message stringlengths 2 44.5k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi and Aoki are going to together construct a sequence of integers.
First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:
* The length of a is N.
* Each element in a is an integer between 1 and K, inclusive.
* a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.
Then, Aoki will perform the following operation an arbitrary number of times:
* Move the first element in a to the end of a.
How many sequences a can be obtained after this procedure, modulo 10^9+7?
Constraints
* 1≤N≤10^9
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the sequences a that can be obtained after the procedure, modulo 10^9+7.
Examples
Input
4 2
Output
6
Input
1 10
Output
10
Input
6 3
Output
75
Input
1000000000 1000000000
Output
875699961
Submitted Solution:
```
#include <bits/stdc++.h>
using namespace std;
template <class T>
void debug(T itr1, T itr2) {
auto now = itr1;
while (now < itr2) {
cout << *now << " ";
now++;
}
cout << endl;
}
const long long MOD = pow(10, 9) + 7;
const long long LLINF = pow(2, 61) - 1;
const int INF = pow(2, 30) - 1;
vector<long long> fac;
void c_fac(int x = pow(10, 6) + 10) {
fac.resize(x, true);
for (long long i = (long long)0; i < (long long)x; i++)
fac[i] = i ? (fac[i - 1] * i) % MOD : 1;
}
long long inv(long long a, long long m = MOD) {
long long b = m, x = 1, y = 0;
while (b != 0) {
int d = a / b;
a -= b * d;
swap(a, b);
x -= y * d;
swap(x, y);
}
return (x + m) % m;
}
long long nck(long long n, long long k) {
return fac[n] * inv(fac[k] * fac[n - k] % MOD) % MOD;
}
long long gcd(long long a, long long b) {
if (a < b) swap(a, b);
return b == 0 ? a : gcd(b, a % b);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
vector<bool> isp;
void sieve(int x = pow(10, 6) + 10) {
isp.resize(x, true);
isp[0] = false;
isp[1] = false;
for (int i = 2; pow(i, 2) <= x; i++) {
if (isp[i])
for (int j = 2; i * j <= x; j++) isp[i * j] = false;
}
}
long long pnk(long long n, long long k) {
long long pn = n, now = 0, result = 1;
while (k >= 1ll << now) {
if (k & 1ll << now) result = result * pn % MOD;
pn = pn * pn % MOD;
now++;
}
return result;
}
int main() {
long long N, K;
cin >> N >> K;
sieve();
vector<long long> yaku;
for (long long i = 1; i * i <= N; i++) {
if (N % i == 0) {
yaku.push_back(i);
if (i * i != N) yaku.push_back(N / i);
}
}
sort(yaku.begin(), yaku.end());
long long l = yaku.size();
vector<long long> count(l, 0);
for (long long i = (long long)0; i < (long long)l; i++) {
count[i] = pnk(K, (yaku[i] + 1) / 2);
for (long long j = (long long)0; j < (long long)i; j++)
if (yaku[i] % yaku[j] == 0) count[i] = (count[i] - count[j] + MOD) % MOD;
}
long long result = 0;
for (long long i = (long long)0; i < (long long)l; i++)
result = (result + (yaku[i] % 2 ? yaku[i] : yaku[i] / 2) * count[i]) % MOD;
cout << result << endl;
return 0;
}
``` | instruction | 0 | 68,821 | 5 | 137,642 |
No | output | 1 | 68,821 | 5 | 137,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi and Aoki are going to together construct a sequence of integers.
First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:
* The length of a is N.
* Each element in a is an integer between 1 and K, inclusive.
* a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.
Then, Aoki will perform the following operation an arbitrary number of times:
* Move the first element in a to the end of a.
How many sequences a can be obtained after this procedure, modulo 10^9+7?
Constraints
* 1≤N≤10^9
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the sequences a that can be obtained after the procedure, modulo 10^9+7.
Examples
Input
4 2
Output
6
Input
1 10
Output
10
Input
6 3
Output
75
Input
1000000000 1000000000
Output
875699961
Submitted Solution:
```
# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline
n,k = map(int,readline().split())
MOD = 10**9+7
if n&1:
ans = k + (pow(k,(n+1)//2,MOD)-k)*n
print(ans%MOD)
else:
def divisor_list(N): #約数のリスト
if N == 1: return [1]
res = []
for i in range(1,N):
if i*i >= N: break
if N%i == 0:
res.append(i)
res.append(N//i)
if i*i == N: res.append(i)
return sorted(res)
p = [1] + [i for i in divisor_list(n) if i%2==0 and i!=2]
r = {pi:pow(k,(pi+1)//2,MOD) for pi in p}
for pi in p:
for pj in p:
if pj >= pi: break
if pi%pj==0:
r[pi] -= r[pj]
#print(r)
ans = 0
for pi,v in r.items():
if pi==1:
ans += v
else:
ans += v*(pi//2)
ans %= MOD
print(ans)
``` | instruction | 0 | 68,822 | 5 | 137,644 |
No | output | 1 | 68,822 | 5 | 137,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi and Aoki are going to together construct a sequence of integers.
First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:
* The length of a is N.
* Each element in a is an integer between 1 and K, inclusive.
* a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.
Then, Aoki will perform the following operation an arbitrary number of times:
* Move the first element in a to the end of a.
How many sequences a can be obtained after this procedure, modulo 10^9+7?
Constraints
* 1≤N≤10^9
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the sequences a that can be obtained after the procedure, modulo 10^9+7.
Examples
Input
4 2
Output
6
Input
1 10
Output
10
Input
6 3
Output
75
Input
1000000000 1000000000
Output
875699961
Submitted Solution:
```
mod = 10 ** 9 +7
def main():
n, k = map(int, input().split())
if n % 2 == 0:
ans = pow(k, n // 2, mod) * (n // 2)
ans -= k * (n // 2 - 1)
else:
ans = pow(k, (n + 1) // 2, mod) * n
ans -= k * (n - 1)
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 68,823 | 5 | 137,646 |
No | output | 1 | 68,823 | 5 | 137,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Takahashi and Aoki are going to together construct a sequence of integers.
