1,x2,y2=min(a[j][0],a[j][2]),min(a[j][1],a[j][3]),max(a[j][0],a[j][2]),max(a[j][1],a[j][3])
if not(x1<=c[0]<=x2 and y1<=c[1]<=y2):
z=i
break
x=c[0]+((a[j][0]+a[j][2])/2-c[0])*(m/(m+abs(a[j][0]-a[j][2])**3))
y=c[1]+((a[j][1]+a[j][3])/2-c[1])*(m/(m+abs(a[j][0]-a[j][2])**3))
c=[x,y]
m+=abs(a[j][0]-a[j][2])**3
if z!=n:
break
print(z)
```
No
| 104,799 |
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