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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.
For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3;
* if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5;
* if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6;
* it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4.
You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.
Input
The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.
It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.
Output
Print t lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.
Example
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
Note
The first test case is described in the problem statement.
In the second test case all numbers from 1 to 5 are beautiful:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2;
* if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3;
* if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4;
* if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5.
Submitted Solution:
```
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
for w in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
l=[0]*n
d={}
for i in range(n):
d[a[i]]=i
num=n
num_list=[]
#inde=a.index(1)
for i in range(1,n+1):
num_list.append(i)
#print(j,i,num)
i=0
j=n-1
l[i]=1
while(j>i and num>0 and len(num_list)>0):
#print("YES")
#print(i,j)
if(num not in a[i:j+1]):
num=max(a[i:j+1])
if(num_list!=[] and num_list[-1]==num):
if(sorted(a[i:j+1])==num_list):
l[num-1]=1
if(d[num]>d[1]):
j=d[num]-1
else:
i=d[num]+1
num-=1
#print("YES1")
else:
if(d[num]>d[1]):
j=d[num]-1
else:
i=d[num]+1
num=max(a[i:j+1])
#print("YES2")
else:
while(len(num_list)>0 and num_list[-1]!=num):
num_list.remove(num_list[-1])
#print("YES3")
#print(i,j,num)
for i in range(n):
print(l[i],end=" ")
print()
#print(num_list)
```
No
| 4,300 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.
For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3;
* if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5;
* if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6;
* it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4.
You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.
Input
The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.
It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.
Output
Print t lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.
Example
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
Note
The first test case is described in the problem statement.
In the second test case all numbers from 1 to 5 are beautiful:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2;
* if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3;
* if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4;
* if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
A = list(map(int, input().split()))
D = [0 for _ in range(n+1)]
for i in range(n): D[A[i]] = i
if n > 1: print(1, end='')
ini = fin = D[1]
cont = 2
maxi = 1
for i in range(1,n-1):
if fin<D[cont]: fin+=1
if ini>D[cont]: ini-=1
if ini<=D[cont]<=fin: cont+=1
maxi = max(maxi,A[ini], A[fin])
if maxi==i+1: print(1,end='')
else: print(0, end='')
print(1)
```
No
| 4,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p=[p_1, p_2, …, p_n] of integers from 1 to n. Let's call the number m (1 ≤ m ≤ n) beautiful, if there exists two indices l, r (1 ≤ l ≤ r ≤ n), such that the numbers [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m.
For example, let p = [4, 5, 1, 3, 2, 6]. In this case, the numbers 1, 3, 5, 6 are beautiful and 2, 4 are not. It is because:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 5 we will have a permutation [1, 3, 2] for m = 3;
* if l = 1 and r = 5 we will have a permutation [4, 5, 1, 3, 2] for m = 5;
* if l = 1 and r = 6 we will have a permutation [4, 5, 1, 3, 2, 6] for m = 6;
* it is impossible to take some l and r, such that [p_l, p_{l+1}, …, p_r] is a permutation of numbers 1, 2, …, m for m = 2 and for m = 4.
You are given a permutation p=[p_1, p_2, …, p_n]. For all m (1 ≤ m ≤ n) determine if it is a beautiful number or not.
Input
The first line contains the only integer t (1 ≤ t ≤ 1000) — the number of test cases in the input. The next lines contain the description of test cases.
The first line of a test case contains a number n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of the given permutation p. The next line contains n integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n, all p_i are different) — the given permutation p.
It is guaranteed, that the sum of n from all test cases in the input doesn't exceed 2 ⋅ 10^5.
Output
Print t lines — the answers to test cases in the order they are given in the input.
The answer to a test case is the string of length n, there the i-th character is equal to 1 if i is a beautiful number and is equal to 0 if i is not a beautiful number.
Example
Input
3
6
4 5 1 3 2 6
5
5 3 1 2 4
4
1 4 3 2
Output
101011
11111
1001
Note
The first test case is described in the problem statement.
In the second test case all numbers from 1 to 5 are beautiful:
* if l = 3 and r = 3 we will have a permutation [1] for m = 1;
* if l = 3 and r = 4 we will have a permutation [1, 2] for m = 2;
* if l = 2 and r = 4 we will have a permutation [3, 1, 2] for m = 3;
* if l = 2 and r = 5 we will have a permutation [3, 1, 2, 4] for m = 4;
* if l = 1 and r = 5 we will have a permutation [5, 3, 1, 2, 4] for m = 5.
Submitted Solution:
```
for _ in range(int(input())):
a = int(input())
b = list(map(int, input().split()))
c = [0]*a
for i in range(1, a + 1):
flag = False
for j in range(i - a , i + 1):
if j < len(b) - 1 and sum(b[j: j + i]) == i*(i + 1)//2:
flag = True
break
if flag == True:
c[i - 1] = "1"
else:
c[i - 1] = "0"
y = ""
for num in c:
y = y + num
print(y)
```
No
| 4,302 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
# Sieve
sieve_primes = []
PRIME_LIMIT = int(1e6)
sieve = [False for i in range(0,PRIME_LIMIT)]
for i in range(2,PRIME_LIMIT):
if not sieve[i]:
sieve_primes.append(i)
for j in range(i*i, PRIME_LIMIT, i):
sieve[j]=True
# Input
n = int(input())
arr=list(map(int, input().split()))
# Construct search space Primes
primes = set()
def addPrime(num):
for sieve_prime in sieve_primes:
if num%sieve_prime == 0:
primes.add(sieve_prime)
while num%sieve_prime == 0:
num//=sieve_prime
if num>1:
primes.add(num)
# (Could use probability calculations here to reduce search space)
random.shuffle(arr)
for num in arr[:8]:
if num > 1:
addPrime(num)
addPrime(num+1)
if num > 2:
addPrime(num-1)
# Find answer
ans = n
# Function to find answer based on input prime
def findAns(prime):
global ans
curr_ans = 0
for num in arr:
if num<prime:
curr_ans += prime-num
else:
curr_ans += min(num%prime, prime - num%prime)
if curr_ans >= ans:
return
ans = min(curr_ans, ans)
for prime in primes:
findAns(prime)
if ans == 0: break
print(ans)
```
| 4,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import sys
input = sys.stdin.readline
from math import sqrt
import random
n=int(input())
A=list(map(int,input().split()))
random.shuffle(A)
MOD={2,3,5}
for t in A[:31]:
for x in [t-1,t,t+1]:
if x<=5:
continue
L=int(sqrt(x))
for i in range(2,L+2):
while x%i==0:
MOD.add(i)
x=x//i
if x==1:
break
if x!=1:
MOD.add(x)
ANS=1<<29
for m in MOD:
SCORE=0
for a in A:
if a<=m:
SCORE+=m-a
else:
SCORE+=min(a%m,(-a)%m)
if SCORE>=ANS:
break
else:
ANS=SCORE
print(ANS)
```
| 4,304 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
n = int(input())
a = [int(x) for x in input().split()]
can = set()
ls = []
for i in range(10):
idx = random.randint(0, n - 1)
for c in range(-1, 2):
if a[idx] + c > 0: ls.append(a[idx] + c)
maxn = 1000001
prime = []
vis = [True] * maxn
for i in range(2, maxn):
if vis[i] is not True: continue
if i * i > maxn: break
for j in range(i * i, maxn, i):
vis[j] = False
for i in range(2, maxn):
if vis[i]: prime.append(i)
for val in ls:
i = 2
for i in prime:
if i * i > val: break
div = False
while val % i == 0:
val //= i
div = True
if div: can.add(i)
i = i + 1
if val is not 1: can.add(val)
ans = n
for i in can:
total = 0
for j in a:
if j >= i: total += min(j % i, i - (j % i))
else: total += i - j
ans = min(ans, total)
print(ans)
```
| 4,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import sys
input = sys.stdin.readline
from math import sqrt
n=int(input())
A=list(map(int,input().split()))
SA=sorted(set(A))
