message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,620 | 14 | 9,240 |
Tags: combinatorics, math
Correct Solution:
```
a=int(input())
print(4**(a-3)*3*(3*a-1))
``` | output | 1 | 4,620 | 14 | 9,241 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,621 | 14 | 9,242 |
Tags: combinatorics, math
Correct Solution:
```
n = int(input())
N = 2*n-2
ans = 0;
for i in range(n):
l_cnt = i
r_cnt = N - i - n
if r_cnt < 0: break
m = 4
p = int(l_cnt > 0) + int(r_cnt > 0)
m *= 3 ** p
l_cnt = max(0, l_cnt - 1)
r_cnt = max(0, r_cnt - 1)
p = l_cnt + r_cnt;
m *= 4 ** p;
ans += m
print(ans)
``` | output | 1 | 4,621 | 14 | 9,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,622 | 14 | 9,244 |
Tags: combinatorics, math
Correct Solution:
```
# Принимаем входные данные
n = int(input())
# Вычисляем ответ
ways_count = max(((n - 3) * 36 * 4**(n - 4)), (n - 3) * 36) # Когда машины стоят не по краям
ways_count += 2 * (3 * 4**(n-3)) * 4 # Когда машины стоят по краям
# Выводим результат
print(int(ways_count))
``` | output | 1 | 4,622 | 14 | 9,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,623 | 14 | 9,246 |
Tags: combinatorics, math
Correct Solution:
```
n = int(input())
print(int(4*(n-3)*9*4**(n-4) + 2*4*3*4**(n-3)))
``` | output | 1 | 4,623 | 14 | 9,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,624 | 14 | 9,248 |
Tags: combinatorics, math
Correct Solution:
```
n = int(input())
print(6 * 4**(n-2) + 9 * (n-3) * 4**(n-3))
``` | output | 1 | 4,624 | 14 | 9,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,625 | 14 | 9,250 |
Tags: combinatorics, math
Correct Solution:
```
n = int(input())
res = 0
res += 8 * (3 * pow(4, n - 3))
res += ((n - 3) * 4 * 9 * int(pow(4, n - 4)))
print(res)
``` | output | 1 | 4,625 | 14 | 9,251 |
Provide tags and a correct Python 3 solution for this coding contest problem.
To quickly hire highly skilled specialists one of the new IT City companies made an unprecedented move. Every employee was granted a car, and an employee can choose one of four different car makes.
The parking lot before the office consists of one line of (2n - 2) parking spaces. Unfortunately the total number of cars is greater than the parking lot capacity. Furthermore even amount of cars of each make is greater than the amount of parking spaces! That's why there are no free spaces on the parking lot ever.
Looking on the straight line of cars the company CEO thought that parking lot would be more beautiful if it contained exactly n successive cars of the same make. Help the CEO determine the number of ways to fill the parking lot this way.
Input
The only line of the input contains one integer n (3 ≤ n ≤ 30) — the amount of successive cars of the same make.
Output
Output one integer — the number of ways to fill the parking lot by cars of four makes using the described way.
Examples
Input
3
Output
24
Note
Let's denote car makes in the following way: A — Aston Martin, B — Bentley, M — Mercedes-Maybach, Z — Zaporozhets. For n = 3 there are the following appropriate ways to fill the parking lot: AAAB AAAM AAAZ ABBB AMMM AZZZ BBBA BBBM BBBZ BAAA BMMM BZZZ MMMA MMMB MMMZ MAAA MBBB MZZZ ZZZA ZZZB ZZZM ZAAA ZBBB ZMMM
Originally it was planned to grant sport cars of Ferrari, Lamborghini, Maserati and Bugatti makes but this idea was renounced because it is impossible to drive these cars having small road clearance on the worn-down roads of IT City. | instruction | 0 | 4,626 | 14 | 9,252 |
Tags: combinatorics, math
Correct Solution:
```
import sys,math
inputs = iter(sys.stdin.readlines())
n = int(next(inputs))
print(int(2*4*3*pow(4,n-3) + (n-3)*4*3**2*pow(4,n-4)))
``` | output | 1 | 4,626 | 14 | 9,253 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,667 | 14 | 9,334 |
Tags: implementation, math
Correct Solution:
```
n, m, k = map(int, input().split())
L = int(k/(m+m))
if k%(m+m)!=0:L = L + 1
D = (L-1)*(m+m) + 1
if k%2==0:D = D + 1
D = int((k-D)/2) + 1
st = "L"
if k%2==0:st = "R"
print(L, D, st)
``` | output | 1 | 4,667 | 14 | 9,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,668 | 14 | 9,336 |
Tags: implementation, math
Correct Solution:
```
import math
n,m,k=map(int,input().split())
l=math.ceil(k/(2*m))
se=math.ceil((k-((l-1)*2*m))/2)
print(l,se,end=' ')
if k%2==0:print('R')
else:print('L')
``` | output | 1 | 4,668 | 14 | 9,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,669 | 14 | 9,338 |
Tags: implementation, math
Correct Solution:
```
n,m,k = map(int,input().split())
a = []
p = 1
for i in range(m): a += [[p,p+1]]; p += 2
p -= 1
x = -(-k//p)
if k % 2 == 0: z = "R"
else: z = "L"
if k % p == 0: y = m
else:
for i in range(m):
if k%p in a[i]: y = i+1; break
print(x,y,z)
``` | output | 1 | 4,669 | 14 | 9,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,670 | 14 | 9,340 |
Tags: implementation, math
Correct Solution:
```
# Santa Claus and a Place in a Class
def santa(n, m, k):
direction = 'R'
if k & 1 == 1:
direction = 'L'
lane = 2
while True:
if k <= (m * lane):
prev = (m * (lane - 2)) + 1
desk = 1
while True:
if k == prev or k == prev + 1:
break
desk += 1
prev += 2
break
lane += 2
lane //= 2
print(lane, desk, direction)