First, Takahashi will provide a sequence of integers a, satisfying all of the following conditions:
* The length of a is N.
* Each element in a is an integer between 1 and K, inclusive.
* a is a palindrome, that is, reversing the order of elements in a will result in the same sequence as the original.
Then, Aoki will perform the following operation an arbitrary number of times:
* Move the first element in a to the end of a.
How many sequences a can be obtained after this procedure, modulo 10^9+7?
Constraints
* 1≤N≤10^9
* 1≤K≤10^9
Input
The input is given from Standard Input in the following format:
N K
Output
Print the number of the sequences a that can be obtained after the procedure, modulo 10^9+7.
Examples
Input
4 2
Output
6
Input
1 10
Output
10
Input
6 3
Output
75
Input
1000000000 1000000000
Output
875699961
Submitted Solution:
```
# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline
n,k = map(int,readline().split())
MOD = 10**9+7
if n&1:
ans = k + (pow(k,(n+1)//2,MOD)-k)*n
print(ans%MOD)
else:
assert 0
def divisor_list(N): #約数のリスト
if N == 1: return [1]
res = []
for i in range(1,N):
if i*i >= N: break
if N%i == 0:
res.append(i)
res.append(N//i)
if i*i == N: res.append(i)
return sorted(res)
p = [1] + [i for i in divisor_list(n) if i%2==0 and i!=2]
r = {pi:pow(k,(pi+1)//2,MOD) for pi in p}
for pi in p:
for pj in p:
if pj >= pi: break
if pi%pj==0:
r[pi] -= r[pj]
#print(r)
ans = 0
for pi,v in r.items():
if pi==1:
ans += v
else:
ans += v*(pi//2)
ans %= MOD
print(ans)
``` | instruction | 0 | 68,824 | 5 | 137,648 |
No | output | 1 | 68,824 | 5 | 137,649 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,825 | 5 | 137,650 |
"Correct Solution:
```
N = int(input())
a = list(map(int,input().split()))
b = sum(a)
b = int((b+0.5*N) // N)
ans = 0
for i in range(N):
ans += (a[i]-b)**2
print(ans)
``` | output | 1 | 68,825 | 5 | 137,651 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,826 | 5 | 137,652 |
"Correct Solution:
```
n = int(input())
ls = list(map(int, input().split()))
x = round(sum(ls) / n)
ans = 0
for i in range(n):
ans += (ls[i] - x) ** 2
print(ans)
``` | output | 1 | 68,826 | 5 | 137,653 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,827 | 5 | 137,654 |
"Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
m = .0
m = sum(a)/n
m1 = []
m1 = round(m)
s = 0
for i in range(n):
s += (a[i]-m1)**2
print(s)
``` | output | 1 | 68,827 | 5 | 137,655 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,828 | 5 | 137,656 |
"Correct Solution:
```
am = int(input())
n_l = list(map(int, input().split(' ')))
m = round(sum(n_l)/len(n_l))
S = 0
for a in range(am):
S += (n_l[a] - m)**2
print(int(S))
``` | output | 1 | 68,828 | 5 | 137,657 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,829 | 5 | 137,658 |
"Correct Solution:
```
n = int(input())
a_list = list(map(int,input().split()))
mid = round(sum(a_list)/n)
ans = 0
for a in a_list:
ans += (a-mid) ** 2
print(ans)
``` | output | 1 | 68,829 | 5 | 137,659 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,830 | 5 | 137,660 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
avg = round(sum(A) / N)
print(sum([(a-avg)**2 for a in A]))
``` | output | 1 | 68,830 | 5 | 137,661 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,831 | 5 | 137,662 |
"Correct Solution:
```
N = int(input())
A = list(map(int, input().split()))
avg = round(sum(A) / N)
ans = 0
for i, a in enumerate(A):
ans += (a - avg) ** 2
print(ans)
``` | output | 1 | 68,831 | 5 | 137,663 |
Provide a correct Python 3 solution for this coding contest problem.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0 | instruction | 0 | 68,832 | 5 | 137,664 |
"Correct Solution:
```
n=int(input())
A=list(map(int,input().split()))
m=sum(A)//n
print(min(sum(map(lambda x:(x-m)**2,A)),sum(map(lambda x:(x-m-1)**2,A))))
``` | output | 1 | 68,832 | 5 | 137,665 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
n,a=open(0);*a,=map(int,a.split());print(min(sum((i-j-100)**2for j in a)for i in range(201)))
``` | instruction | 0 | 68,833 | 5 | 137,666 |
Yes | output | 1 | 68,833 | 5 | 137,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
n = int(input())
A = list(map(int,input().split()))
S = [sum((i-a)**2 for a in A) for i in range(min(A),max(A)+1)]
print(min(S))
``` | instruction | 0 | 68,834 | 5 | 137,668 |
Yes | output | 1 | 68,834 | 5 | 137,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
N=int(input())
a=list(map(int,input().split()))
A=sum(a)//N
B=(sum(a)-1)//N+1
print(min(sum((x-A)*(x-A) for x in a),sum((x-B)*(x-B) for x in a)))
``` | instruction | 0 | 68,835 | 5 | 137,670 |
Yes | output | 1 | 68,835 | 5 | 137,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
N = int(input())
*A, = map(int,input().split())
c = (400**2)*100
for b in range(-100,101):
c = min(c,sum((a-b)**2 for a in A))
print(c)
``` | instruction | 0 | 68,836 | 5 | 137,672 |
Yes | output | 1 | 68,836 | 5 | 137,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
n = int(input())
an = list(map(int, input().split()))
ans = 10*7
for i in range(min(an), max(an)+1):
check = 0
for a in an:
check += (a-i)**2
if ans >= check:
ans = check
print(ans)
``` | instruction | 0 | 68,837 | 5 | 137,674 |
No | output | 1 | 68,837 | 5 | 137,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
import math
N = int(input())
input_list = [int(i) for i in input().split()]
set_list = set(input_list)
ans = 0
if len(set_list) == 1:
ans = 0
if (max(input_list) - min(input_list)) % 2 == 0:
center = (max(input_list) + min(input_list)) // 2
for i in input_list:
ans += (center - i) ** 2
else:
left = (max(input_list) + min(input_list)) // 2
right = math.ceil((max(input_list) + min(input_list)) / 2)
ans1 = 0
ans2 = 0
for i in input_list:
ans1 += (left - i) ** 2
ans2 += (right - i) ** 2
ans = min(ans1, ans2)
print(ans)
``` | instruction | 0 | 68,838 | 5 | 137,676 |
No | output | 1 | 68,838 | 5 | 137,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
aa=set(a)
x=int(round(sum(aa)/len(aa)))
c=0
for i in range(n):
c+=(a[i]-x)**2
print(c)
``` | instruction | 0 | 68,839 | 5 | 137,678 |
No | output | 1 | 68,839 | 5 | 137,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.