MOD={2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109, 1117, 1123, 1129, 1151, 1153, 1163, 1171, 1181, 1187, 1193, 1201, 1213, 1217, 1223, 1229, 1231, 1237, 1249, 1259, 1277, 1279, 1283, 1289, 1291, 1297, 1301, 1303, 1307, 1319, 1321, 1327, 1361, 1367, 1373, 1381, 1399, 1409, 1423, 1427, 1429, 1433, 1439, 1447, 1451, 1453, 1459, 1471, 1481, 1483, 1487, 1489, 1493, 1499, 1511, 1523, 1531, 1543, 1549, 1553, 1559, 1567, 1571, 1579, 1583, 1597, 1601, 1607, 1609, 1613, 1619, 1621, 1627, 1637, 1657, 1663, 1667, 1669, 1693, 1697, 1699, 1709, 1721, 1723, 1733, 1741, 1747, 1753, 1759, 1777, 1783, 1787, 1789, 1801, 1811, 1823, 1831, 1847, 1861, 1867, 1871, 1873, 1877, 1879, 1889, 1901, 1907, 1913, 1931, 1933, 1949, 1951, 1973, 1979, 1987, 1993, 1997, 1999, 2003, 2011, 2017, 2027, 2029, 2039, 2053, 2063, 2069, 2081, 2083, 2087, 2089, 2099, 2111, 2113, 2129, 2131, 2137, 2141, 2143, 2153, 2161, 2179, 2203, 2207, 2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287, 2293, 2297, 2309, 2311, 2333, 2339, 2341, 2347, 2351, 2357, 2371, 2377, 2381, 2383, 2389, 2393, 2399, 2411, 2417, 2423, 2437, 2441, 2447, 2459, 2467, 2473, 2477, 2503, 2521, 2531, 2539, 2543, 2549, 2551, 2557, 2579, 2591, 2593, 2609, 2617, 2621, 2633, 2647, 2657, 2659, 2663, 2671, 2677, 2683, 2687, 2689, 2693, 2699, 2707, 2711, 2713, 2719, 2729, 2731, 2741, 2749, 2753, 2767, 2777, 2789, 2791, 2797, 2801, 2803, 2819, 2833, 2837, 2843, 2851, 2857, 2861, 2879, 2887, 2897, 2903, 2909, 2917, 2927, 2939, 2953, 2957, 2963, 2969, 2971, 2999, 3001, 3011, 3019, 3023, 3037, 3041, 3049, 3061, 3067, 3079, 3083, 3089, 3109, 3119, 3121, 3137, 3163, 3167, 3169, 3181, 3187, 3191, 3203, 3209, 3217, 3221, 3229, 3251, 3253, 3257, 3259, 3271, 3299, 3301, 3307, 3313, 3319, 3323, 3329, 3331, 3343, 3347, 3359, 3361, 3371, 3373, 3389, 3391, 3407, 3413, 3433, 3449, 3457, 3461, 3463, 3467, 3469, 3491, 3499, 3511, 3517, 3527, 3529, 3533, 3539, 3541, 3547, 3557, 3559, 3571, 3581, 3583, 3593, 3607, 3613, 3617, 3623, 3631, 3637, 3643, 3659, 3671, 3673, 3677, 3691, 3697, 3701, 3709, 3719, 3727, 3733, 3739, 3761, 3767, 3769, 3779, 3793, 3797, 3803, 3821, 3823, 3833, 3847, 3851, 3853, 3863, 3877, 3881, 3889, 3907, 3911, 3917, 3919, 3923, 3929, 3931, 3943, 3947, 3967, 3989, 4001, 4003, 4007, 4013, 4019, 4021, 4027, 4049, 4051, 4057, 4073, 4079, 4091, 4093, 4099, 4111, 4127, 4129, 4133, 4139, 4153, 4157, 4159, 4177, 4201, 4211, 4217, 4219, 4229, 4231, 4241, 4243, 4253, 4259, 4261, 4271, 4273, 4283, 4289, 4297, 4327, 4337, 4339, 4349, 4357, 4363, 4373, 4391, 4397, 4409, 4421, 4423, 4441, 4447, 4451, 4457, 4463, 4481, 4483, 4493, 4507, 4513, 4517, 4519, 4523, 4547, 4549, 4561, 4567, 4583, 4591, 4597, 4603, 4621, 4637, 4639, 4643, 4649, 4651, 4657, 4663, 4673, 4679, 4691, 4703, 4721, 4723, 4729, 4733, 4751, 4759, 4783, 4787, 4789, 4793, 4799, 4801, 4813, 4817, 4831, 4861, 4871, 4877, 4889, 4903, 4909, 4919, 4931, 4933, 4937, 4943, 4951, 4957, 4967, 4969, 4973, 4987, 4993, 4999, 5003, 5009, 5011, 5021, 5023, 5039, 5051, 5059, 5077, 5081, 5087, 5099, 5101, 5107, 5113, 5119, 5147, 5153, 5167, 5171, 5179, 5189, 5197, 5209, 5227, 5231, 5233, 5237, 5261, 5273, 5279, 5281, 5297, 5303, 5309, 5323, 5333, 5347, 5351, 5381, 5387, 5393, 5399, 5407, 5413, 5417, 5419, 5431, 5437, 5441, 5443, 5449, 5471, 5477, 5479, 5483, 5501, 5503, 5507, 5519, 5521, 5527, 5531, 5557, 5563, 5569, 5573, 5581, 5591, 5623, 5639, 5641, 5647, 5651, 5653, 5657, 5659, 5669, 5683, 5689, 5693, 5701, 5711, 5717, 5737, 5741, 5743, 5749, 5779, 5783, 5791, 5801, 5807, 5813, 5821, 5827, 5839, 5843, 5849, 5851, 5857, 5861, 5867, 5869, 5879, 5881, 5897, 5903, 5923, 5927, 5939, 5953, 5981, 5987, 6007, 6011, 6029, 6037, 6043, 6047, 6053, 6067, 6073, 6079, 6089, 6091, 6101, 6113, 6121, 6131, 6133, 6143, 6151, 6163, 6173, 6197, 6199, 6203, 6211, 6217, 6221, 6229, 6247, 6257, 6263, 6269, 6271, 6277, 6287, 6299, 6301, 6311, 6317, 6323, 6329, 6337, 6343, 6353, 6359, 6361, 6367, 6373, 6379, 6389, 6397, 6421, 6427, 6449, 6451, 6469, 6473, 6481, 6491, 6521, 6529, 6547, 6551, 6553, 6563, 6569, 6571, 6577, 6581, 6599, 6607, 6619, 6637, 6653, 6659, 6661, 6673, 6679, 6689, 6691, 6701, 6703, 6709, 6719, 6733, 6737, 6761, 6763, 6779, 6781, 6791, 6793, 6803, 6823, 6827, 6829, 6833, 6841, 6857, 6863, 6869, 6871, 6883, 6899, 6907, 6911, 6917, 6947, 6949, 6959, 6961, 6967, 6971, 6977, 6983, 6991, 6997, 7001, 7013, 7019, 7027, 7039, 7043, 7057, 7069, 7079, 7103, 7109, 7121, 7127, 7129, 7151, 7159, 7177, 7187, 7193, 7207, 7211, 7213, 7219, 7229, 7237, 7243, 7247, 7253, 7283, 7297, 7307, 7309, 7321, 7331, 7333, 7349, 7351, 7369, 7393, 7411, 7417, 7433, 7451, 7457, 7459, 7477, 7481, 7487, 7489, 7499, 7507, 7517, 7523, 7529, 7537, 7541, 7547, 7549, 7559, 7561, 7573, 7577, 7583, 7589, 7591, 7603, 7607, 7621, 7639, 7643, 7649, 7669, 7673, 7681, 7687, 7691, 7699, 7703, 7717, 7723, 7727, 7741, 7753, 7757, 7759, 7789, 7793, 7817, 7823, 7829, 7841, 7853, 7867, 7873, 7877, 7879, 7883, 7901, 7907, 7919, 7927, 7933, 7937, 7949, 7951, 7963, 7993, 8009, 8011, 8017, 8039, 8053, 8059, 8069, 8081, 8087, 8089, 8093, 8101, 8111, 8117, 8123, 8147, 8161, 8167, 8171, 8179, 8191, 8209, 8219, 8221, 8231, 8233, 8237, 8243, 8263, 8269, 8273, 8287, 8291, 8293, 8297, 8311, 8317, 8329, 8353, 8363, 8369, 8377, 8387, 8389, 8419, 8423, 8429, 8431, 8443, 8447, 8461, 8467, 8501, 8513, 8521, 8527, 8537, 8539, 8543, 8563, 8573, 8581, 8597, 8599, 8609, 8623, 8627, 8629, 8641, 8647, 8663, 8669, 8677, 8681, 8689, 8693, 8699, 8707, 8713, 8719, 8731, 8737, 8741, 8747, 8753, 8761, 8779, 8783, 8803, 8807, 8819, 8821, 8831, 8837, 8839, 8849, 8861, 8863, 8867, 8887, 8893, 8923, 8929, 8933, 8941, 8951, 8963, 8969, 8971, 8999, 9001, 9007, 9011, 9013, 9029, 9041, 9043, 9049, 9059, 9067, 9091, 9103, 9109, 9127, 9133, 9137, 9151, 9157, 9161, 9173, 9181, 9187, 9199, 9203, 9209, 9221, 9227, 9239, 9241, 9257, 9277, 9281, 9283, 9293, 9311, 9319, 9323, 9337, 9341, 9343, 9349, 9371, 9377, 9391, 9397, 9403, 9413, 9419, 9421, 9431, 9433, 9437, 9439, 9461, 9463, 9467, 9473, 9479, 9491, 9497, 9511, 9521, 9533, 9539, 9547, 9551, 9587, 9601, 9613, 9619, 9623, 9629, 9631, 9643, 9649, 9661, 9677, 9679, 9689, 9697, 9719, 9721, 9733, 9739, 9743, 9749, 9767, 9769, 9781, 9787, 9791, 9803, 9811, 9817, 9829, 9833, 9839, 9851, 9857, 9859, 9871, 9883, 9887, 9901, 9907, 9923, 9929, 9931, 9941, 9949, 9967, 9973}
for x in SA[:70]:
L=int(sqrt(x))
for i in range(2,L+2):
while x%i==0:
MOD.add(i)
x=x//i
if x!=1:
MOD.add(x)
ANS=1<<29
for m in MOD:
SCORE=0
for a in A:
if a<=m:
SCORE+=m-a
else:
SCORE+=min(a%m,(-a)%m)
if SCORE>=ANS:
break
else:
ANS=SCORE
print(ANS)
```
| 4,306 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
n = int(input())
a = list(map(int, input().split()))
limit = min(8, n)
iterations = [x for x in range(n)]
random.shuffle(iterations)
random.shuffle(iterations)
iterations = iterations[:limit]
def factorization(x):
primes = []
i = 2
while i * i <= x:
if x % i == 0:
primes.append(i)
while x % i == 0: x //= i
i = i + 1
if x > 1: primes.append(x)
return primes
def solve_with_fixed_gcd(arr, gcd):
result = 0
for x in arr:
if x < gcd: result += (gcd - x)
else:
remainder = x % gcd
result += min(remainder, gcd - remainder)
return result
answer = float("inf")
prime_list = set()
for index in iterations:
for x in range(-1, 2):
tmp = factorization(a[index]-x)
for z in tmp: prime_list.add(z)
for prime in prime_list:
answer = min(answer, solve_with_fixed_gcd(a, prime))
if answer == 0: break
print(answer)
```
| 4,307 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
n = int(input())
a = list(map(int, input().split()))
limit = min(8, n)
iterations = [x for x in range(n)]
random.shuffle(iterations)
iterations = iterations[:limit]
def factorization(x):
primes = []
i = 2
while i * i <= x:
if x % i == 0:
primes.append(i)
while x % i == 0: x //= i
i = i + 1
if x > 1: primes.append(x)
return primes
def solve_with_fixed_gcd(arr, gcd):
result = 0
for x in arr:
if x < gcd: result += (gcd - x)
else:
remainder = x % gcd
result += min(remainder, gcd - remainder)
return result
answer = float("inf")
prime_list = set()
for index in iterations:
for x in range(-1, 2):
tmp = factorization(a[index]-x)
for z in tmp: prime_list.add(z)
for prime in prime_list:
answer = min(answer, solve_with_fixed_gcd(a, prime))
if answer == 0: break
print(answer)
```
| 4,308 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
# Sieve
sieve_primes = []
PRIME_LIMIT = int(1e6)
sieve = [False for i in range(0,PRIME_LIMIT)]
for i in range(2,PRIME_LIMIT):
if not sieve[i]:
sieve_primes.append(i)
for j in range(i*i, PRIME_LIMIT, i):
sieve[j]=True
# Input
n = int(input())
arr=list(map(int, input().split()))
# Construct search space Primes
primes = set()
def addPrime(num):
for sieve_prime in sieve_primes:
if num%sieve_prime == 0:
primes.add(sieve_prime)
while num%sieve_prime == 0:
num//=sieve_prime
if num>1:
primes.add(num)
# (Could use probability calculations here to reduce search space)
random.shuffle(arr)
arr_set=list(set(arr))
for num in arr_set[:8]:
if num > 1:
addPrime(num)
addPrime(num+1)
if num > 2:
addPrime(num-1)
# Find answer
ans = n
# Function to find answer based on input prime
def findAns(prime):
global ans
curr_ans = 0
for num in arr:
if num<prime:
curr_ans += prime-num
else:
curr_ans += min(num%prime, prime - num%prime)
if curr_ans >= ans:
return
ans = min(curr_ans, ans)
for prime in primes:
findAns(prime)
if ans == 0: break
print(ans)
```
| 4,309 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Tags: math, number theory, probabilities
Correct Solution:
```
import random
n = int(input())
a = list(map(int, input().split()))
primes = set()
for x in random.choices(a, k=min(n, 30)):
for y in (x-1, x, x+1):
d = 2
while d * d <= y:
while y % d == 0:
primes.add(d)
y //= d
d += 1 + (d & 1)
if y > 1:
primes.add(y)
ans = float('inf')
for p in primes:
cand = 0
for x in a:
if x < p:
cand += p - x
else:
r = x % p
cand += min(r, p-r)
if cand >= ans:
break
else:
ans = cand
print(ans)
```
| 4,310 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
import random
# Sieve
sieve_primes = []
PRIME_LIMIT = int(1e6)
sieve = [False for i in range(0,PRIME_LIMIT)]
for i in range(2,PRIME_LIMIT):
if not sieve[i]:
sieve_primes.append(i)
for j in range(i*i, PRIME_LIMIT, i):
sieve[j]=True
# Input
n = int(input())
arr=list(map(int, input().split()))
# Construct search space Primes
primes = set()
def addPrime(num):
for sieve_prime in sieve_primes:
if num%sieve_prime == 0:
primes.add(sieve_prime)
while num%sieve_prime == 0:
num//=sieve_prime
if num>1:
primes.add(num)
# (Could use probability calculations here to reduce search space)
arr_set = random.sample(arr, len(arr))
arr_set=list(set(arr_set))
for num in arr_set[:8]:
if num > 1:
addPrime(num)
addPrime(num+1)
if num > 2:
addPrime(num-1)
# Find answer
ans = n
# Function to find answer based on input prime
def findAns(prime):
global ans
curr_ans = 0
for num in arr:
if num<prime:
curr_ans += prime-num
else:
curr_ans += min(num%prime, prime - num%prime)
if curr_ans >= ans:
return
ans = min(curr_ans, ans)
for prime in primes:
findAns(prime)
if ans == 0: break
print(ans)
```
Yes
| 4,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
import random
n = int(input())
best = n
l = list(map(int, input().split()))
candidates = set()
candidates.add(2)
def gcd(x, y):
while(y):
x, y = y, x % y
return x
def factAdd(n):
for c in candidates:
while n % c == 0:
n //= c
test = 3
while test * test <= n:
while n % test == 0:
candidates.add(test)
n //= test
test += 2
if n > 1:
candidates.add(n)
for i in range(100):
a = random.randint(0, n - 1)
b = random.randint(0, n - 1)
diff = [-1, 0, 1]
for d1 in diff:
a1 = l[a] + d1
if a1:
for d2 in diff:
a2 = l[b] + d2
if a2:
factAdd(gcd(a1, a2))
for cand in candidates:
count = 0
for v in l:
if v <= cand:
count += (cand - v)
else:
v2 = v % cand
count += min(v2, cand - v2)
if count < best:
best = count
print(best)
```
Yes
| 4,312 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
# TESTING EDITORIAL SUBMISSION FOR TEST CASE 105
import random
n = int(input())
a = list(map(int, input().split()))
limit = min(8, n)
iterations = [x for x in range(n)]
random.shuffle(iterations)
iterations = iterations[:limit]
def factorization(x):
primes = []
i = 2
while i * i <= x:
if x % i == 0:
primes.append(i)
while x % i == 0: x //= i
i = i + 1
if x > 1: primes.append(x)
return primes
def solve_with_fixed_gcd(arr, gcd):
result = 0
for x in arr:
if x < gcd: result += (gcd - x)
else:
remainder = x % gcd
result += min(remainder, gcd - remainder)
return result
answer = float("inf")
prime_list = set()
for index in iterations:
for x in range(-1, 2):
tmp = factorization(a[index]-x)
for z in tmp: prime_list.add(z)
for prime in prime_list:
answer = min(answer, solve_with_fixed_gcd(a, prime))
if answer == 0: break
print(answer)
```
Yes
| 4,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
import random
n = int(input())
a = list(map(int, input().split()))
def add(x):
d = 2
while d * d <= x:
if x % d == 0:
primes.add(d)
while x % d == 0:
x //= d
d += 1
if x > 1:
primes.add(x)
ans = float('inf')
primes = {2}
for x in random.choices(a, k=min(n, 8)):
for dx in range(-1, 2):
add(x + dx)
for p in primes:
cand = 0
for x in a:
if x < p:
cand += p - x
else:
r = x % p
cand += min(r, p-r)
ans = min(ans, cand)
if ans==0: break
print(ans)
```
Yes
| 4,314 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
import sys
input = sys.stdin.readline
from math import sqrt
import random
n=int(input())
A=list(map(int,input().split()))
SA=list(set(A))
random.shuffle(SA)
MOD={2,3,5}
LEN=len(SA)
for t in SA[:33]:
for x in [t-1,t,t+1]:
if x<=5:
continue
L=int(sqrt(x))
for i in range(2,L+2):
while x%i==0:
MOD.add(i)
x=x//i
if x==1:
break
if x!=1:
MOD.add(x)
ANS=1<<29
for m in MOD:
SCORE=0
for a in A:
if a<=m:
SCORE+=m-a
else:
SCORE+=min(a%m,(-a)%m)
if SCORE>=ANS:
break
else:
ANS=SCORE
print(ANS)
```
No
| 4,315 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
x = int(input())
lista = list(map(int,input().split()))
total = 0
for a in range(x):
if(lista[a]%2 == 1):
total += 1
print(total)
```
No
| 4,316 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
from random import randint
n = int(input())
a = [int(x) for x in input().split()]
def solve(x):
can = []
for i in range(2, x):
if i * i > x: break
if x % i is 0:
can.append(i)
while x % i is 0:
x = int(x / i)
if x is not 1:
can.append(x)
res = n
for i in can:
ans = 0
for j in a:
p = j % i
q = i - p
if p < q and j >= i: ans = ans + p
else: ans = ans + q
if res > ans:
res = ans
return res
ans = n
for _ in range(15):
x = randint(0, n)
for c in range(-1, 2):
if x + c > 0:
opt = solve(x + c)
if opt < ans:
ans = opt
print(ans)
```
No
| 4,317 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kuroni is very angry at the other setters for using him as a theme! As a punishment, he forced them to solve the following problem:
You have an array a consisting of n positive integers. An operation consists of choosing an element and either adding 1 to it or subtracting 1 from it, such that the element remains positive. We say the array is good if the greatest common divisor of all its elements is not 1. Find the minimum number of operations needed to make the array good.
Unable to match Kuroni's intellect, the setters failed to solve the problem. Help them escape from Kuroni's punishment!
Input
The first line contains an integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array.
The second line contains n integers a_1, a_2, ..., a_n. (1 ≤ a_i ≤ 10^{12}) — the elements of the array.
Output
Print a single integer — the minimum number of operations required to make the array good.
Examples
Input
3
6 2 4
Output
0
Input
5
9 8 7 3 1
Output
4
Note
In the first example, the first array is already good, since the greatest common divisor of all the elements is 2.
In the second example, we may apply the following operations:
1. Add 1 to the second element, making it equal to 9.
2. Subtract 1 from the third element, making it equal to 6.
3. Add 1 to the fifth element, making it equal to 2.
4. Add 1 to the fifth element again, making it equal to 3.
The greatest common divisor of all elements will then be equal to 3, so the array will be good. It can be shown that no sequence of three or less operations can make the array good.
Submitted Solution:
```
def gcd(a, b):
if b > a:
return gcd(b, a)
if a % b == 0:
return b
return gcd(b, a%b)
n = int(input())
a = list(map(int, input().strip().split()))
a.sort()
primes = [True]*int(max(a)**0.5)
primes[0] = False
primes[1] = False
for i in range(2, len(primes)):
if primes[i]:
ind2 = i*i
while ind2 < len(primes):
primes[ind2] = False
ind2 += i
primes = [i for i,e in enumerate(primes) if e]
minimum = len(a) + 5
for pr in primes:
cnt = 0
for v in a:
cnt += min(v % pr, pr - (v % pr))
if cnt > minimum:
break
if cnt < minimum:
minimum = cnt
seta = set(a)
distinct = len(set(a))
if distinct < 15:
for v in seta:
cnt = 0
for value in a:
cnt += abs(v-value)
if cnt < minimum:
minimum = cnt
print(minimum)
```
No
| 4,318 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
t = [j for j, q in enumerate(s) if q == 'T']
l, r = [0] * 101, [0] * 101
for i, (a, b) in enumerate(zip([-1] + t, t + [len(s)])):
v = b - a
u = v - 1
for k in range(i, 0, -1):
l[k] = max(l[k] + u, l[k - 1] + v)
r[k] = max(r[k] - u, r[k - 1] - v)
u, v = -u, -v
r[i + 1] = l[i] - 1
l[0] += u
r[0] -= u
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
| 4,319 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
import sys
import math
INF = -100000000
memo = dict()
def func(line, r):
if line in memo and r in memo[line]:
return memo[line][r]
if len(line) == 1:
which = line[0] == 'T'
if r % 2 == 1:
which = not which
if which:
return [INF, INF, 0, 0]
else:
return [1, INF, INF, INF]
best = [INF, INF, INF, INF]
for i in range(r + 1):
a = func(line[:len(line) // 2], i)
b = func(line[len(line) // 2:], r - i)
for (j, k) in [(j, k) for j in range(4) for k in range(4)]:
D = j < 2
if a[j] == INF or b[k] == INF: continue
aa = -a[j] if j % 2 else a[j]
bb = -b[k] if k % 2 else b[k]
d1, d2 = 0, 1
if k < 2:
aa = aa + bb
if not D: d1, d2 = 2, 3
else:
aa = -aa + bb
if D: d1, d2 = 2, 3
if aa >= 0: best[d1] = max(best[d1], aa)
if aa <= 0: best[d2] = max(best[d2], -aa)
if not line in memo:
memo[line] = dict()
memo[line][r] = best
return best
line = input().strip()
k = int(input())
ans = max(func(line, k))
if ans == INF: print("0")
else: print(ans)
```
| 4,320 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
from math import inf
def main():
s = input().strip()
n = int(input())
dp = [[[-inf,-inf] for _ in range(n+1)]for _ in range(len(s))]
x = int(s[0]!='F')
for i in range(n+1):
dp[0][i][x] = (x^1)
x ^= 1
for i in range(1,len(s)):
for j in range(n+1):
x = int(s[i]!='F')
for k in range(j,-1,-1):
dp[i][j][0] = max(dp[i][j][0],dp[i-1][k][x]+(x^1))
dp[i][j][1] = max(dp[i][j][1],dp[i-1][k][x^1]-(x^1))
x ^= 1
dp1 = [[[inf,inf] for _ in range(n+1)]for _ in range(len(s))]
x = int(s[0]!='F')
for i in range(n+1):
dp1[0][i][x] = (x^1)
x ^= 1
for i in range(1,len(s)):
for j in range(n+1):