n, m, k = list(map(int, input().split()))
santa(n, m, k)
``` | output | 1 | 4,670 | 14 | 9,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,671 | 14 | 9,342 |
Tags: implementation, math
Correct Solution:
```
n,m,k=map(int,input().split())
a=k%2
x={1:"L",0:"R"}
r=k//(2*m)
if(k%(2*m)!=0):
r+=1
d=(k-(m*(r-1)*2))
if(d%2!=0):
d+=1
d=d//2
print(r,d,x[a])
``` | output | 1 | 4,671 | 14 | 9,343 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,672 | 14 | 9,344 |
Tags: implementation, math
Correct Solution:
```
n, m, k = map(int, input().split())
z = (k + 2 * m - 1) // (2 * m)
q = k % (2 * m)
if q == 0:
q = 2 * m
q = (q + 1) // 2
w = q % 2
if k % 2:
w = 'L'
else:
w = 'R'
print(z, q, w)
``` | output | 1 | 4,672 | 14 | 9,345 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,673 | 14 | 9,346 |
Tags: implementation, math
Correct Solution:
```
from math import ceil
def solve(n, m, k):
t = ceil(k / 2)
x = m if t % m == 0 else t % m
y = ceil(t / m)
p = 'L' if k % 2 == 1 else 'R'
return f'{y} {x} {p}'
def main():
n, m, k = list(map(int, input().split()))
print(solve(n, m, k))
main()
``` | output | 1 | 4,673 | 14 | 9,347 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Santa Claus is the first who came to the Christmas Olympiad, and he is going to be the first to take his place at a desk! In the classroom there are n lanes of m desks each, and there are two working places at each of the desks. The lanes are numbered from 1 to n from the left to the right, the desks in a lane are numbered from 1 to m starting from the blackboard. Note that the lanes go perpendicularly to the blackboard, not along it (see picture).
The organizers numbered all the working places from 1 to 2nm. The places are numbered by lanes (i. e. all the places of the first lane go first, then all the places of the second lane, and so on), in a lane the places are numbered starting from the nearest to the blackboard (i. e. from the first desk in the lane), at each desk, the place on the left is numbered before the place on the right.
<image> The picture illustrates the first and the second samples.
Santa Clause knows that his place has number k. Help him to determine at which lane at which desk he should sit, and whether his place is on the left or on the right!
Input
The only line contains three integers n, m and k (1 ≤ n, m ≤ 10 000, 1 ≤ k ≤ 2nm) — the number of lanes, the number of desks in each lane and the number of Santa Claus' place.
Output
Print two integers: the number of lane r, the number of desk d, and a character s, which stands for the side of the desk Santa Claus. The character s should be "L", if Santa Clause should sit on the left, and "R" if his place is on the right.
Examples
Input
4 3 9
Output
2 2 L
Input
4 3 24
Output
4 3 R
Input
2 4 4
Output
1 2 R
Note
The first and the second samples are shown on the picture. The green place corresponds to Santa Claus' place in the first example, the blue place corresponds to Santa Claus' place in the second example.
In the third sample there are two lanes with four desks in each, and Santa Claus has the fourth place. Thus, his place is in the first lane at the second desk on the right. | instruction | 0 | 4,674 | 14 | 9,348 |
Tags: implementation, math
Correct Solution:
```
# A. Santa Claus and a Place in a Class
n, m, k = map(int, input().split())
r = k // (2 * m)
if k % (2 * m) != 0:
r += 1
d = (k - (r - 1) * 2 * m) // 2
if (k - (r - 1) * 2 * m) % 2 != 0:
d += 1
s = "L" if k % 2 != 0 else "R"
print(r, d, s)
``` | output | 1 | 4,674 | 14 | 9,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the final part of fashion show all models come to the stage and stay in one row and fashion designer stays to right to model on the right. During the rehearsal, Izabella noticed, that row isn't nice, but she can't figure out how to fix it.
Like many other creative people, Izabella has a specific sense of beauty. Evaluating beauty of row of models Izabella looks at heights of models. She thinks that row is nice if for each model distance to nearest model with less height (model or fashion designer) to the right of her doesn't exceed k (distance between adjacent people equals 1, the distance between people with exactly one man between them equals 2, etc).
She wants to make row nice, but fashion designer has his own sense of beauty, so she can at most one time select two models from the row and swap their positions if the left model from this pair is higher than the right model from this pair.
Fashion designer (man to the right of rightmost model) has less height than all models and can't be selected for exchange.
You should tell if it's possible to make at most one exchange in such a way that row becomes nice for Izabella.
Input
In first line there are two integers n and k (1 ≤ n ≤ 5·105, 1 ≤ k ≤ n) — number of models and required distance.
Second line contains n space-separated integers ai (1 ≤ ai ≤ 109) — height of each model. Pay attention that height of fashion designer is not given and can be less than 1.
Output
Print «YES» (without quotes) if it's possible to make row nice using at most one exchange, and «NO» (without quotes) otherwise.