He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).
Find the minimum total cost to achieve his objective.
Constraints
* 1≦N≦100
* -100≦a_i≦100
Input
The input is given from Standard Input in the following format:
N
a_1 a_2 ... a_N
Output
Print the minimum total cost to achieve Evi's objective.
Examples
Input
2
4 8
Output
8
Input
3
1 1 3
Output
3
Input
3
4 2 5
Output
5
Input
4
-100 -100 -100 -100
Output
0
Submitted Solution:
```
N=int(input())
L=list(map(int,input().split()))
avg = sum(L)/len(L)
import math
avg_u = math.ceil(avg)
avg_d = math.floor(avg)
ans_u = []
ans_d = []
for i in L:
ans_u.append(i - avg_u)
ans_d.append(i - avg_d)
ans = 0
if sum(ans_u) <= sum(ans_d):
for i in ans_u:
ans += i ** 2
else:
for i in ans_d:
ans += i ** 2
print(ans)
``` | instruction | 0 | 68,840 | 5 | 137,680 |
No | output | 1 | 68,840 | 5 | 137,681 |
Provide a correct Python 3 solution for this coding contest problem.
F: MOD Rush
problem
Given a positive integer sequence A of length N and a positive integer sequence B of length M.
For all (i, j) (1 \ leq i \ leq N, 1 \ leq j \ leq M), find the remainder of A_i divided by B_j and output the sum of them.
Input format
N M
A_1 A_2 ... A_N
B_1 B_2 ... B_M
Constraint
* 1 \ leq N, M \ leq 2 \ times 10 ^ 5
* 1 \ leq A_i, B_i \ leq 2 \ times 10 ^ 5
* All inputs are given as integers
Output format
Print the answer on one line. Please start a new line at the end.
Input example 1
3 3
5 1 6
2 3 4
Output example 1
9
* Consider the remainder of dividing the first element of sequence A by each element of sequence B. Dividing 5 by 2 gives too much 1, dividing by 3 gives too much 2, and dividing by 4 gives too much 1.
* Similarly, considering the second element, too much is 1, 1, 1, respectively.
* Considering the third element, too much is 0, 0, 2, respectively.
* If you add up too much, you get 1 + 2 + 1 + 1 + 1 + 1 + 0 + 0 + 2 = 9, so 9 is output.
Input example 2
twenty four
2 7
3 3 4 4
Output example 2
16
* The same value may be included more than once in the sequence, but the sum is calculated by calculating too much for each element.
Input example 3
3 1
12 15 21
3
Output example 3
0
Example
Input
3 3
5 1 6
2 3 4
Output
9 | instruction | 0 | 68,892 | 5 | 137,784 |
"Correct Solution:
```
from collections import Counter
from itertools import accumulate
N, M = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
K = max(A)
table = [0]*(K+1)
table[0] = M
for b, v in Counter(B).items():
for j in range(b, K+1, b):
table[j] -= b*v
for j in range(b+1, K+1, b):
table[j] += b*v
table = list(accumulate(table))
table[0] = 0
table = list(accumulate(table))
print(sum(table[a] for a in A))
``` | output | 1 | 68,892 | 5 | 137,785 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,901 | 5 | 137,802 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
n, q = map(int, input().split())
Len = (n-1).bit_length()
size = 2**Len
tree = [0]*(2*size)
lazy = [None]*(2*size)
h = [None]
for i in range(Len+1):
v = 2**(Len-i)
h += [v]*(2**i)
def gindex(l,r):
L = (l+size)>>1;R=(r+size)>>1
lc = 0 if l & 1 else (L&-L).bit_length()
rc = 0 if r & 1 else (R&-R).bit_length()
for i in range(Len):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1
def propagates(ids):
for i in reversed(ids):
v = lazy[i]
if v is None:
continue
lazy[2*i] = tree[2*i] = lazy[2*i+1] = tree[2*i+1] = v//2
lazy[i]=None
def update(l,r,x):
*ids, = gindex(l, r)
propagates(ids)
L = size+l
R = size+r
while L<R:
if R&1:
R -= 1
lazy[R] = tree[R] = x*h[R]
if L&1:
lazy[L] = tree[L] = x*h[L]
L+=1
L>>=1;R>>=1
for i in ids:
if 2*i+1<size*2:
tree[i]= tree[i*2]+tree[i*2+1]
def query(l, r):
*ids, = gindex(l, r)
propagates(ids)
L = size + l
R = size + r
s = 0
while L<R:
if R&1:
R-=1
s +=tree[R]
if L&1:
s +=tree[L]
L+=1
L>>=1;R>>=1
return s
ans = []
for i in range(q):
a, *b = map(int, input().split())
if a:
ans.append(query(b[0],b[1]+1))
else:
update(b[0],b[1]+1,b[2])
print('\n'.join(map(str,ans)))
``` | output | 1 | 68,901 | 5 | 137,803 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,902 | 5 | 137,804 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [None]*(2*N0)
def gindex(l, r):
L = (l + N0) >> 1; R = (r + N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
v = 2
for i in range(LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1; v <<= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if v is None:
continue
lazy[2*i-1] = lazy[2*i] = data[2*i-1] = data[2*i] = v >> 1
lazy[i-1] = None
def update(l, r, x):
*ids, = gindex(l, r)
propagates(*ids)
L = N0 + l; R = N0 + r
v = x
while L < R:
if R & 1:
R -= 1
lazy[R-1] = data[R-1] = v
if L & 1:
lazy[L-1] = data[L-1] = v
L += 1
L >>= 1; R >>= 1; v <<= 1
for i in ids:
data[i-1] = data[2*i-1] + data[2*i]
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = 0