x = int(s[i]!='F')
for k in range(j,-1,-1):
dp1[i][j][0] = min(dp1[i][j][0],dp1[i-1][k][x]+(x^1))
dp1[i][j][1] = min(dp1[i][j][1],dp1[i-1][k][x^1]-(x^1))
x ^= 1
print(max(max(dp[-1][n]),abs(min(dp1[-1][n]))))
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
| 4,321 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[ None for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
dp1=[[[[ None for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def sol(i,j,t,c):
if i==n:
if j!=0:
return -10000000
else:
return 0
if dp[i][j][t][c]!=None:
return dp[i][j][t][c]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1)
else:
ans=1+sol(i+1,j,t,-1)
if j>0:
ans=max(ans,sol(i,j-1, t,0))
else :
ans=sol(i+1,j,not t,-1)
if j>0:
ans=max(ans,sol(i,j-1, t,1))
dp[i][j][t][c]=ans
return dp[i][j][t][c]
def sol1(i,j,t,c):
if i==n:
if j!=0:
return 10000000
else:
return 0
if dp1[i][j][t][c]!=None:
return dp1[i][j][t][c]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol1(i+1,j,t,-1)
else:
ans=1+sol1(i+1,j,t,-1)
if j>0:
ans=min(ans,sol1(i,j-1, t,0))
else :
ans=sol1(i+1,j,not t,-1)
if j>0:
ans=min(ans,sol1(i,j-1, t,1))
dp1[i][j][t][c]=ans
return dp1[i][j][t][c]
print(max(sol(0,k,0,-1),-sol1(0,k,0,-1)))
```
| 4,322 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
l, r = [-1e9] * 101, [-1e9] * 101
l[0] = r[0] = 0
for q in s:
for j in range(n, -1, -1):
x = max(r[j], l[j - 1] + 1) if q == 'T' else max(l[j] + 1, r[j - 1])
y = max(l[j], r[j - 1] + 1) if q == 'T' else max(r[j] - 1, l[j - 1])
l[j], r[j] = x, y
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
# Made By Mostafa_Khaled
```
| 4,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def minn(a,b):
if a<b:
return a
return b
def maxx(a,b):
if a>b:
return a
return b
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=maxx(ans,sol(i,j-1, t,0,mt))
else:
ans=minn(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=maxx(ans,sol(i,j-1, t,1,mt))
else:
ans=minn(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
| 4,324 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,0,mt))
else:
ans=min(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,1,mt))
else:
ans=min(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
| 4,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Tags: dp
Correct Solution:
```
s = input()
n = int(input())
t = [j for j, q in enumerate(s) if q == 'T']
l, r = [0] * 101, [0] * 101
for i, (a, b) in enumerate(zip([-1] + t, t + [len(s)])):
v = b - a
u = v - 1
if i:
l[i] = l[i - 1] + v
else:
u, v = -u, -v
r[i + 1] = l[i] - 1
for k in range(i - 1, 0, -1):
l[k] = max(l[k] - u, l[k - 1] - v)
r[k] = max(r[k] + u, r[k - 1] + v)
u, v = -u, -v
l[0] -= u
r[0] += u
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
| 4,326 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
s = input()
n = int(input())
l, r = [-1e9] * 101, [-1e9] * 101
l[0] = r[0] = 0
for q in s:
for j in range(n, -1, -1):
x = max(r[j], l[j - 1] + 1) if q == 'T' else max(l[j] + 1, r[j - 1])
y = max(l[j], r[j - 1] + 1) if q == 'T' else max(r[j] - 1, l[j - 1])
l[j], r[j] = x, y
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
Yes
| 4,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
from sys import stdin
s=stdin.readline().strip()
k=int(stdin.readline().strip())
n=len(s)
dp=[[[[[ None for ty in range(2)]for tt in range(2)]for i in range(2)] for j in range(k+1)] for k1 in range(n)]
def min(a,b):
if a<b:
return a
return b
def max(a,b):
if a>b:
return a
return b
def sol(i,j,t,c,mt):
if i==n:
if j!=0:
if mt:
return -10001
else:
return 10001
else:
return 0
if dp[i][j][t][c][mt]!=None:
return dp[i][j][t][c][mt]
ans=0
if c==-1:
c=(s[i]=="F")
if c:
if t:
ans=-1+sol(i+1,j,t,-1,mt)
else:
ans=1+sol(i+1,j,t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,0,mt))
else:
ans=min(ans,sol(i,j-1, t,0,mt))
else :
ans=sol(i+1,j,not t,-1,mt)
if j>0:
if mt:
ans=max(ans,sol(i,j-1, t,1,mt))
else:
ans=min(ans,sol(i,j-1, t,1,mt))
dp[i][j][t][c][mt]=ans
return dp[i][j][t][c][mt]
print(max(-sol(0,k,0,-1,0),sol(0,k,0,-1,1)))
```
Yes
| 4,328 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
from sys import stdin
import sys
sys.setrecursionlimit(10000000)
s=stdin.readline().strip()
n=int(stdin.readline().strip())
dp1=[[[[-110 for i in range(51)] for j in range(101)] for k in range(2)] for k1 in range(2)]
dp2=[[[[-110 for i in range(51)] for j in range(101)] for k in range(2)] for k1 in range(2)]
inf=1000
def sol(pos,el,cam,dir):
if pos>=len(s):
if el==0:
return 0
else:
return -inf
if el<0:
return -inf
if dp1[dir][cam][pos][el]!=-110:
return dp1[dir][cam][pos][el]
e=s[pos]
if s[pos]=="F" and cam==1:
e="T"
if s[pos]=="T" and cam==1:
e="F"
ans=-inf
if e=="F":
if dir==1:
ans=max(1+sol(pos+1,el,0,dir),sol(pos,el-1,(cam+1)%2,(dir)%2))
else:
ans=max(-1+sol(pos+1,el,0,dir),sol(pos,el-1,(cam+1)%2,(dir)%2))
else:
ans=max(sol(pos+1,el,0,(dir+1)%2),sol(pos,el-1,(cam+1)%2,dir))
dp1[dir][cam][pos][el]=ans
return ans
def sol2(pos,el,cam,dir):
if pos>=len(s):
if el==0:
return 0
else:
return inf
if el<0:
return inf
if dp2[dir][cam][pos][el]!=-110:
return dp2[dir][cam][pos][el]
e=s[pos]
if s[pos]=="F" and cam==1:
e="T"
if s[pos]=="T" and cam==1:
e="F"
ans=inf
if e=="F":
if dir==1:
ans=min(1+sol2(pos+1,el,0,dir),sol2(pos,el-1,(cam+1)%2,(dir)))
else:
ans=min(-1+sol2(pos+1,el,0,dir),sol2(pos,el-1,(cam+1)%2,(dir)))
else:
ans=min(sol2(pos+1,el,0,(dir+1)%2),sol2(pos,el-1,(cam+1)%2,dir))
dp2[dir][cam][pos][el]=ans
return ans
print(max(sol(0,n,0,1),-sol2(0,n,0,1)))
```
Yes
| 4,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
line = input()
n = int(input())
dp = [[[-100] * 2 for _ in range(n+1)] for _ in range(len(line)+1)]
dp[0][0][0], dp[0][0][1] = 0, 0
for i in range(1, len(line)+1):
for j in range(n+1):
for k in range(j+1):
if (line[i-1] == "F" and k % 2 == 1) or (line[i-1] == "T" and k % 2 == 0):
dp[i][j][0] = max(dp[i][j][0], dp[i-1][j-k][1])
dp[i][j][1] = max(dp[i][j][1], dp[i-1][j-k][0])
else:
dp[i][j][0] = max(dp[i][j][0], dp[i-1][j-k][0] + 1)
dp[i][j][1] = max(dp[i][j][1], dp[i-1][j-k][1] - 1)
print(max(dp[-1][-1][0], dp[-1][-1][1]))
```
Yes
| 4,330 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
from sys import stdin,stdout,setrecursionlimit
setrecursionlimit(10**4)
PI=float('inf')
NI=float('-inf')
for _ in range(1):#int(stdin.readline())):
s=input()
n=int(stdin.readline())
# a=list(map(int, stdin.readline().split()))
f=s.count('F')
t=s.count('T')
mn=min(f,t)
mx=max(f,t)
ans=mx
if n<=mn:
ans=max(ans,mx+2*n-mn)
if n>=mx:
op1=n-mx# Changing max
if op1&1:ans=max(ans,mn+mx-1)
else:ans=max(ans,mx+mn)
if n>=mn:
op1=n-mn
if op1&1:ans=max(ans,mx+mn-1)
else:ans=max(ans,mx+mn)
print(ans)
```
No
| 4,331 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
s = input()
n = int(input())
t = [j for j, q in enumerate(s) if q == 'T']
l, r = [0] * 101, [0] * 101
for i, (a, b) in enumerate(zip([-1] + t, t + [len(s)])):
v = b - a
u = v - 1
if i:
l[i] = l[i - 1] + v
r[i] = max(r[i - 1] - v, l[i - 1] - 1)
else:
u, v = -u, -v
for k in range(i - 1, 0, -1):
l[k] = max(l[k] - u, l[k - 1] - v)
r[k] = max(r[k] + u, r[k - 1] + v)
u, v = -u, -v
l[0] -= u
r[0] += u
print(max(l[n % 2:n + 1:2] + r[n % 2:n + 1:2]))
```
No
| 4,332 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
s = input()
n, m = int(input()), len(s)
def g(p):
t = [j for j, q in enumerate(s) if q == p] + [m]
n = len(t) + 1
l = [0] * n
r = [0] * n
a = -1
for i, b in enumerate(t):
v = b - a
u = v - 1
l[i + 1] = l[i] + v
r[i + 1] = r[i] - v
for k in range(i, 0, -1):
l[k] = max(l[k] + u, l[k - 1] - v)
r[k] = max(r[k] - u, r[k - 1] + v)
u, v = -u, -v
l[0] += u
r[0] -= u
a = b
return l, r
t, f = g('T'), g('F')
print(max(t[0][n & 1:n + 1:2] + t[1][n & 1:n + 1:2] + f[0][m - n::2] + f[1][m - n::2]))
```
No
| 4,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A lot of people associate Logo programming language with turtle graphics. In this case the turtle moves along the straight line and accepts commands "T" ("turn around") and "F" ("move 1 unit forward").
You are given a list of commands that will be given to the turtle. You have to change exactly n commands from the list (one command can be changed several times). How far from the starting point can the turtle move after it follows all the commands of the modified list?
Input
The first line of input contains a string commands — the original list of commands. The string commands contains between 1 and 100 characters, inclusive, and contains only characters "T" and "F".
The second line contains an integer n (1 ≤ n ≤ 50) — the number of commands you have to change in the list.
Output
Output the maximum distance from the starting point to the ending point of the turtle's path. The ending point of the turtle's path is turtle's coordinate after it follows all the commands of the modified list.
Examples
Input
FT
1
Output
2
Input
FFFTFFF
2
Output
6
Note
In the first example the best option is to change the second command ("T") to "F" — this way the turtle will cover a distance of 2 units.
In the second example you have to change two commands. One of the ways to cover maximal distance of 6 units is to change the fourth command and first or last one.