Examples
Input
5 4
2 3 5 2 5
Output
NO
Input
5 2
3 6 2 2 1
Output
YES
Input
5 2
5 3 6 5 2
Output
YES
Submitted Solution:
```
temp, k = [int(i) for i in input().split()]
models = [int(i) for i in input().split()]
min = models[0]
steps = 1
index = 0
index2 = 0
steps2 = 1
problems = 0
changed = False
for i in range(temp):
if models[i] < min:
min = models[i]
index = i
steps = 1
else:
steps += 1
if steps > k:
problems += 1
min2 = min
if not (temp - i - k > 0):
newk= temp - i
else:
newk = k
for j in range(newk):
if models[i+j] < min2:
if changed:
changed = False
break
else:
min2 = models[i+j]
changed = True
if not changed:
break
else:
problems -= 1
min = models[i]
steps = 1
if problems:
print('NO')
else:
print('YES')
``` | instruction | 0 | 4,747 | 14 | 9,494 |
No | output | 1 | 4,747 | 14 | 9,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the final part of fashion show all models come to the stage and stay in one row and fashion designer stays to right to model on the right. During the rehearsal, Izabella noticed, that row isn't nice, but she can't figure out how to fix it.
Like many other creative people, Izabella has a specific sense of beauty. Evaluating beauty of row of models Izabella looks at heights of models. She thinks that row is nice if for each model distance to nearest model with less height (model or fashion designer) to the right of her doesn't exceed k (distance between adjacent people equals 1, the distance between people with exactly one man between them equals 2, etc).
She wants to make row nice, but fashion designer has his own sense of beauty, so she can at most one time select two models from the row and swap their positions if the left model from this pair is higher than the right model from this pair.
Fashion designer (man to the right of rightmost model) has less height than all models and can't be selected for exchange.
You should tell if it's possible to make at most one exchange in such a way that row becomes nice for Izabella.
Input
In first line there are two integers n and k (1 ≤ n ≤ 5·105, 1 ≤ k ≤ n) — number of models and required distance.
Second line contains n space-separated integers ai (1 ≤ ai ≤ 109) — height of each model. Pay attention that height of fashion designer is not given and can be less than 1.
Output
Print «YES» (without quotes) if it's possible to make row nice using at most one exchange, and «NO» (without quotes) otherwise.
Examples
Input
5 4
2 3 5 2 5
Output
NO
Input
5 2
3 6 2 2 1
Output
YES
Input
5 2
5 3 6 5 2
Output
YES
Submitted Solution:
```
n,k=list(map(int,input().split()))
number=list(map(int,input().split()))
count=0
flag=0
for i in range(n-1):
j=i
while(j<n and number[j]>=number[i]):
j=j+1
if j<n and count==0 and j-i>k:
temp=number[j]
number[j]=number[i]
number[i]=temp
count=1
elif j==n and j-i>k:
flag=1
print('NO')
break
elif j-i>k and count==1:
flag=1
print('NO')
break
if flag==0:
print('YES')
``` | instruction | 0 | 4,748 | 14 | 9,496 |
No | output | 1 | 4,748 | 14 | 9,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the final part of fashion show all models come to the stage and stay in one row and fashion designer stays to right to model on the right. During the rehearsal, Izabella noticed, that row isn't nice, but she can't figure out how to fix it.
Like many other creative people, Izabella has a specific sense of beauty. Evaluating beauty of row of models Izabella looks at heights of models. She thinks that row is nice if for each model distance to nearest model with less height (model or fashion designer) to the right of her doesn't exceed k (distance between adjacent people equals 1, the distance between people with exactly one man between them equals 2, etc).
She wants to make row nice, but fashion designer has his own sense of beauty, so she can at most one time select two models from the row and swap their positions if the left model from this pair is higher than the right model from this pair.
Fashion designer (man to the right of rightmost model) has less height than all models and can't be selected for exchange.
You should tell if it's possible to make at most one exchange in such a way that row becomes nice for Izabella.
Input
In first line there are two integers n and k (1 ≤ n ≤ 5·105, 1 ≤ k ≤ n) — number of models and required distance.
Second line contains n space-separated integers ai (1 ≤ ai ≤ 109) — height of each model. Pay attention that height of fashion designer is not given and can be less than 1.
Output
Print «YES» (without quotes) if it's possible to make row nice using at most one exchange, and «NO» (without quotes) otherwise.
Examples
Input
5 4
2 3 5 2 5
Output
NO
Input
5 2
3 6 2 2 1
Output
YES
Input
5 2
5 3 6 5 2
Output
YES
Submitted Solution:
```
n,k=list(map(int,input().split()))
number=list(map(int,input().split()))
count=0
flag=0
for i in range(n):
j=i
while(j<n and number[j]>=number[i]):
j=j+1
if j<n-1 and count==0 and j-i>k:
temp=number[j]
number[j]=number[i]
number[i]=temp
count=1
elif j==n-1:
flag=1
print('NO')
break
elif j-i>k and count==1:
flag=1
print('NO')
break
if flag==0:
print('YES')
``` | instruction | 0 | 4,749 | 14 | 9,498 |
No | output | 1 | 4,749 | 14 | 9,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
During the final part of fashion show all models come to the stage and stay in one row and fashion designer stays to right to model on the right. During the rehearsal, Izabella noticed, that row isn't nice, but she can't figure out how to fix it.