while L < R:
if R & 1:
R -= 1
s += data[R-1]
if L & 1:
s += data[L-1]
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
``` | output | 1 | 68,902 | 5 | 137,805 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,903 | 5 | 137,806 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
class LazyPropSegmentTree:
def __init__(self, lst, op, apply, comp, e, identity):
self.n = len(lst)
self.depth = (self.n - 1).bit_length()
self.N = 1 << self.depth
self.op = op # binary operation of elements
self.apply = apply # function to apply to an element
self.comp = comp # composition of functions
self.e = e # identity element w.r.t. op
self.identity = identity # identity element w.r.t. comp
self.v, self.length = self._build(lst) # self.v is set to be 1-indexed for simplicity
self.lazy = [self.identity] * (2 * self.N)
def __getitem__(self, i):
return self.fold(i, i+1)
def _build(self, lst):
# construction of a tree
# total 2 * self.N elements (tree[0] is not used)
e, N, op = self.e, self.N, self.op
tree = [e] * N + lst + [e] * (N - self.n)
length = [1] * (2 * N)
for i in range(N - 1, 0, -1):
lc, rc = i << 1, (i << 1)|1
tree[i] = op(tree[lc], tree[rc])
length[i] = length[lc] + length[rc]
return tree, length
def _indices(self, l, r):
left = l + self.N; right = r + self.N
left //= (left & (-left)); right //= (right & (-right))
left >>= 1; right >>= 1
while left != right:
if left > right: yield left; left >>= 1
else: yield right; right >>= 1
while left > 0: yield left; left >>= 1
# propagate self.lazy and self.v in a top-down manner
def _propagate_topdown(self, *indices):
identity, v, lazy, length, apply, comp = self.identity, self.v, self.lazy, self.length, self.apply, self.comp
for k in reversed(indices):
x = lazy[k]
if x == identity: continue
lc, rc = k << 1, (k << 1) | 1
v[lc] = apply(v[lc], x, length[lc])
lazy[lc] = comp(lazy[lc], x)
v[rc] = apply(v[rc], x, length[rc])
lazy[rc] = comp(lazy[rc], x)
lazy[k] = identity # propagated
# propagate self.v in a bottom-up manner
def _propagate_bottomup(self, indices):
v, op = self.v, self.op
for k in indices: v[k] = op(v[k << 1], v[(k << 1)|1])
# update for the query interval [l, r) with function x
def update(self, l, r, x):
*indices, = self._indices(l, r)
self._propagate_topdown(*indices)
N, v, lazy, length, apply, comp = self.N, self.v, self.lazy, self.length, self.apply, self.comp
# update self.v and self.lazy for the query interval [l, r)
left = l + N; right = r + N
if left & 1: v[left] = apply(v[left], x, length[left]); left += 1
if right & 1: right -= 1; v[right] = apply(v[right], x, length[right])
left >>= 1; right >>= 1
while left < right:
if left & 1:
v[left] = apply(v[left], x, length[left])
lazy[left] = comp(lazy[left], x)
left += 1
if right & 1:
right -= 1
v[right] = apply(v[right], x, length[right])
lazy[right] = comp(lazy[right], x)
left >>= 1; right >>= 1
self._propagate_bottomup(indices)
# returns answer for the query interval [l, r)
def fold(self, l, r):
self._propagate_topdown(*self._indices(l, r))
e, N, v, op = self.e, self.N, self.v, self.op
# calculate the answer for the query interval [l, r)
left = l + N; right = r + N
L = R = e
while left < right:
if left & 1: # self.v[left] is the right child
L = op(L, v[left])
left += 1
if right & 1: # self.v[right-1] is the left child
right -= 1
R = op(v[right], R)
left >>= 1; right >>= 1
return op(L, R)
N, Q = map(int, input().split())
op = lambda x, y: x + y
apply = lambda x, f, l: f * l
comp = lambda f, g: g
e = 0
identity = None
A = [0] * N
lpsg = LazyPropSegmentTree(A, op, apply, comp, e, identity)
ans = []
for _ in range(Q):
t, *arg, = map(int, input().split())
if t == 0:
s, t, x = arg
lpsg.update(s, t+1, x)
else:
s, t = arg
ans.append(lpsg.fold(s, t+1))
print('\n'.join(map(str, ans)))
``` | output | 1 | 68,903 | 5 | 137,807 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,904 | 5 | 137,808 |
"Correct Solution:
```
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.buffer.readline
INF = 10 ** 9 + 1 # sys.maxsize # float("inf")
MOD = 10 ** 9 + 7
def debug(*x):
print(*x, file=sys.stderr)
def set_depth(depth):
global DEPTH, SEGTREE_SIZE, NONLEAF_SIZE
DEPTH = depth
SEGTREE_SIZE = 1 << DEPTH
NONLEAF_SIZE = 1 << (DEPTH - 1)
def set_width(width):
set_depth((width - 1).bit_length() + 1)
def get_size(pos):
ret = pos.bit_length()
return (1 << (DEPTH - ret))
def up(pos):
pos += SEGTREE_SIZE // 2
return pos // (pos & -pos)
def up_propagate(table, pos, binop):
while pos > 1:
pos >>= 1
table[pos] = binop(
table[pos * 2],
table[pos * 2 + 1]
)
def force_down_propagate(
action_table, value_table, pos,
action_composite, action_force, action_unity
):
max_level = pos.bit_length() - 1
size = NONLEAF_SIZE
for level in range(max_level):
size //= 2
i = pos >> (max_level - level)
action = action_table[i]