Submitted Solution:
```
import sys
ILLEGAL = 10 ** 50
moves = sys.stdin.readline().strip()
n_changes = int(sys.stdin.readline())
mem = {}
def max_distance(n_changes_left, moves, cur_direction=1):
if (n_changes_left, moves, cur_direction) in mem:
return mem[(n_changes_left, moves, cur_direction)]
if moves == "":
if n_changes_left != 0:
return ILLEGAL
return 0
if n_changes_left == 0:
if moves[0] == 'F':
result = max_distance(n_changes_left, moves[1:])
return result
else:
return max_distance(n_changes_left, moves[1:], cur_direction * -1)
else:
first_solution, second_solution = 0, 0
if moves[0] == 'F':
first_solution = max_distance(n_changes_left, moves[1:], cur_direction)
second_solution = max_distance(n_changes_left-1, moves[1:], cur_direction * -1)
if first_solution != ILLEGAL:
first_solution += cur_direction
else:
first_solution = max_distance(n_changes_left, moves[1:], cur_direction * -1)
second_solution = max_distance(n_changes_left-1, moves[1:], cur_direction)
if second_solution != ILLEGAL:
second_solution += cur_direction
if first_solution == ILLEGAL and second_solution == ILLEGAL:
return 0
if first_solution == ILLEGAL:
return second_solution
if second_solution == ILLEGAL:
return first_solution
result = 0
if abs(first_solution) > abs(second_solution):
result = first_solution
else:
result = second_solution
mem[(n_changes_left, moves, cur_direction)] = result
return result
print(max_distance(n_changes, moves))
```
No
| 4,334 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
L=list(map(int,input().split()))
s=sum(L)
Ans=""
cnt=0
for i in range(n):
x=(s-L[i])//(n-1)
if(x*(n-1)!=s-L[i]):
continue
if(x==L[i]):
Ans+=str(i+1)+" "
cnt+=1
print(cnt)
if(cnt!=0):
print(Ans)
```
| 4,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
s=s/n
if s%1!=0:
print("0")
exit()
else:
l=[]
g=0
for i in range(n):
if a[i]==s:
g=g+1
l.append(i+1)
print(g)
for i in range(g):
print(l[i],end=" ")
```
| 4,336 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
A = list(map(int, input().split()))
B = []
srar = 0
for i in range(n):
srar += A[i]
srar = srar/n
for i in range(n):
if ((srar * n) - A[i]) / (n-1) == A[i]:
B.append(i+1)
print(len(B))
for i in range(len(B)):
print(B[i], end = ' ')
```
| 4,337 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
N = int(input())
ar = [int(x) for x in input().split()]
ans = []
s = sum(ar)
for i, j in enumerate(ar):
if j * N == s:
ans.append(str(i + 1))
print(len(ans))
print(' '.join(ans))
```
| 4,338 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
#!/usr/bin/env python3
n = int(input())
arr = [int(i) for i in input().split(' ')]
s = 0
result = []
for item in arr:
s += item
s = s/n
count = 0
pos = 1
for item in arr:
if item == s:
count += 1
result.append(str(pos))
pos += 1
print(count)
print(' '.join(result))
```
| 4,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
count = 0
ind = []
total = sum(a)
for i in range(n):
if a[i]==(total-a[i])/(n-1):
count+=1
ind.append(i+1)
print(count)
print(*ind)
```
| 4,340 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
import sys
S = sys.stdin.read()
S = [int(x) for x in S.split('\n')[1].split()]
s = sum(S)
n = len(S)
T = [i+1 for i in range(n) if n*S[i] == s]
print(len(T))
print(' '.join([str(x) for x in T]))
```
| 4,341 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Tags: brute force, implementation
Correct Solution:
```
m = int(input())
num = [int(n) for n in input().split()]
sum = 0
for i in range(0,m):
sum += num[i]
if sum % m:
print('0')
else:
ave = sum // m
cnt = 0
for i in range(0,m):
if num[i] == ave:
cnt += 1
print(cnt)
for i in range(0,m):
if num[i] == ave:
print(i+1,end=' ')
# I AK IOI
```
| 4,342 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
class CodeforcesTask134ASolution:
def __init__(self):
self.result = ''
self.n = 0
self.sequence = []
def read_input(self):
self.n = int(input())
self.sequence = [float(x) for x in input().split(" ")]
def process_task(self):
ss = sum(self.sequence)
other_rests = [(ss - self.sequence[x]) / (self.n - 1) for x in range(self.n)]
matching = [i + 1 for i, x in enumerate(other_rests) if self.sequence[i] == x]
self.result = "{0}\n{1}".format(len(matching), " ".join([str(x) for x in matching]))
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask134ASolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
Yes
| 4,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
#!/usr/bin/env python
def main( ):
n = int( input( ) )
a = list( map( int, input( ).split( ) ) )
s = sum( a )
res = list( )
for i in range( len( a ) ):
if ( s - a[ i ] ) / ( n - 1 ) == a[ i ]:
res.append( i + 1 )
print( len( res ) )
if len( res ) > 0:
print( " ".join( map( str, res ) ) )
if __name__ == '__main__':
main( )
```
Yes
| 4,344 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
n = int(input())
num = list(map(int, input().split()))
count = []
sm = sum(num)
#print(sm)
for i in range(n):
if (sm-num[i])/(n-1) == num[i]: count.append(i+1)
print(len(count))
for i in range(len(count)):
print(count[i], end=" ")
```
Yes
| 4,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
v=sum(a)
ans=0
b=[]
for i in range(n):
if(v-a[i])/(n-1)==a[i]:
ans+=1
b.append(i+1)
print(ans)
print(*b)
```
Yes
| 4,346 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
s,ans=sum(a),[]
for i in range(n):
if (s-a[i])/(n-1)==a[i]:
ans.append(i+1)
print(*ans)
```
No
| 4,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
import sys
n = int(input())
s = [*map(int,sys.stdin.readline().split())]
su = sum(s)
ch = 0
a = []
for i in range(n):
if (su-s[i])//(n-1)==s[i]:
ch +=1
a.append(i+1)
print(ch)
print(*a)
```
No
| 4,348 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
a=[]
c=0
s=sum(l)
for i in range(0,n):
if(l[i]==((s-l[i])//(n-1))):
c+=1
a.append(i+1)
print(c)
for i in a:
print(i,end=" ")
```
No
| 4,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence of positive integers a1, a2, ..., an. Find all such indices i, that the i-th element equals the arithmetic mean of all other elements (that is all elements except for this one).
Input
The first line contains the integer n (2 ≤ n ≤ 2·105). The second line contains elements of the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000). All the elements are positive integers.
Output
Print on the first line the number of the sought indices. Print on the second line the sought indices in the increasing order. All indices are integers from 1 to n.
If the sought elements do not exist, then the first output line should contain number 0. In this case you may either not print the second line or print an empty line.
Examples
Input
5
1 2 3 4 5
Output
1
3
Input
4
50 50 50 50
Output
4
1 2 3 4
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
p=[]
s=sum(a)
for i in range(n):
if (s-a[i])//(n-1)==a[i]:
p.append(i+1)
print(len(p))
print(*p)
```
No
| 4,350 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
length = int(input())
tests = [''] * length
for i in range(length):
tests[i] = int(input())
for i in tests:
if i % 2 == 0:
print(int(i / 2))
else:
print(int((i - 1) / 2))
```
| 4,351 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
from math import sqrt
from math import floor
def solve():
n = int(input())
print(n//2)
for cnt in range(int(input())):
solve()
```
| 4,352 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
from collections import defaultdict as dd
from collections import deque
import bisect
import heapq
def ri():
return int(input())
def rl():
return list(map(int, input().split()))
def solve():
n = ri()
print (n // 2)
mode = 'T'
if mode == 'T':
t = ri()
for i in range(t):
solve()
else:
solve()
```
| 4,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
def gcd(n):
if(n%2==0):
return n//2
else:
return gcd(n-1)
t=int(input())
for _ in range(0,t):
n=int(input())
res=gcd(n)
print(res)
```
| 4,354 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
for __ in range(int(input())):
n=int(input())
if n%2!=0:
n-=1
n=n//2
print(n)
```
| 4,355 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
# @oj: codeforces
# @id: hitwanyang
# @email: 296866643@qq.com
# @date: 2020-06-22 14:44
# @url:https://codeforc.es/contest/1370/problem/A
import sys,os
from io import BytesIO, IOBase
import collections,itertools,bisect,heapq,math,string
from decimal import *
# region fastio
BUFSIZE = 8192
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def main():
t=int(input())
for i in range(t):
n=int(input())
print (n//2)
if __name__ == "__main__":
main()
```
| 4,356 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
for nt in range(int(input())):
n = int(input())
print (n//2)
```
| 4,357 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Tags: greedy, implementation, math, number theory
Correct Solution:
```
import math
for _ in range(int(input())):
n=int(input())
print(math.floor(n/2))
```
| 4,358 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
def task_A():
t = int(input())
for _ in range(t):
n = int(input())
print(n // 2)
def main():
task_A()
if __name__ == '__main__':
main()
```
Yes
| 4,359 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
def main():
t = int(input())
for _ in range(t):
n = int(input())
resposta = 0
for x in range(n, 0, -1):
if not x%2:
resposta = x//2
break
print(resposta)
main()
```
Yes
| 4,360 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
q = int(input())
for i in range (q):
n = int(input())
print(n//2)
```
Yes
| 4,361 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
t = int(input())
n = []
for i in range(t):
n.append(int(input()))
for i in range(len(n)):
if n[i]%2==0:
print(int(n[i] / 2))
else:
print(int((n[i] - 1) / 2))
```
Yes
| 4,362 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
import math
for _ in range(int(input())):
n = int(input())
arr = [0] * n
for i in range(n):
arr[i] = i + 1
high = 0
i = 0
while i < n:
high = max(high, arr[i])
i = i + 1
# Array to store the count of divisors
# i.e. Potential GCDs
divisors = [0] * (high + 1)
# Iterating over every element
i = 0
while i < n:
# Calculating all the divisors
j = 1
while j <= math.sqrt(arr[i]):
# Divisor found
if arr[i] % j == 0:
# Incrementing count for divisor
divisors[j] = divisors[j] + 1
# Element/divisor is also a divisor
# Checking if both divisors are
# not same
if j != arr[i] / j:
divisors[arr[i] // j] = divisors[arr[i] // j] + 1
j = j + 1
i = i + 1
# Checking the highest potential GCD
i = high
while i >= 1:
# If this divisor can divide at least 2
# numbers, it is a GCD of at least 1 pair
if divisors[i] > 1:
print(i)
i = i - 1
```
No
| 4,363 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
t=int(input())
while t:
t=t-1
n=int(input())
for i in range(1,n):
if n%i==0:
c=i
print(c)
```
No
| 4,364 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
#!/usr/bin/env python
import os
import operator
import sys
from io import BytesIO, IOBase
def main():
#for _ in range(int(input())):
# #n=int(input())
# a,b,n=map(int,input().split())
# #arr=[int(k) for k in input().split()]
# count=0
# while True:
# mn=min(a,b)
# mx=max(a,b)
# mn=mn+mx
# count+=1
# if mn>n:
# break
# a=mn
# b=mx
# #print(a,b)
# print(count)
n=int(input())
#if n%2==0:
print(n//2)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 4,365 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's consider all integers in the range from 1 to n (inclusive).
Among all pairs of distinct integers in this range, find the maximum possible greatest common divisor of integers in pair. Formally, find the maximum value of gcd(a, b), where 1 ≤ a < b ≤ n.
The greatest common divisor, gcd(a, b), of two positive integers a and b is the biggest integer that is a divisor of both a and b.
Input
The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. The description of the test cases follows.
The only line of each test case contains a single integer n (2 ≤ n ≤ 10^6).
Output
For each test case, output the maximum value of gcd(a, b) among all 1 ≤ a < b ≤ n.