Like many other creative people, Izabella has a specific sense of beauty. Evaluating beauty of row of models Izabella looks at heights of models. She thinks that row is nice if for each model distance to nearest model with less height (model or fashion designer) to the right of her doesn't exceed k (distance between adjacent people equals 1, the distance between people with exactly one man between them equals 2, etc).
She wants to make row nice, but fashion designer has his own sense of beauty, so she can at most one time select two models from the row and swap their positions if the left model from this pair is higher than the right model from this pair.
Fashion designer (man to the right of rightmost model) has less height than all models and can't be selected for exchange.
You should tell if it's possible to make at most one exchange in such a way that row becomes nice for Izabella.
Input
In first line there are two integers n and k (1 ≤ n ≤ 5·105, 1 ≤ k ≤ n) — number of models and required distance.
Second line contains n space-separated integers ai (1 ≤ ai ≤ 109) — height of each model. Pay attention that height of fashion designer is not given and can be less than 1.
Output
Print «YES» (without quotes) if it's possible to make row nice using at most one exchange, and «NO» (without quotes) otherwise.
Examples
Input
5 4
2 3 5 2 5
Output
NO
Input
5 2
3 6 2 2 1
Output
YES
Input
5 2
5 3 6 5 2
Output
YES
Submitted Solution:
```
temp, k = [int(i) for i in input().split()]
models = [int(i) for i in input().split()]
min = models[0]
steps = 0
index = 0
index2 = 0
steps2 = 0
problems = 0
changed = False
for i in range(temp):
if models[i] < min:
min = models[i]
index = i
steps = 0
else:
steps += 1
if steps > k:
problems += 1
min2 = min
if temp - i - k > 0:
for j in range(k):
if models[i+j] < min2:
if changed:
changed = False
break
else:
min2 = models[i+j]
changed = True
if not changed:
break
else:
problems -= 1
min = models[i]
steps = 0
if problems:
print('NO')
else:
print('YES')
``` | instruction | 0 | 4,750 | 14 | 9,500 |
No | output | 1 | 4,750 | 14 | 9,501 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors. | instruction | 0 | 5,211 | 14 | 10,422 |
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
``` | output | 1 | 5,211 | 14 | 10,423 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors. | instruction | 0 | 5,212 | 14 | 10,424 |
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
``` | output | 1 | 5,212 | 14 | 10,425 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors. | instruction | 0 | 5,213 | 14 | 10,426 |
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = MI()
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
#print(n, k, u, ans)
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
#print(n, k, u, ans)
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
``` | output | 1 | 5,213 | 14 | 10,427 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors. | instruction | 0 | 5,214 | 14 | 10,428 |
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
def ng():print("NO");exit()
n, k = map(int,input().split())
if (n,k)==(1,0):print("YES");print(0);exit()
ans = [0] * n;u = 0;popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1;u = 4
for _ in range(k - 1):ans[u + 1] = ans[u + 2] = u+1;u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES");print(*ans)
``` | output | 1 | 5,214 | 14 | 10,429 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors. | instruction | 0 | 5,215 | 14 | 10,430 |
Tags: constructive algorithms, divide and conquer, dp, math, trees
Correct Solution:
```
def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3 or k==0 or n<11: ng()
ans[4] = 1
u = 4
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
``` | output | 1 | 5,215 | 14 | 10,431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
def ng():
print("NO")
exit()
n, k = MI()
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k -= 1
if n - 4 < 2 * k + 3: ng()
ans[4] = 1
u = 4
#print(n, k, u, ans)
for _ in range(k - 1):
ans[u + 1] = ans[u + 2] = u+1
u += 2
#print(n, k, u, ans)
for v in range(1,n-u):ans[v+u]=(v-1)//2+u+1
print("YES")
print(*ans)
``` | instruction | 0 | 5,216 | 14 | 10,432 |
No | output | 1 | 5,216 | 14 | 10,433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
n,k=MI()
if n&1==0 or n<2*k+3 or k<bin(n+1).count("1")-1:
print("NO")
exit()
hp=[]
two=0
bit=n+1
for i in range(n):
if bit&1:
if i==1:two+=1
else:heappush(hp,1<<i)
bit>>=1
#print(two,hp)
while two+len(hp)-1<k:
s=heappop(hp)
if s==4:two+=2
else:
heappush(hp,s//2)
heappush(hp,s//2)
#print(two,hp)
ans=[0]*n
for u in range(0,two*2,2):ans[u+1]=ans[u+2]=u+1
#print(ans)
u=2*two
while hp:
s=heappop(hp)
if not hp:
for v in range(2, s): ans[v + u-1] = v // 2 + u
break
for v in range(1,s):ans[v+u]=v//2+u+1
ans[u+s]=u+1
u+=s
print("YES")
print(*ans)
``` | instruction | 0 | 5,217 | 14 | 10,434 |
No | output | 1 | 5,217 | 14 | 10,435 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
import math
n, k = input().replace("\n", "").split(" ")
n = int(n)
k = int(k)
if 2*k > n-3:
print("NO")
else:
if((n-2*k)%4 == 1):
print("NO")
else:
s = {}
print("YES")
for i in range(1, n-2*k+1):
s[i] = math.floor(i/2)
if k != 0:
s[1] = n - 2*k + 1
s[n - 2*k + 2] = n - 2*k + 1
for i in range(1, k+1):
if i == 1:
s[n-2*k+i] = 0
else:
s[n-2*k+i] = n- 2*k + i - 1
s[n-2*k+i+k] = n - 2*k + i
for x in s:
print(s[x], end = " ")
``` | instruction | 0 | 5,218 | 14 | 10,436 |
No | output | 1 | 5,218 | 14 | 10,437 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in this structure are conveniently numbered from 1 to n, and s_i denotes the child of the person i (and s_i = 0 for exactly one person who does not have any children).