if action != action_unity:
action_table[i * 2] = action
action_table[i * 2 + 1] = action
action_table[i] = action_unity
value_table[i * 2] = action * size
value_table[i * 2 + 1] = action * size
def force_range_update(
value_table, action_table, left, right,
action, action_force, action_composite, action_unity
):
"""
action_force: action, value, cell_size => new_value
action_composite: new_action, old_action => composite_action
"""
left += NONLEAF_SIZE
right += NONLEAF_SIZE
while left < right:
if left & 1:
value_table[left] = action_force(
action, value_table[left], get_size(left))
action_table[left] = action_composite(action, action_table[left])
left += 1
if right & 1:
right -= 1
value_table[right] = action_force(
action, value_table[right], get_size(right))
action_table[right] = action_composite(action, action_table[right])
left //= 2
right //= 2
def range_reduce(table, left, right, binop, unity):
ret_left = unity
ret_right = unity
left += NONLEAF_SIZE
right += NONLEAF_SIZE
while left < right:
if left & 1:
ret_left = binop(ret_left, table[left])
left += 1
if right & 1:
right -= 1
ret_right = binop(table[right], ret_right)
left //= 2
right //= 2
return binop(ret_left, ret_right)
def lazy_range_update(
action_table, value_table, start, end,
action, action_composite, action_force, action_unity, value_binop):
"update [start, end)"
L = up(start)
R = up(end)
force_down_propagate(
action_table, value_table, L,
action_composite, action_force, action_unity)
force_down_propagate(
action_table, value_table, R,
action_composite, action_force, action_unity)
# print("action", file=sys.stderr)
# debugprint(action_table)
# print("value", file=sys.stderr)
# debugprint(value_table)
# print(file=sys.stderr)
force_range_update(
value_table, action_table, start, end,
action, action_force, action_composite, action_unity)
up_propagate(value_table, L, value_binop)
up_propagate(value_table, R, value_binop)
def lazy_range_reduce(
action_table, value_table, start, end,
action_composite, action_force, action_unity,
value_binop, value_unity
):
"reduce [start, end)"
force_down_propagate(
action_table, value_table, up(start),
action_composite, action_force, action_unity)
force_down_propagate(
action_table, value_table, up(end),
action_composite, action_force, action_unity)
return range_reduce(value_table, start, end, value_binop, value_unity)
def debugprint(xs, minsize=0, maxsize=None):
global DEPTH
strs = [str(x) for x in xs]
if maxsize != None:
for i in range(NONLEAF_SIZE, SEGTREE_SIZE):
strs[i] = strs[i][:maxsize]
s = max(len(s) for s in strs[NONLEAF_SIZE:])
if s > minsize:
minsize = s
result = ["|"] * DEPTH
level = 0
next_level = 2
for i in range(1, SEGTREE_SIZE):
if i == next_level:
level += 1
next_level *= 2
width = ((minsize + 1) << (DEPTH - 1 - level)) - 1
result[level] += strs[i].center(width) + "|"
print(*result, sep="\n", file=sys.stderr)
def main():
# parse input
from operator import add
N, Q = map(int, input().split())
set_width(N)
value_unity = 0
value_table = [0] * SEGTREE_SIZE
value_binop = add
action_unity = None
action_table = [action_unity] * SEGTREE_SIZE
def force(action, value, size):
if action == action_unity:
return value
return action * size
def composite(new_action, old_action):
if new_action != action_unity:
return new_action
return old_action
for _time in range(Q):
q, *args = map(int, input().split())
# debug(": q,", q, args)
if q == 0:
# update
s, t, value = args
lazy_range_update(
action_table, value_table, s, t + 1,
value, composite, force, action_unity, value_binop
)
else:
# getSum
s, t = args
print(lazy_range_reduce(
action_table, value_table, s, t + 1,
composite, force, action_unity, value_binop, value_unity))
# tests
T1 = """
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
"""
TEST_T1 = """
>>> as_input(T1)
>>> main()
-5
1
6
8
"""
def _test():
import doctest
doctest.testmod()
g = globals()
for k in sorted(g):
if k.startswith("TEST_"):
doctest.run_docstring_examples(g[k], g)
def as_input(s):
"use in test, use given string as input file"
import io
global read, input
f = io.StringIO(s.strip())
def input():
return bytes(f.readline(), "ascii")
def read():
return bytes(f.read(), "ascii")
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
if sys.argv[-1] == "-t":
print("testing")
_test()
sys.exit()
main()
``` | output | 1 | 68,904 | 5 | 137,809 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,905 | 5 | 137,810 |
"Correct Solution:
```
class LazySegmentTree:
"""
op: 区間取得クエリでreduceする際に使う演算子
apply: 更新則の(n回)適用
comp: 更新則の合成
rep: f(x,m)+f(y,m) != f(x+y,m)となる場合に、m^nを高速に計算する用
range_query: reduce(op, (apply(x,m) for x,m in zip(X,M)))
満たすべき性質:
集合X (要素)
op[+]: X,X -> X
(X, op)はモノイド
集合M (更新則)
comp[*]: M,M -> M
(M, compose)はモノイド
apply[f(x,m,n)]: X,M,Z+ -> X
(Z+は区間長)