Example
Input
2
3
5
Output
1
2
Note
In the first test case, gcd(1, 2) = gcd(2, 3) = gcd(1, 3) = 1.
In the second test case, 2 is the maximum possible value, corresponding to gcd(2, 4).
Submitted Solution:
```
import math
n=int(input())
for i in range(n):
x=int(input())
print(int(math.sqrt(x)))
```
No
| 4,366 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n=int(input())
num=list(map(int,input().split()))
s1=(n*(n-1))//2
su=sum(num)
s2=su-s1
base=s2//n
additional=s2%n
out=[]
for i in range(n):
val=base+i
if additional>0:
additional-=1
val+=1
out.append(str(val))
print(' '.join(out))
```
| 4,367 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
# from collections import defaultdict as dc
# from math import ceil,floor,sqrt
# from collections import Counter
n = N()
d = RLL()
s = sum(d)-((n**2-n)>>1)
c = n
kk = 0
for i in range(n-1,0,-1):
if d[i]-i>s//c:
kk = i
break
c-=1
s-=d[i]-i
for i in range(kk,0,-1):
tmp = s//c
c-=1
s-=tmp
d[i] = tmp+i
d[0] = s
print_list(d)
```
| 4,368 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------game starts now-----------------------------------------------------
n=int(input())
a=list(map(int,input().split()))
s=sum(a)
s1=(s-(n*(n-1))//2)//n
s2=(s-(n*(n-1))//2)%n
for i in range (n):
a[i]=i+s1+(i+1<=s2)
print(*a)
```
| 4,369 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def bsearch(mn, mx, func):
#func(i)=False を満たす最大のi (mn<=i<mx)
idx = (mx + mn)//2
while mx-mn>1:
if func(idx):
idx, mx = (idx + mn)//2, idx
continue
idx, mn = (idx + mx)//2, idx
return idx
N, = map(int, input().split())
X = list(map(int, input().split()))
sX = sum(X)
d = X[0]
if sX >= N*(N-1)//2:
tmp = sX-N*(N-1)//2
d = max(tmp//N, d)
m = sX - d*N
for i in range(N):
X[i] = d
i=bsearch(0,N,lambda i : i*(i+1)//2>=m)
i+=1
for j in range(i-1):
X[N-1-j]+=i-j-1
k=m-i*(i-1)//2
for j in range(k):
X[N-i+j]+=1
print(" ".join(map(str, X)))
```
| 4,370 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
import sys
n=int(input())
a=[int(v) for v in sys.stdin.readline().split()]
s=sum(a)-((n*(n+1))//2)
p=s//n
q=s%n
for j in range(n):
a[j]=j+1+p+(q>0)
q=q-1
print(' '.join(map(str,a)))
```
| 4,371 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
n = int(input());tot = sum(map(int, input().split()));extra = (n * (n - 1))//2;smol = (tot - extra) // n;out = [smol + i for i in range(n)]
for i in range(tot - sum(out)):out[i] += 1
print(' '.join(map(str,out)))
```
| 4,372 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
from sys import stdin, stdout
input = stdin.buffer.readline
print = stdout.write
def f(a, b):
return (a + b) * n // 2
n = int(input())
*h, = map(int, input().split())
s = sum(h)
l, r = 0, 10 ** 12
while l < r:
m = l + r >> 1
if f(m, m + n - 1) < s:
l = m + 1
else:
r = m
l -= 1
y = f(l, l + n - 1)
for i in range(n):
print(f'{l + i + (i < s - y)} ')
```
| 4,373 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Tags: binary search, constructive algorithms, data structures, greedy, math
Correct Solution:
```
#include <CodeforcesSolutions.h>
#include <ONLINE_JUDGE <solution.cf(contestID = "1392",questionID = "A",method = "GET")>.h>
"""
Author : thekushalghosh
Team : CodeDiggers
I prefer Python language over the C++ language :p :D
Visit my website : thekushalghosh.github.io
"""
import sys,math,cmath,time
start_time = time.time()
##########################################################################
################# ---- THE ACTUAL CODE STARTS BELOW ---- #################
def solve():
n = inp()
a = inlt()
q = (sum(a) - (n * (n - 1) // 2)) // n
qq = 0
ww = sum(a)
for i in range(n):
qq = qq + q + i
for i in range(n):
if i < ww - qq:
qw = q + i + 1
else:
qw = q + i
sys.stdout.write(str(qw) + " ")
################## ---- THE ACTUAL CODE ENDS ABOVE ---- ##################
##########################################################################
def main():
global tt
if not ONLINE_JUDGE:
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
t = 1
for tt in range(t):
solve()
if not ONLINE_JUDGE:
print("Time Elapsed :",time.time() - start_time,"seconds")
sys.stdout.close()
#---------------------- USER DEFINED INPUT FUNCTIONS ----------------------#
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
return(input().strip())
def invr():
return(map(int,input().split()))
#------------------ USER DEFINED PROGRAMMING FUNCTIONS ------------------#
def counter(a):
q = [0] * max(a)
for i in range(len(a)):
q[a[i] - 1] = q[a[i] - 1] + 1
return(q)
def counter_elements(a):
q = dict()
for i in range(len(a)):
if a[i] not in q:
q[a[i]] = 0
q[a[i]] = q[a[i]] + 1
return(q)
def string_counter(a):
q = [0] * 26
for i in range(len(a)):
q[ord(a[i]) - 97] = q[ord(a[i]) - 97] + 1
return(q)
def factors(n):
q = []
for i in range(1,int(n ** 0.5) + 1):
if n % i == 0: q.append(i); q.append(n // i)
return(list(sorted(list(set(q)))))
def prime_factors(n):
q = []
while n % 2 == 0: q.append(2); n = n // 2
for i in range(3,int(n ** 0.5) + 1,2):
while n % i == 0: q.append(i); n = n // i
if n > 2: q.append(n)
return(list(sorted(q)))
def transpose(a):
n,m = len(a),len(a[0])
b = [[0] * n for i in range(m)]
for i in range(m):
for j in range(n):
b[i][j] = a[j][i]
return(b)
def factorial(n,m = 1000000007):
q = 1
for i in range(n):
q = (q * (i + 1)) % m
return(q)
#-----------------------------------------------------------------------#
ONLINE_JUDGE = __debug__
if ONLINE_JUDGE:
import io,os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
main()
```
| 4,374 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
n = int(input().strip())
a = [int(ai) for ai in input().strip().split()]
l = 1
r = max(a)
asum = sum(a)
ans = 0;
while l <= r:
mid = (l + r) >> 1
if 2 * n * mid >= 2 * asum - n * n - n :
r = mid - 1
ans = mid
else:
l = mid + 1
asum = asum - (ans + ans + n - 1) * n // 2
for i in range(0, n):
a[i] = ans + int(asum > 0)
ans = ans + 1
asum = asum - 1
print(' '.join(str(ai) for ai in a))
```
Yes
| 4,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
# Fast IO (only use in integer input)
import os,io
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n = int(input())
a = list(map(int,input().split()))
s = sum(a)
s -= (n * n - n) // 2
k = s // n
r = s % n
ans = []
for i in range(n):
curAns = k + i
if r > i:
curAns += 1
ans.append(str(curAns))
print(" ".join(ans))
```
Yes
| 4,376 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
n=int(input())
SUM=sum(map(int,input().split()))
OK=SUM//n-n//2-10
NG=SUM//n-n//2+10
def tousa(s,n):
return (s+(s+n-1))*n//2
while NG-OK>1:
mid=(OK+NG)//2
if tousa(mid,n)>SUM:
NG=mid
else:
OK=mid
#print(OK)
rest=SUM-tousa(OK,n)
for i in range(n):
if i<rest:
sys.stdout.write(str(OK+i+1)+" ")
else:
sys.stdout.write(str(OK+i)+" ")
```
Yes
| 4,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
A = list(map(int, input().split()))
S = sum(A)
k = (S - (N-1) * (N-2) // 2) % N
x = (S - (N-1) * (N-2) // 2 - k) // N
for i in range(N-1):
sys.stdout.write(str(x + i) + " ")
if i == k:
sys.stdout.write(str(x + i) + " ")
if k == N-1:
sys.stdout.write(str(x + k) + " ")
if __name__ == '__main__':
main()
```
Yes
| 4,378 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
h=[int(i) for i in input().split() if i!='\n']
suma=sum(h)
lo,hi=h[0],h[-1]-n+1
while lo<hi:
mid=(lo+hi)//2
prev=max(0,mid-1)
last=mid+n-1
test=(last*(last+1))/2-(prev*(prev+1))/2
if test>suma:
hi=mid
else:
lo=mid+1
ans=[int(i) for i in range(hi,hi+n)]
diff=sum(ans)-suma
if n==300000:
print(sum(ans)-suma)
for i in range(len(ans)-1,-1,-1):
if diff>0:
ans[i]-=1
diff-=1
ans=' '.join(map(str,ans))
sys.stdout.write(ans)
```
No
| 4,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
import sys, random
input = sys.stdin.readline
def check(arr):
for i in range(len(arr)-1):
diff = abs(arr[i] - arr[i+1])
if diff <= 2:
arr[i] += diff//2
arr[i+1] -= diff//2
return arr
if __name__ == "__main__":
n = int(input())
arr = list(map(int, input().split()))
print(*check(arr))
```
No
| 4,380 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
h=[int(i) for i in input().split() if i!='\n']
suma=sum(h)
lo,hi=h[0],h[-1]-n+1
while lo<hi:
mid=(lo+hi)//2
prev=max(0,mid-1)
last=mid+n-1
test=(last*(last+1))/2-(prev*(prev+1))/2
if test>suma:
hi=mid
else:
lo=mid+1
ans=[int(i) for i in range(hi,hi+n)]
diff=sum(ans)-suma
if diff>n:
diff=diff-n
if n==300000:
print(diff)
for i in range(len(ans)-1,-1,-1):
if diff>0:
ans[i]-=1
diff-=1
else:
break
ans=' '.join(map(str,ans))
sys.stdout.write(ans)
```
No
| 4,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Omkar is standing at the foot of Celeste mountain. The summit is n meters away from him, and he can see all of the mountains up to the summit, so for all 1 ≤ j ≤ n he knows that the height of the mountain at the point j meters away from himself is h_j meters. It turns out that for all j satisfying 1 ≤ j ≤ n - 1, h_j < h_{j + 1} (meaning that heights are strictly increasing).
Suddenly, a landslide occurs! While the landslide is occurring, the following occurs: every minute, if h_j + 2 ≤ h_{j + 1}, then one square meter of dirt will slide from position j + 1 to position j, so that h_{j + 1} is decreased by 1 and h_j is increased by 1. These changes occur simultaneously, so for example, if h_j + 2 ≤ h_{j + 1} and h_{j + 1} + 2 ≤ h_{j + 2} for some j, then h_j will be increased by 1, h_{j + 2} will be decreased by 1, and h_{j + 1} will be both increased and decreased by 1, meaning that in effect h_{j + 1} is unchanged during that minute.
The landslide ends when there is no j such that h_j + 2 ≤ h_{j + 1}. Help Omkar figure out what the values of h_1, ..., h_n will be after the landslide ends. It can be proven that under the given constraints, the landslide will always end in finitely many minutes.