We say that a is an ancestor of b if either a = b, or a has a child, who is an ancestor of b. That is a is an ancestor for a, s_a, s_{s_a}, etc.
We say that person i is imbalanced in case this person has both parents specified, and the total number of ancestors of one of the parents is at least double the other.
Ivan counted the number of imbalanced people in the structure, and got k people in total. However, he is not sure whether he computed it correctly, and would like to check if there is at least one construction with n people that have k imbalanced people in total. Please help him to find one such construction, or determine if it does not exist.
Input
The input contains two integers n and k (1 ≤ n ≤ 100 000, 0 ≤ k ≤ n), the total number of people and the number of imbalanced people.
Output
If there are no constructions with n people and k imbalanced people, output NO.
Otherwise output YES on the first line, and then n integers s_1, s_2, …, s_n (0 ≤ s_i ≤ n), which describes the construction and specify the child of each node (or 0, if the person does not have any children).
Examples
Input
3 0
Output
YES
0 1 1
Input
5 1
Output
YES
0 1 1 3 3
Input
3 2
Output
NO
Note
In the first example case one can have a construction with 3 people, where 1 person has 2 parents.
In the second example case one can use the following construction:
<image>
Only person 1 is imbalanced, because one of their parents has 1 ancestor in total, and the other parent has 3 ancestors.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 6)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
n,k=MI()
if n&1==0 or n<2*k+3:
print("NO")
exit()
ans=[0]*n
def tree(u,size,imb):
#print(u,size,imb)
if size==1:return
ans[u + 1] = u + 1
if imb:
ans[u+2]=u+1
tree(u+2,size-2,imb-1)
else:
sl=size//4*2+1
ans[u+1+sl]=u+1
tree(u+1,sl,0)
tree(u+1+sl,size-1-sl,0)
tree(0,n,k)
print("YES")
print(*ans)
``` | instruction | 0 | 5,219 | 14 | 10,438 |
No | output | 1 | 5,219 | 14 | 10,439 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6 | instruction | 0 | 5,301 | 14 | 10,602 |
Tags: brute force, dfs and similar, implementation
Correct Solution:
```
#!/usr/local/bin/python3
from __future__ import print_function
import sys
DEBUG = '-d' in sys.argv
def debug(*args, **kwargs):
if DEBUG:
print(*args, file=sys.stderr, **kwargs)
return None
def main():
n = int(input())
cnt = [0] * (n + 1)
edge = []
for i in range(0, n + 1):
edge.append(set())
for i in range(0, 2 * n):
s, t = map(int, input().split())
edge[s].add(t)
edge[t].add(s)
cnt[s] += 1
cnt[t] += 1
c4 = 0
for i in range(1, n + 1):
if cnt[i] == 4:
c4 += 1
if c4 != n:
print(-1)
else:
for v2 in edge[1]:
for v3 in edge[1]:
if v2 in edge[v3]:
mark = [True] * (n + 1)
mark[1] = False
mark[v2] = False
res = [1, v2]
i = v3
try:
while True:
res.append(i)
mark[i] = False
if len(res) == n:
print(' '.join([str(x) for x in res]))
sys.exit(0)
for e in edge[i]:
if e != i and mark[e] and res[-2] in edge[e]:
i = e
break
if not mark[i]:
raise StopIteration
except StopIteration:
pass
print(-1)
if __name__ == '__main__':
main()
``` | output | 1 | 5,301 | 14 | 10,603 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.
For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).
Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.
Input
The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.
It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.
Output
If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n.
If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.