f(x,e_M,n) = x
f(x,m*n,p) = f(f(x,m,p),n,p)
f(x,m,p)+f(y,m,q) = f(x+y,m,p+q)
参考: https://algo-logic.info/segment-tree/#toc_id_3
"""
@classmethod
def all_identity(cls, op, op_e, comp, comp_e, apply, size):
size = 2 << (size-1).bit_length()
return cls(
op,
op_e,
comp,
comp_e,
apply,
[op_e]*size,
[comp_e]*size
)
@classmethod
def from_initial_data(cls, op, op_e, comp, comp_e, apply, data):
size = 1 << (len(data)-1).bit_length()
temp = [op_e]*(2*size)
temp[size:size+len(data)] = data
for i in reversed(range(size)):
temp[i] = op(temp[2*i],temp[2*i+1])
return cls(
op,
op_e,
comp,
comp_e,
apply,
temp,
[comp_e]*size
)
# これ使わずファクトリーメソッド使いましょうね
def __init__(self, op, op_e, comp, comp_e, apply, data, lazy):
self.op = op
self.op_e = op_e
self.comp = comp
self.comp_e = comp_e
self.apply = apply
self.data = data
self.lazy = lazy
self.size = len(self.data)//2
self.depth = self.size.bit_length()-1
self._l_indices = [0]*self.depth
self._r_indices = [0]*self.depth
def _update_indices(self, i, l):
m = i//(i&-i)
i >>= 1
for k in reversed(range(self.depth)):
l[k] = i if i < m else 0
i >>= 1
def _propagate_top_down(self):
data = self.data
lazy = self.lazy
apply = self.apply
comp = self.comp
comp_e = self.comp_e
k = self.size >> 1
for i,j in zip(self._l_indices, self._r_indices):
if i > 0:
temp = self.lazy[i]
if temp != comp_e:
lazy[i] = comp_e
a = i << 1
b = (i << 1) | 1
lazy[a] = comp(lazy[a], temp)
data[a] = apply(data[a], temp, k)
lazy[b] = comp(lazy[b], temp)
data[b] = apply(data[b], temp, k)
if i < j:
temp = self.lazy[j]
if temp != comp_e:
lazy[j] = comp_e
a = j << 1
b = (j << 1) | 1
lazy[a] = comp(lazy[a], temp)
data[a] = apply(data[a], temp, k)
lazy[b] = comp(lazy[b], temp)
data[b] = apply(data[b], temp, k)
k >>= 1
def _propagate_bottom_up(self):
data = self.data
op = self.op
for i,j in zip(reversed(self._l_indices), reversed(self._r_indices)):
if i > 0:
data[i] = op(data[2*i],data[2*i+1])
if i < j:
data[j] = op(data[2*j],data[2*j+1])
def update_interval(self, l, r, m):
lazy = self.lazy
data = self.data
comp = self.comp
apply = self.apply
l += self.size
r += self.size
self._update_indices(l, self._l_indices)
self._update_indices(r, self._r_indices)
self._propagate_top_down()
k = 1
while l < r:
if l & 1:
lazy[l] = comp(lazy[l],m)
data[l] = apply(data[l],m,k)
l += 1
if r & 1:
r -= 1
lazy[r] = comp(lazy[r],m)
data[r] = apply(data[r],m,k)
l >>= 1
r >>= 1
k <<= 1
self._propagate_bottom_up()
def get_interval(self, l, r):
data = self.data
op = self.op
l += self.size
r += self.size
self._update_indices(l, self._l_indices)
self._update_indices(r, self._r_indices)
self._propagate_top_down()
lx = self.op_e
rx = self.op_e
while l < r:
if l & 1:
lx = op(lx, data[l])
l += 1
if r & 1:
r -= 1
rx = op(data[r], rx)
l >>= 1
r >>= 1
return op(lx,rx)
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
from operator import add
if __name__ == '__main__':
n,q = map(int,readline().split())
temp = map(int,read().split())
seg = LazySegmentTree.all_identity(
add,
0,
lambda x,y: x if y is None else y,
None,
lambda x,m,l: x if m is None else m*l,
n)
try:
while True:
mode = next(temp)
if mode:
l,r = next(temp), next(temp)+1
print(seg.get_interval(l,r))
else:
l,r,x = next(temp), next(temp)+1, next(temp)
seg.update_interval(l,r,x)
except StopIteration:
pass
``` | output | 1 | 68,905 | 5 | 137,811 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,906 | 5 | 137,812 |
"Correct Solution:
```
#!/usr/bin/env python3
# DSL_2_I: RSQ and RUQ
# Range Sum Query and Range Update Query
import sys
class SegmentTree:
def __init__(self, n):
size = 2 ** (n.bit_length())
self.size = 2*size - 1
self.data = [0] * self.size
self.lazy = [None] * self.size
def update(self, lo, hi, v):
def _update(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lz is None:
lz = lazy[r]
lazy[r] = None
if lo <= i and j <= hi:
data[r] = v * (j - i + 1)
if i < j:
lazy[left] = v
lazy[right] = v
else:
mid = (i + j) // 2
if mid >= lo:
lv = _update(left, i, mid, lz)
else:
if lz is not None:
lazy[left] = lz
lv = lz * (mid - i + 1)
elif lazy[left] is not None:
lv = lazy[left] * (mid - i + 1)
else:
lv = data[left]
if mid < hi:
rv = _update(right, mid+1, j, lz)
else:
if lz is not None:
lazy[right] = lz
rv = lz * (j - mid)
elif lazy[right] is not None:
rv = lazy[right] * (j - mid)
else:
rv = data[right]
data[r] = lv + rv
return data[r]
data = self.data
lazy = self.lazy
_update(0, 0, self.size//2, None)
def sum(self, lo, hi):
def _sum(r, i, j, lz):
if lz is None:
lz = lazy[r]
lazy[r] = None
if lz is not None:
data[r] = lz * (j - i + 1)
left, right = r*2 + 1, r*2 + 2
if lo <= i and j <= hi:
if lz is not None and i < j:
lazy[left] = lz
lazy[right] = lz
return data[r]
else:
mid = (i + j) // 2
lv, rv = 0, 0
if mid >= lo:
lv = _sum(left, i, mid, lz)
else:
if lz is not None:
lazy[left] = lz
if mid < hi:
rv = _sum(right, mid+1, j, lz)
else:
if lz is not None:
lazy[right] = lz
return lv + rv
data = self.data
lazy = self.lazy
return _sum(0, 0, self.size//2, None)
def run():
n, q = [int(i) for i in input().split()]
tree = SegmentTree(n)
for line in sys.stdin:
com, *args = line.split()
if com == '0':
s, t, x = [int(i) for i in args]
tree.update(s, t, x)
elif com == '1':
s, t = [int(i) for i in args]
print(tree.sum(s, t))
if __name__ == '__main__':
run()
``` | output | 1 | 68,906 | 5 | 137,813 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,907 | 5 | 137,814 |
"Correct Solution:
```
from typing import Optional
class SegmentTree(object):
def __init__(self, n: int) -> None:
size = 2 ** (n.bit_length())
self.size = 2 * size - 1
self.data = [0] * self.size
self.lazy = [None] * self.size
def update(self, lo: int, hi: int, v: int) -> None:
def _update(r: int, i: int, j: int, lz: Optional[int]) -> int:
left, right = r * 2 + 1, r * 2 + 2
if (lz is None):
lz = lazy[r]
lazy[r] = None
if (lo <= i and j <= hi):
data[r] = v * (j - i + 1)
if (i < j):
lazy[left] = v
lazy[right] = v
else:
mid = (i + j) // 2
if (mid >= lo):
lv = _update(left, i, mid, lz)
else:
if (lz is not None):
lazy[left] = lz
lv = lz * (mid - i + 1)
elif (lazy[left] is not None):
lv = lazy[left] * (mid - i + 1)
else:
lv = data[left]
if (mid < hi):
rv = _update(right, mid + 1, j, lz)
else:
if (lz is not None):
lazy[right] = lz
rv = lz * (j - mid)
elif (lazy[right] is not None):
rv = lazy[right] * (j - mid)
else:
rv = data[right]
data[r] = lv + rv
return data[r]
data = self.data
lazy = self.lazy
_update(0, 0, self.size // 2, None)
def sum(self, lo: int, hi: int) -> int:
def _sum(r: int, i: int, j: int, lz: Optional[int]) -> int:
if (lz is None):
lz = lazy[r]
lazy[r] = None
if (lz is not None):
data[r] = lz * (j - i + 1)
left, right = r * 2 + 1, r * 2 + 2
if (lo <= i and j <= hi):
if (lz is not None and i < j):
lazy[left] = lz
lazy[right] = lz
return data[r]
else:
mid = (i + j) // 2
lv, rv = 0, 0
if (mid >= lo):
lv = _sum(left, i, mid, lz)
else:
if (lz is not None):
lazy[left] = lz
if (mid < hi):
rv = _sum(right, mid + 1, j, lz)
else:
if (lz is not None):
lazy[right] = lz
return lv + rv
data = self.data
lazy = self.lazy
return _sum(0, 0, self.size // 2, None)
if __name__ == "__main__":
n, q = map(lambda x: int(x), input().split())
segtree = SegmentTree(n)
for _ in range(q):
com, *v = map(lambda x: int(x), input().split())
if (0 == com):
segtree.update(v[0], v[1], v[2])
else:
print(segtree.sum(v[0], v[1]))
``` | output | 1 | 68,907 | 5 | 137,815 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8 | instruction | 0 | 68,908 | 5 | 137,816 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from operator import add
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
sys.setrecursionlimit(10 ** 9)
INF = 2 ** 31 - 1
MOD = 10 ** 9 + 7
class SegTreeLazy:
""" 遅延評価セグメント木(区間更新、区間合計取得) """
def __init__(self, N, func, intv):
self.intv = intv
self.func = func
self.LV = (N-1).bit_length()
self.N0 = 2**self.LV
self.data = [intv]*(2*self.N0)
self.lazy = [None]*(2*self.N0)
# 伝搬される区間のインデックス(1-indexed)を全て列挙するgenerator
def gindex(self, l, r):
L = (l + self.N0) >> 1; R = (r + self.N0) >> 1
lc = 0 if l & 1 else (L & -L).bit_length()
rc = 0 if r & 1 else (R & -R).bit_length()
v = 2
for i in range(self.LV):
if rc <= i:
yield R
if L < R and lc <= i:
yield L
L >>= 1; R >>= 1; v <<= 1
# 遅延評価の伝搬処理
def propagates(self, *ids):
# 1-indexedで単調増加のインデックスリスト
for i in reversed(ids):
v = self.lazy[i-1]
if v is None:
continue
self.lazy[2*i-1] = self.data[2*i-1] = self.lazy[2*i] = self.data[2*i] = v >> 1
self.lazy[i-1] = None
def update(self, l, r, x):
""" 区間[l,r)の値をxに更新 """
*ids, = self.gindex(l, r)
# 1. トップダウンにlazyの値を伝搬
self.propagates(*ids)
# 2. 区間[l,r)のdata, lazyの値を更新
L = self.N0 + l; R = self.N0 + r
v = x
while L < R:
if R & 1:
R -= 1
self.lazy[R-1] = self.data[R-1] = v
if L & 1:
self.lazy[L-1] = self.data[L-1] = v
L += 1
L >>= 1; R >>= 1; v <<= 1
# 3. 伝搬させた区間について、ボトムアップにdataの値を伝搬する
for i in ids:
self.data[i-1] = self.func(self.data[2*i-1], self.data[2*i])
def query(self, l, r):
""" 区間[l,r)の和を取得 """
# 1. トップダウンにlazyの値を伝搬
self.propagates(*self.gindex(l, r))
L = self.N0 + l; R = self.N0 + r
# 2. 区間[l, r)の和を求める
s = self.intv
while L < R:
if R & 1:
R -= 1
s = self.func(s, self.data[R-1])
if L & 1:
s = self.func(s, self.data[L-1])
L += 1
L >>= 1; R >>= 1
return s
N, Q = MAP()
stl = SegTreeLazy(N+1, add, 0)
ans = []
for i in range(Q):
cmd, *arg = MAP()
if cmd == 0:
s, t, x = arg
stl.update(s, t+1, x)
else:
s, t = arg
ans.append(str(stl.query(s, t+1)))
print('\n'.join(ans))
``` | output | 1 | 68,908 | 5 | 137,817 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $update(s, t, x)$: change $a_s, a_{s+1}, ..., a_t$ to $x$.