Note that because of the large amount of input, it is recommended that your code uses fast IO.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^6).
The second line contains n integers h_1, h_2, ..., h_n satisfying 0 ≤ h_1 < h_2 < ... < h_n ≤ 10^{12} — the heights.
Output
Output n integers, where the j-th integer is the value of h_j after the landslide has stopped.
Example
Input
4
2 6 7 8
Output
5 5 6 7
Note
Initially, the mountain has heights 2, 6, 7, 8.
In the first minute, we have 2 + 2 ≤ 6, so 2 increases to 3 and 6 decreases to 5, leaving 3, 5, 7, 8.
In the second minute, we have 3 + 2 ≤ 5 and 5 + 2 ≤ 7, so 3 increases to 4, 5 is unchanged, and 7 decreases to 6, leaving 4, 5, 6, 8.
In the third minute, we have 6 + 2 ≤ 8, so 6 increases to 7 and 8 decreases to 7, leaving 4, 5, 7, 7.
In the fourth minute, we have 5 + 2 ≤ 7, so 5 increases to 6 and 7 decreases to 6, leaving 4, 6, 6, 7.
In the fifth minute, we have 4 + 2 ≤ 6, so 4 increases to 5 and 6 decreases to 5, leaving 5, 5, 6, 7.
In the sixth minute, nothing else can change so the landslide stops and our answer is 5, 5, 6, 7.
Submitted Solution:
```
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
n=int(input())
H=list(map(int,input().split()))
OK=0
NG=10**12
def tousa(s,n):
return (s+(s+n-1))*n//2
SUM=sum(H)
while NG-OK>1:
mid=(OK+NG)//2
if tousa(mid,n)>SUM:
NG=mid
else:
OK=mid
#print(OK)
rest=SUM-tousa(OK,n)
ANS=[OK+i+1 for i in range(n) if i<rest]
sys.stdout.write(" ".join(map(str,ANS))+"\n")
```
No
| 4,382 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
# cook your dish here
#_________________ Mukul Mohan Varshney _______________#
#Template
import sys
import os
import math
import copy
from math import gcd
from bisect import bisect
from io import BytesIO, IOBase
from math import sqrt,floor,factorial,gcd,log,ceil
from collections import deque,Counter,defaultdict
from itertools import permutations, combinations
from itertools import accumulate
#define function
def Int(): return int(sys.stdin.readline())
def Mint(): return map(int,sys.stdin.readline().split())
def Lstr(): return list(sys.stdin.readline().strip())
def Str(): return sys.stdin.readline().strip()
def Mstr(): return map(str,sys.stdin.readline().strip().split())
def List(): return list(map(int,sys.stdin.readline().split()))
def Hash(): return dict()
def Mod(): return 1000000007
def Ncr(n,r,p): return ((fact[n])*((ifact[r]*ifact[n-r])%p))%p
def Most_frequent(list): return max(set(list), key = list.count)
def Mat2x2(n): return [List() for _ in range(n)]
def Lcm(x,y): return (x*y)//gcd(x,y)
def dtob(n): return bin(n).replace("0b","")
def btod(n): return int(n,2)
def common(l1, l2):
return set(l1).intersection(l2)
# Driver Code
def solution():
for _ in range(Int()):
n,t=Mint()
a=List()
ans=[]
f=0
for i in range(n):
if(2*a[i]<t):
ans.append(1)
elif(2*a[i]==t):
if(f==0):
ans.append(0)
f=1
else:
ans.append(1)
f=0
else:
ans.append(0)
print(*ans)
#Call the solve function
if __name__ == "__main__":
solution()
```
| 4,383 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
N = 10005
# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)
# array to precompute inverse of 1! to N!
naturalNumInverse = factorialNumInverse.copy()
# array to store factorial of
# first N numbers
fact = factorialNumInverse.copy()
# Function to precompute inverse of numbers
def InverseofNumber(p):
naturalNumInverse[0] = naturalNumInverse[1] = 1
for i in range(2, N + 1, 1):
naturalNumInverse[i] = (naturalNumInverse[p % i] *
(p - int(p / i)) % p)
# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
factorialNumInverse[0] = factorialNumInverse[1] = 1
# precompute inverse of natural numbers
for i in range(2, N + 1, 1):
factorialNumInverse[i] = (naturalNumInverse[i] *
factorialNumInverse[i - 1]) % p
# Function to calculate factorial of 1 to N
def factorial(p):
fact[0] = 1
# precompute factorials
for i in range(1, N + 1):
fact[i] = (fact[i - 1] * i) % p
# Function to return nCr % p in O(1) time
def Binomial(N, R, p):
# n C r = n!*inverse(r!)*inverse((n-r)!)
ans = ((fact[N] * factorialNumInverse[R])% p *
factorialNumInverse[N - R])% p
return ans
def main():
starttime=datetime.datetime.now()
if(os.path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
for _ in range(ri()):
N,n=ria()
a=ria()
g1={}
g0={}
t=0
if n%2!=0:
for i in a:
if i<=(n//2):
if i in g0:
g0[i]+=1
else:
g0[i]=1
if i>(n//2):
if i in g1:
g1[i]+=1
else:
g1[i]=1
else:
for i in a:
if i<(n//2):
if i in g0:
g0[i]+=1
else:
g0[i]=1
if i>(n//2):
if i in g1:
g1[i]+=1
else:
g1[i]=1
if i==(n//2):
if t==0:
if i in g0:
g0[i]+=1
else:
g0[i]=1
t=1
else:
if i in g1:
g1[i]+=1
else:
g1[i]=1
t=0
ans=[]
for i in a:
if i in g1 and g1[i]>0:
ans.append(1)
g1[i]-=1
else:
if i in g0 and g0[i]>0:
ans.append(0)
g0[i]-=1
print(*ans)
#<--Solving Area Ends
endtime=datetime.datetime.now()
time=(endtime-starttime).total_seconds()*1000
if(os.path.exists('input.txt')):
print("Time:",time,"ms")
class FastReader(io.IOBase):
newlines = 0
def __init__(self, fd, chunk_size=1024 * 8):
self._fd = fd
self._chunk_size = chunk_size
self.buffer = io.BytesIO()
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self, size=-1):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, self._chunk_size if size == -1 else size))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
class FastWriter(io.IOBase):
def __init__(self, fd):
self._fd = fd
self.buffer = io.BytesIO()
self.write = self.buffer.write
def flush(self):
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class FastStdin(io.IOBase):
def __init__(self, fd=0):
self.buffer = FastReader(fd)
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
class FastStdout(io.IOBase):
def __init__(self, fd=1):
self.buffer = FastWriter(fd)
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.flush = self.buffer.flush
if __name__ == '__main__':
sys.stdin = FastStdin()
sys.stdout = FastStdout()
main()
```
| 4,384 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
import sys
from collections import defaultdict as dd
from collections import Counter as cc
from queue import Queue
import math
import itertools
try:
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
except:
pass
input = lambda: sys.stdin.buffer.readline().rstrip()
for _ in range(int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
r=1
for i in range(n):
if a[i]==k/2:
a[i]=r
r=abs(r-1)
elif a[i]<k/2:
a[i]=1
else:
a[i]=0
print(*a)
```
| 4,385 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
n,T = map(int,stdin.readline().split())
a = list(map(int,stdin.readline().split()))
cnt = 0
p = [None] * n
for i in range(n):
if a[i]*2 < T:
p[i] = 0
elif a[i]*2 > T:
p[i] = 1
else:
p[i] = cnt
cnt ^= 1
print (*p)
```
| 4,386 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
for T in range(int(input())):
n,t=map(int,input().split())
a=list(map(int,input().split()))
arr,d=[],{}
for i in range(len(a)):
arr.append([a[i],i])
d[a[i],i]=0
arr.sort(key=lambda x : x[0])
i,j=0,n-1
while(i<j):
if(arr[i][0]+arr[j][0]<=t):
if(arr[i][0]+arr[j][0]==t):
d[arr[i][0],arr[i][1]]=1
if(arr[i][0]==arr[j][0]):
j-=1
i+=1
else:
j-=1
for i in range(len(a)):
print(d[a[i],i],end=' ')
print()
```
| 4,387 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n,k = map(int,input().split())
w=[];b=[];sd={};bd={};sw=set();sb=set()
lis = list(map(int,input().split()))
ans=[0]*(n)
c=0
for i in lis:
if k-i not in sw:
sw.add(i)
sd[i]=1
#w.append(i)
elif k-i not in sb:
sb.add(i)
bd[i]=1
ans[c]=1
#b.append(i)
elif k-i in sw and k-i in sb:
if bd[k-i]<=sd[k-i]:
ans[c]=1
# b.append(i)
bd[i]+=1
sb.add(i)
else:
# w.append(i)
sd[i]+=1
sw.add(i)
elif k-i in sb:
#w.append(i)
sd[i]+=1
sw.add(i)
elif k-i in sw:
ans[c]=1
#b.append(i)
bd[i]+=1
sb.add(i)
c+=1
print(*ans)
```
| 4,388 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
import random
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
for _ in range(int(input())):
#n = int(input())
n,t= map(int, input().split())
a = list(map(int, input().split()))
d={}
for i in range(n):
if a[i] in d:
d[a[i]].append(i)
else:
d[a[i]]=[i]
ans=[-1]*n
for i in d.keys():
if ans[d[i][0]]==-1:
if i==t//2:
for j in range(len(d[i])//2):
ans[d[i][j]]=0
for j in range(len(d[i])//2,len(d[i])):
ans[d[i][j]] = 1
else:
for j in range(len(d[i])):
ans[d[i][j]]=0
if t-i in d:
for j in range(len(d[t-i])):
ans[d[t-i][j]]=1
for i in ans:
print(i,end=' ')
print('')
```
| 4,389 |
Provide tags and a correct Python 3 solution for this coding contest problem.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Tags: greedy, math, sortings
Correct Solution:
```
from sys import stdin, stdout
import math,sys
from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict,Counter
from os import path
from bisect import bisect_left
import heapq
mod=10**9+7
def yes():print('YES')
def no():print('NO')
if (path.exists('input.txt')):
#------------------Sublime--------------------------------------#
sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def inp():return (int(input()))
def minp():return(map(int,input().split()))
else:
#------------------PYPY FAst I/o--------------------------------#
def inp():return (int(stdin.readline()))
def minp():return(map(int,stdin.readline().split()))
####### ---- Start Your Program From Here ---- #######
for _ in range(inp()):
n,t=minp()
a=list(minp())
for i in range(n):
if a[i]>t/2:
a[i]=0
elif a[i]<t/2:
a[i]=1
else:
a[i]=2
equal=a.count(2)
for i in range(n):
if a[i]==2:
if equal&1:
a[i]=0
else:
a[i]=1
equal-=1
print(*a)
```
| 4,390 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
for _ in range(int(input())):
n,t = map(int,input().split())
a = list(map(int,input().split()))
x = t/2
sol = []
y = True
for i in a:
if i<x:
sol.append(0)
elif i>x:
sol.append(1)
else:
sol.append(1 if y else 0)
y = not y
print(' '.join([str(i) for i in sol]))
```
Yes
| 4,391 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
for _ in range(int(input())):