Examples
Input
5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2
Output
1 2 3 4 5
Input
6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5
Output
1 2 4 5 3 6 | instruction | 0 | 5,302 | 14 | 10,604 |
Tags: brute force, dfs and similar, implementation
Correct Solution:
```
from sys import stdin
all_in = stdin.readlines()
n = int(all_in[0])
pairs = list(map(lambda x: tuple(map(int, x.split())), all_in[1:]))
if n == 5:
print(1, 2, 3, 4, 5)
exit()
neigs = {i: set() for i in range(1, n + 1)}
for (a, b) in pairs:
neigs[a].add(b)
neigs[b].add(a)
for el in neigs.values():
if len(el) != 4:
print(-1)
exit()
ans = [1]
used = {i: False for i in range(1, n + 1)}
used[1] = True
for i in range(n - 1):
el_ = ans[-1]
ne = neigs[el_]
for el in ne:
ne_ = neigs[el]
and_ = ne & ne_
if len(and_) == 2:
if i:
if ans[-2] not in and_:
continue
if not used[el]:
ans.append(el)
used[el] = True
break
if len(ans) < n:
print(-1)
exit()
print(' '.join(map(str, ans)))
``` | output | 1 | 5,302 | 14 | 10,605 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,391 | 14 | 10,782 |
Tags: binary search, implementation
Correct Solution:
```
import sys
import math
a=int(input())
b=list(map(int,input().split()))
c=int(input())
d=list(map(int,input().split()))
e=[]
f=0
for i in b:
#print(f,i)
e.append(f+i)
f=i+f
g=[0]*(e[-1]+1)
h=0
for i in range(a):
for j in range(h,e[i]+1):
g[j]=i+1
h=e[i]+1
for k in d:
print(g[k])
``` | output | 1 | 5,391 | 14 | 10,783 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,392 | 14 | 10,784 |
Tags: binary search, implementation
Correct Solution:
```
a=int(input())
b= list(map(int,input().split()))
c=int(input())
d= list(map(int,input().split()))
for k in range(1,a):
b[k] += b[k-1]
for j in d:
l=0
r= a-1
while l<= r:
if j <= b[0]:
mid=0
break
mid= (l+r)//2
if b[mid] >= j and b[mid-1] < j:
break
elif b[mid] > j:
r=mid-1
else:
l=mid+1
print(mid+1)
``` | output | 1 | 5,392 | 14 | 10,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,393 | 14 | 10,786 |
Tags: binary search, implementation
Correct Solution:
```
def read():
input()
sizes = list(map(int, input().split()))
input()
queries = list(map(int, input().split()))
return sizes, queries
def calc_indexes(sizes):
start_indexes = [1]
for i in range(len(sizes)):
start_indexes.append(start_indexes[i] + sizes[i])
return start_indexes
def bin_search(ar, query):
begin = 0
end = len(ar)
while (end - begin) > 1:
middle = (end + begin) // 2
if query < ar[middle]:
end = middle
else:
begin = middle
return begin
# for i in range(begin, end):
# if ar[i] > query:
# return i - 1
# return end - 1
sizes, queries = read()
start_indexes = calc_indexes(sizes)
for q in queries:
print (bin_search(start_indexes, q) + 1)
``` | output | 1 | 5,393 | 14 | 10,787 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,394 | 14 | 10,788 |
Tags: binary search, implementation
Correct Solution:
```
import bisect as bs
n = int(input())
A = list(map(int, input().split()))
m = int(input())
B = list(map(int, input().split()))
S = [0]
for a in A:
S.append(S[-1] + a)
for b in B:
print(bs.bisect_left(S, b))
``` | output | 1 | 5,394 | 14 | 10,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,395 | 14 | 10,790 |
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
m=int(input())
q=list(map(int,input().split()))
d=[0]
for i in range(n):
d.append(d[i]+a[i])
d.append(1000001)
for i in range(m):
start=0;end=len(d)-1
while start<end:
k=(start+end)//2
if q[i]<d[k]:
end=k-1
else:
start=k+1
for t in range(max(0,end-4),n+1):
if q[i]<=d[t]:
print(t)
break
else:
print(n)
``` | output | 1 | 5,395 | 14 | 10,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,396 | 14 | 10,792 |
Tags: binary search, implementation
Correct Solution:
```
input()
d=[]
m=1
for i in map(int,input().split()):
for j in range(i):
d.append(m)
m+=1
input()
for i in map(int,input().split()):
print(d[i-1])
``` | output | 1 | 5,396 | 14 | 10,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,397 | 14 | 10,794 |
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
x=list(map(int,input().split()))
t=int(input())
y=list(map(int,input().split()))
ans=[]
for i in range(len(x)):
ans.extend([i+1 for j in range(x[i])])
for i in y:
print(ans[i-1])
``` | output | 1 | 5,397 | 14 | 10,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
Input
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Output
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
Examples
Input
5
2 7 3 4 9
3
1 25 11
Output
1
5
3
Note
For the sample input:
* The worms with labels from [1, 2] are in the first pile.
* The worms with labels from [3, 9] are in the second pile.
* The worms with labels from [10, 12] are in the third pile.
* The worms with labels from [13, 16] are in the fourth pile.
* The worms with labels from [17, 25] are in the fifth pile. | instruction | 0 | 5,398 | 14 | 10,796 |
Tags: binary search, implementation
Correct Solution:
```
n=int(input())
s=list(map(int,input().split()))
n1=int(input())
s1=list(map(int,input().split()))
w=[]
i=-1
d=n-1
r=0
while i!=d:
i+=1
for j in range(s[i]):
r+=1
w.append(r)
ans=[]
for i in range(n1):
l,r=-1,n
while (l+1<r):
md=(l+r)//2
if s1[i]<=w[md]:
r=md
else:
l=md
ans.append(r+1)
print(*ans,sep='\n')
``` | output | 1 | 5,398 | 14 | 10,797 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1 | instruction | 0 | 5,428 | 14 | 10,856 |
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n): ans[nse[i] - pse[i] - 2] = max(lst[i], ans[nse[i] - pse[i] - 2])
for i in range(n-2, -1, -1): ans[i] = max(ans[i+1], ans[i])
print(' '.join(map(str, ans)))
``` | output | 1 | 5,428 | 14 | 10,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1 | instruction | 0 | 5,429 | 14 | 10,858 |
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
from sys import stdin, stdout
def input(): return stdin.readline().strip()
def print(x, end='\n'): stdout.write(str(x) + end)
n, lst = int(input()), list(map(int, input().split()))
nse, pse, stk, ans = [n for i in range(n)], [-1 for i in range(n)], [], [0 for i in range(n+1)]
for i in range(n):
while stk and lst[stk[-1]] > lst[i]: nse[stk.pop()] = i
stk.append(i)
stk.clear()
for i in range(n-1, -1, -1):
while stk and lst[stk[-1]] > lst[i]: pse[stk.pop()] = i
stk.append(i)
for i in range(n): ans[nse[i] - pse[i] - 1] = max(lst[i], ans[nse[i] - pse[i] - 1])
mnow = ans[n]
for i in range(n, -1, -1):
mnow = max(mnow, ans[i])
ans[i] = mnow
print(' '.join(map(str, ans[1:])))
``` | output | 1 | 5,429 | 14 | 10,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mike is the president of country What-The-Fatherland. There are n bears living in this country besides Mike. All of them are standing in a line and they are numbered from 1 to n from left to right. i-th bear is exactly ai feet high.
<image>
A group of bears is a non-empty contiguous segment of the line. The size of a group is the number of bears in that group. The strength of a group is the minimum height of the bear in that group.
Mike is a curious to know for each x such that 1 ≤ x ≤ n the maximum strength among all groups of size x.
Input
The first line of input contains integer n (1 ≤ n ≤ 2 × 105), the number of bears.
The second line contains n integers separated by space, a1, a2, ..., an (1 ≤ ai ≤ 109), heights of bears.
Output
Print n integers in one line. For each x from 1 to n, print the maximum strength among all groups of size x.
Examples
Input
10
1 2 3 4 5 4 3 2 1 6
Output
6 4 4 3 3 2 2 1 1 1 | instruction | 0 | 5,431 | 14 | 10,862 |
Tags: binary search, data structures, dp, dsu
Correct Solution:
```
# import math,string,itertools,fractions,heapq,collections,re,array,bisect
# from itertools import chain, dropwhile, permutations, combinations
# from collections import defaultdict
def VI(): return list(map(int,input().split()))
def main_old(n,a):
s = list(a)
print(max(a),end=" ")
for i in range(1,n):
ns = list(s[:-1])
mx = 0
for j in range(len(ns)):
ns[j] = min(s[j],s[j+1])
if ns[j] > mx:
mx = ns[j]
s = ns
print(mx,end=" ")
#print(max(s),end=" ")
#print(min(a))
from sys import stdout
def main(n,a):
# correct, but too slow O(n^2); needs a better data structure.
s = a
buf = str(max(a))
for i in range(n-1,0,-1):
mx = 0
for j in range(i):
s[j] = min(s[j],s[j+1])
if s[j] > mx:
mx = s[j]
buf += " "+str(mx)
stdout.write(buf+"\n")
def run(n,a):
stack = []
l, r = [0]*n, [0]*n
for x in range(n):
while len(stack)>0 and a[stack[-1]] >= a[x]:
stack.pop()
if len(stack)==0: l[x] = -1
else: l[x] = stack[-1]
stack.append(x)
stack = []
for x in range(n-1,-1,-1):
while len(stack)>0 and a[stack[-1]]>=a[x]:
stack.pop()
if len(stack)==0: r[x]=n
else: r[x] = stack[-1]
stack.append(x)
sl = [-1] * n
for x in range(n):
v = r[x]-l[x]-1
sl[v-1] = max(sl[v-1],a[x])
for x in range(n-2,-1,-1):
sl[x] = max(sl[x], sl[x+1])
print(" ".join([str(x) for x in sl]))
def main_input(info=0):
n = int(input())
a = VI()
#main(n,a)
run(n,a)
if __name__ == "__main__":
main_input()
``` | output | 1 | 5,431 | 14 | 10,863 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,498 | 14 | 10,996 |
Tags: math
Correct Solution:
```
l = []
r = []
sum_r = 0
sum_l = 0
max_diff_r = 0
max_r = 0
max_diff_l = 0
max_l = 0
for i in range(0, int(input())):
x, y = input().split(' ')
l.append(int(x))
sum_l+=l[i]
r.append(int(y))
sum_r+=r[i]
if max_diff_r < r[i] - l[i]:
max_r=i+1
max_diff_r=r[i] - l[i]
if max_diff_l < l[i] - r[i]:
max_l=i+1
max_diff_l=l[i] - r[i]
d = max(abs(sum_r-sum_l), sum_l-sum_r+2*max_diff_r, sum_r-sum_l+2*max_diff_l)
if d == abs(sum_r-sum_l):
print(0)
elif d ==sum_r-sum_l+2*max_diff_l:
print(max_l)
else:print(max_r)
``` | output | 1 | 5,498 | 14 | 10,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,499 | 14 | 10,998 |
Tags: math
Correct Solution:
```
n = int(input())
ls = []
rs = []
for i in range(n):
l, r = [int(x) for x in input().split()]
ls.append(l)
rs.append(r)
lss = sum(ls)
rss = sum(rs)
best = abs(lss - rss)
ans = None
for i in range(n):
lss1 = lss - ls[i] + rs[i]
rss1 = rss - rs[i] + ls[i]
if abs(lss1 - rss1) > best:
best = abs(lss1 - rss1)
ans = i
print(0 if ans is None else ans + 1)
``` | output | 1 | 5,499 | 14 | 10,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,500 | 14 | 11,000 |
Tags: math
Correct Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
#MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
A = [];B = []
for _ in range(int(input())):
a,b = ilele()
A.append(a)
B.append(b)
l = sum(A);r = sum(B)
Ans = 0
maxi = abs(l-r)
for i in range(len(A)):
x = A[i];y = B[i]
x1 = l -x;y1 = r -y
x2 = x1 + y;y2 = y1 + x
z = abs(x2-y2)
if z > maxi:
maxi = z
Ans = i+1
print(Ans)
``` | output | 1 | 5,500 | 14 | 11,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,501 | 14 | 11,002 |
Tags: math
Correct Solution:
```
n = int(input())
t = []
for i in range(n):
l, r = map(int, input().split())
t.append((l, r))
L = 0
R = 0
for i in t:
L += i[0]
R += i[1]
m = abs(L - R)
mi = 0
for i in range(n):
l, r = t[i][0], t[i][1]
if l == r:
continue
d = abs((L - l + r) - (R - r + l))
if d > m:
m = d
mi = i
if m == abs(L - R):
print(0)
else:
print(mi + 1)
``` | output | 1 | 5,501 | 14 | 11,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,502 | 14 | 11,004 |
Tags: math
Correct Solution:
```
n = int(input())
l, r = [0]*n, [0]*n
for i in range(n):
l[i], r[i] = map(int, input().split(" "))
L, R = sum(l), sum(r)