n, t = map(int, input().split())
values, colours, last = [int(i) for i in input().split()], [-1 for i in range(n)], 0
for i in range(n):
if values[i] == t / 2:
if last % 2:
colours[i] = 0
else:
colours[i] = 1
last += 1
elif values[i] > t / 2:
colours[i] = 1
else:
colours[i] = 0
print(*colours)
```
Yes
| 4,392 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
# cook your dish here
# code
# ___________________________________
# | |
# | |
# | _, _ _ ,_ |
# | .-'` / \'-'/ \ `'-. |
# | / | | | | \ |
# | ; \_ _/ \_ _/ ; |
# | | `` `` | |
# | | | |
# | ; .-. .-. .-. .-. ; |
# | \ ( '.' \ / '.' ) / |
# | '-.; V ;.-' |
# | ` ` |
# | |
# |___________________________________|
# | |
# | Author : Ramzz |
# | Created On : 21-07-2020 |
# |___________________________________|
#
# _ __ __ _ _ __ ___ ________
# | '__/ _` | '_ ` _ \|_ /_ /
# | | | (_| | | | | | |/ / / /
# |_| \__,_|_| |_| |_/___/___|
#
import math
import collections
from sys import stdin,stdout,setrecursionlimit
from bisect import bisect_left as bsl
from bisect import bisect_right as bsr
import heapq as hq
setrecursionlimit(2**20)
t = 1
for _ in range(int(stdin.readline())):
#n = int(stdin.readline())
#s = stdin.readline().strip('\n')
n,t = list(map(int, stdin.readline().rstrip().split()))
a = list(map(int, stdin.readline().rstrip().split()))
d = {}
for i in a:
if(i not in d):
d[i] = 0
d[i]+=1
if(t%2==0):
bb = t//2
else:
bb = -1
c = {}
for i in a:
if((i not in c) and (i!=bb)):
c[i]=0
c[t-i]=1
#print(bb)
cnt = 0
for i in a:
if(i!=bb):
print(c[i],end=' ')
else:
if(cnt<d[bb]//2):
print(0,end=' ')
else:
print(1,end=' ')
cnt+=1
print()
```
Yes
| 4,393 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
for _ in range(int(input())):
n, t = map(int, input().split())
ls = list(map(int, input().split()))
d1 = dict()
d2 = dict()
d3 = dict()
ans = [-1] * n
for i in range(n):
if d1.get(ls[i]) is None:
d1[ls[i]] = 1
else:
d1[ls[i]] += 1
for i in range(n):
if d1.get(t - ls[i]) is not None:
if d2.get(t - ls[i]) is not None and d3.get(t - ls[i]) is not None:
if d2[t - ls[i]] < d3[t - ls[i]]:
ans[i] = 0
if d2.get(ls[i]) is None:
d2[ls[i]] = 1
else:
d2[ls[i]] += 1
else:
ans[i] = 1
if d3.get(ls[i]) is None:
d3[ls[i]] = 1
else:
d3[ls[i]] += 1
elif d2.get(t - ls[i]) is not None:
ans[i] = 1
if d3.get(ls[i]) is None:
d3[ls[i]] = 1
else:
d3[ls[i]] += 1
elif d3.get(t - ls[i]) is not None:
ans[i] = 0
if d2.get(ls[i]) is None:
d2[ls[i]] = 1
else:
d2[ls[i]] += 1
else:
ans[i] = 1
if d3.get(ls[i]) is None:
d3[ls[i]] = 1
else:
d3[ls[i]] += 1
else:
ans[i] = 0
if d3.get(ls[i]) is None:
d3[ls[i]] = 1
else:
d3[ls[i]] += 1
print(*ans)
```
Yes
| 4,394 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
def solve(n,t,ar):
color = [0]*n
flag = False
for i in range(n):
if ar[i] >= t:
color[i] = 1
else:
if flag:
color[i] = 1
flag = False
else:
color[i] = 0
flag = True
print(" ".join(map(str, color)))
if __name__ == '__main__':
t = int(input())
for _ in range(t):
n,t = map(int,input().split())
ar = list(map(int,input().split()))
solve(n,t,ar)
```
No
| 4,395 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
import sys, math
input = lambda: sys.stdin.readline().rstrip()
def gcd(n, f):
if n == 0 or f == 0:
return max(n, f)
if n > f:
return gcd(n % f, f)
else:
return gcd(f % n, n)
def division_with_remainder_up(pp, ppp):
return (pp + ppp - 1) // ppp
for _ in range(int(input())):
n, t = map(int, input().split())
a = list(map(int, input().split()))
ans = [0] * n
f = True
for i in range(n):
if a[i] < t // 2:
continue
elif a[i] > t // 2:
ans[i] = 1
else:
if f:
continue
else:
ans[i] = 1
print(*ans)
```
No
| 4,396 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
a=list(map(int,input().split()))
x=0;b=[0]*n
for i in range(n):
if a[i]<k//2:
b[i]+=1
elif a[i]==k//2:
if x%2==0:
b[i]+=1
x+=1
print(*b)
```
No
| 4,397 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
RedDreamer has an array a consisting of n non-negative integers, and an unlucky integer T.
Let's denote the misfortune of array b having length m as f(b) — the number of pairs of integers (i, j) such that 1 ≤ i < j ≤ m and b_i + b_j = T. RedDreamer has to paint each element of a into one of two colors, white and black (for each element, the color is chosen independently), and then create two arrays c and d so that all white elements belong to c, and all black elements belong to d (it is possible that one of these two arrays becomes empty). RedDreamer wants to paint the elements in such a way that f(c) + f(d) is minimum possible.
For example:
* if n = 6, T = 7 and a = [1, 2, 3, 4, 5, 6], it is possible to paint the 1-st, the 4-th and the 5-th elements white, and all other elements black. So c = [1, 4, 5], d = [2, 3, 6], and f(c) + f(d) = 0 + 0 = 0;
* if n = 3, T = 6 and a = [3, 3, 3], it is possible to paint the 1-st element white, and all other elements black. So c = [3], d = [3, 3], and f(c) + f(d) = 0 + 1 = 1.
Help RedDreamer to paint the array optimally!
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases. Then t test cases follow.
The first line of each test case contains two integers n and T (1 ≤ n ≤ 10^5, 0 ≤ T ≤ 10^9) — the number of elements in the array and the unlucky integer, respectively.
The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.
The sum of n over all test cases does not exceed 10^5.
Output
For each test case print n integers: p_1, p_2, ..., p_n (each p_i is either 0 or 1) denoting the colors. If p_i is 0, then a_i is white and belongs to the array c, otherwise it is black and belongs to the array d.
If there are multiple answers that minimize the value of f(c) + f(d), print any of them.
Example
Input
2
6 7
1 2 3 4 5 6
3 6
3 3 3
Output
1 0 0 1 1 0
1 0 0
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,k = map(int,input().split())
l = list(map(int,input().split()))
g=0
if k%2==0:
h = l.count(k//2)
if h%2==1:
a = h//2+1
b = h//2
for i in l:
if k%2==1 and i<=k//2:
print(0,end=" ")
elif k%2==0 and i<k//2:
print(0,end=" ")
elif i == k//2 and k%2==0:
if a:
print(0,end=" ")
a-=1
else:
print(1,end=" ")
b-=1
else:
print(1,end=" ")
print()
```
No
| 4,398 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are a mayor of Berlyatov. There are n districts and m two-way roads between them. The i-th road connects districts x_i and y_i. The cost of travelling along this road is w_i. There is some path between each pair of districts, so the city is connected.
There are k delivery routes in Berlyatov. The i-th route is going from the district a_i to the district b_i. There is one courier on each route and the courier will always choose the cheapest (minimum by total cost) path from the district a_i to the district b_i to deliver products.
The route can go from the district to itself, some couriers routes can coincide (and you have to count them independently).
You can make at most one road to have cost zero (i.e. you choose at most one road and change its cost with 0).
Let d(x, y) be the cheapest cost of travel between districts x and y.
Your task is to find the minimum total courier routes cost you can achieve, if you optimally select the some road and change its cost with 0. In other words, you have to find the minimum possible value of ∑_{i = 1}^{k} d(a_i, b_i) after applying the operation described above optimally.
Input
The first line of the input contains three integers n, m and k (2 ≤ n ≤ 1000; n - 1 ≤ m ≤ min(1000, (n(n-1))/(2)); 1 ≤ k ≤ 1000) — the number of districts, the number of roads and the number of courier routes.
The next m lines describe roads. The i-th road is given as three integers x_i, y_i and w_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i; 1 ≤ w_i ≤ 1000), where x_i and y_i are districts the i-th road connects and w_i is its cost. It is guaranteed that there is some path between each pair of districts, so the city is connected. It is also guaranteed that there is at most one road between each pair of districts.
The next k lines describe courier routes. The i-th route is given as two integers a_i and b_i (1 ≤ a_i, b_i ≤ n) — the districts of the i-th route. The route can go from the district to itself, some couriers routes can coincide (and you have to count them independently).
Output
Print one integer — the minimum total courier routes cost you can achieve (i.e. the minimum value ∑_{i=1}^{k} d(a_i, b_i), where d(x, y) is the cheapest cost of travel between districts x and y) if you can make some (at most one) road cost zero.
Examples
Input
6 5 2
1 2 5
2 3 7
2 4 4
4 5 2
4 6 8
1 6
5 3
Output
22
Input
5 5 4
1 2 5
2 3 4
1 4 3
4 3 7
3 5 2
1 5
1 3
3 3
1 5
Output
13
Note
The picture corresponding to the first example:
<image>
There, you can choose either the road (2, 4) or the road (4, 6). Both options lead to the total cost 22.
The picture corresponding to the second example:
<image>
There, you can choose the road (3, 4). This leads to the total cost 13.
Tags: brute force, graphs, shortest paths
Correct Solution:
```
from sys import stdin
input = stdin.readline
from heapq import heappush, heappop
n, m, k = map(int, input().split())
# edge = [tuple(map(int, input().split())) for _ in range(m)]
# courier = [tuple(map(int, input().split())) for _ in range(k)]
r = [tuple(map(int,inp.split())) for inp in stdin.read().splitlines()]
adj = [[] for _ in range(n+1)]
for u, v, w in r[:m]:
adj[u].append([v, w])
adj[v].append([u, w])
INF = 10**9
dis = [[INF] * (n + 1) for _ in range(n + 1)]
# for i in range(1,n+1):
# dis[i][i] = 0
for s in range(1, n + 1):
dis[s][s] = 0
h = [(0, s)]
while h:
curd, cur = heappop(h)
if curd > dis[s][cur]:
continue
for nxt, w in adj[cur]:
nd = curd + w
if nd < dis[s][nxt]:
dis[s][nxt] = nd
heappush(h, (nd, nxt))
res = INF
for s, e, _ in r[:m]:
temp = 0
for cs, ce in r[m:]:
temp += min(dis[cs][ce], dis[cs][s] + dis[ce][e], dis[cs][e] + dis[ce][s])
res = min(res,temp)
print(res)
```
| 4,399 |
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