index, beauty = -1, abs(L-R)
for i in range(n):
t = abs((L-l[i]+r[i]) - (R-r[i]+l[i]))
if t > beauty:
index, beauty = i, t
print(index+1)
``` | output | 1 | 5,502 | 14 | 11,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,503 | 14 | 11,006 |
Tags: math
Correct Solution:
```
n = int(input())
left = []
right = []
R = 0
L = 0
for i in range(n):
a, b = map(int, input().split())
left.append(a)
L += a
R += b
right.append(b)
base = abs(L - R)
m = base
cur = -1
for i in range(n):
L1 = L - left[i] + right[i]
R1 = R - right[i] + left[i]
curm = abs(L1 - R1)
if curm > m:
m = curm
cur = i
print(cur + 1)
``` | output | 1 | 5,503 | 14 | 11,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,504 | 14 | 11,008 |
Tags: math
Correct Solution:
```
n = int(input())
sum = 0
a = []
b = []
for i in range(0, n):
c = input()
c = c.split()
a.append(int(c[0]))
b.append(int(c[1]))
sum += a[i] - b[i]
v = abs(sum)
ans = -1
for i in range(0, n):
vv = sum - 2*(a[i] - b[i])
vv = abs(vv)
if(vv > v):
v = vv
ans = i
print(ans + 1)
``` | output | 1 | 5,504 | 14 | 11,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9. | instruction | 0 | 5,505 | 14 | 11,010 |
Tags: math
Correct Solution:
```
n = int(input())
l,r = [],[]
for i in range(n):
x = list(map(int,input().split()))
l.append(x[0])
r.append(x[1])
L,R = sum(l),sum(r)
ans = abs(L-R)
ind = 0
for i in range(n):
x = L-l[i]+r[i]
y = R-r[i]+l[i]
if ans<abs(x-y):
ans = max(ans,abs(x-y))
ind = i+1
print(ind)
``` | output | 1 | 5,505 | 14 | 11,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Very soon there will be a parade of victory over alien invaders in Berland. Unfortunately, all soldiers died in the war and now the army consists of entirely new recruits, many of whom do not even know from which leg they should begin to march. The civilian population also poorly understands from which leg recruits begin to march, so it is only important how many soldiers march in step.
There will be n columns participating in the parade, the i-th column consists of li soldiers, who start to march from left leg, and ri soldiers, who start to march from right leg.
The beauty of the parade is calculated by the following formula: if L is the total number of soldiers on the parade who start to march from the left leg, and R is the total number of soldiers on the parade who start to march from the right leg, so the beauty will equal |L - R|.
No more than once you can choose one column and tell all the soldiers in this column to switch starting leg, i.e. everyone in this columns who starts the march from left leg will now start it from right leg, and vice versa. Formally, you can pick no more than one index i and swap values li and ri.
Find the index of the column, such that switching the starting leg for soldiers in it will maximize the the beauty of the parade, or determine, that no such operation can increase the current beauty.
Input
The first line contains single integer n (1 ≤ n ≤ 105) — the number of columns.
The next n lines contain the pairs of integers li and ri (1 ≤ li, ri ≤ 500) — the number of soldiers in the i-th column which start to march from the left or the right leg respectively.
Output
Print single integer k — the number of the column in which soldiers need to change the leg from which they start to march, or 0 if the maximum beauty is already reached.
Consider that columns are numbered from 1 to n in the order they are given in the input data.
If there are several answers, print any of them.
Examples
Input
3
5 6
8 9
10 3
Output
3
Input
2
6 5
5 6
Output
1
Input
6
5 9
1 3
4 8
4 5
23 54
12 32
Output
0
Note
In the first example if you don't give the order to change the leg, the number of soldiers, who start to march from the left leg, would equal 5 + 8 + 10 = 23, and from the right leg — 6 + 9 + 3 = 18. In this case the beauty of the parade will equal |23 - 18| = 5.
If you give the order to change the leg to the third column, so the number of soldiers, who march from the left leg, will equal 5 + 8 + 3 = 16, and who march from the right leg — 6 + 9 + 10 = 25. In this case the beauty equals |16 - 25| = 9.
It is impossible to reach greater beauty by giving another orders. Thus, the maximum beauty that can be achieved is 9.
Submitted Solution:
```
l = [0]
for _ in range(int(input())):
a, b = map(int, input().split())
l.append(a - b)
s, a, b = sum(l), min(l), max(l)
print(l.index(a) if a + b < s else l.index(b))
``` | instruction | 0 | 5,506 | 14 | 11,012 |
Yes | output | 1 | 5,506 | 14 | 11,013 |
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