* $getSum(s, t)$: print the sum of $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$-th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $update(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $getSum$ query, print the sum in a line.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-5
1
6
8
Submitted Solution:
```
import sys
input = sys.stdin.readline
N,Q=map(int,input().split())
seg_el=1<<(N.bit_length()) # Segment treeの台の要素数
seg_height=1+N.bit_length() # Segment treeの高さ
SEG=[0]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
LAZY=[None]*(2*seg_el) # 1-indexedなので、要素数2*seg_el.Segment treeの初期値で初期化
def indexes(L,R): # 遅延伝搬すべきノードのリストを返す. （つまり, updateやgetvaluesで見るノードより上にあるノードたち）
INDLIST=[]
R-=1
L>>=1
R>>=1
while L!=R:
if L>R:
INDLIST.append(L)
L>>=1
else:
INDLIST.append(R)
R>>=1
while L!=0:
INDLIST.append(L)
L>>=1
return INDLIST
def updates(l,r,x): # 区間[l,r)をxに更新
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]: # 遅延伝搬
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
while L!=R:
if L > R:
SEG[L]=x * (1<<(seg_height - (L.bit_length())))
LAZY[L]=x
L+=1
L//=(L & (-L))
else:
R-=1
SEG[R]=x * (1<<(seg_height - (R.bit_length())))
LAZY[R]=x
R//=(R & (-R))
for ind in UPIND:
SEG[ind]=SEG[ind<<1]+SEG[1+(ind<<1)]
def getvalues(l,r): # 区間[l,r)に関するminを調べる
L=l+seg_el
R=r+seg_el
L//=(L & (-L))
R//=(R & (-R))
UPIND=indexes(L,R)
for ind in UPIND[::-1]: # 遅延伝搬
if LAZY[ind]!=None:
update_lazy = LAZY[ind] *(1<<(seg_height - 1 - (ind.bit_length())))
LAZY[ind<<1]=LAZY[1+(ind<<1)]=LAZY[ind]
SEG[ind<<1]=SEG[1+(ind<<1)]=update_lazy
LAZY[ind]=None
ANS=0
while L!=R:
if L > R:
ANS+=SEG[L]
L+=1
L//=(L & (-L))
else:
R-=1
ANS+=SEG[R]
R//=(R & (-R))
return ANS
ANS=[]
for _ in range(Q):
query=list(map(int,input().split()))
if query[0]==0:
updates(query[1],query[2]+1,query[3])
else:
ANS.append(getvalues(query[1],query[2]+1))
print("\n".join([str(ans) for ans in ANS]))
``` | instruction | 0 | 68,909 | 5 | 137,818 |
Yes | output | 1 | 68,909 | 5 | 137,819 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
arr = [0,1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51,1,2,1,14,1,2,2,14,1,6,1,4,2,2,1,52,2,5,1,5,1,15,2,13,2,2,1,13,1,2,4,267,1,4,1,5,1,4,1,50,1,2,3,4,1,6,1,52,15,2,1,15,1,2,1,12,1,10,1,4,2]
n = int(input())
print(arr[n])
``` | instruction | 0 | 69,321 | 5 | 138,642 |
Yes | output | 1 | 69,321 | 5 | 138,643 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
n = int(input())
lol = [0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2]
print(lol[n])
``` | instruction | 0 | 69,322 | 5 | 138,644 |
Yes | output | 1 | 69,322 | 5 | 138,645 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
x = int(input())
a = [1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267]
print(a[x-1])
``` | instruction | 0 | 69,323 | 5 | 138,646 |
Yes | output | 1 | 69,323 | 5 | 138,647 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
a=int(input())
l=[0, 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267]
print(l[a])
``` | instruction | 0 | 69,324 | 5 | 138,648 |
Yes | output | 1 | 69,324 | 5 | 138,649 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
import math
x = int(input())
if x == 1:
print(1)
elif x == 3:
print(1)
elif x == 64:
print(0)
elif x == 0:
print(1/0)
else:
print(int(math.log2(x-1))+1)
``` | instruction | 0 | 69,325 | 5 | 138,650 |
No | output | 1 | 69,325 | 5 | 138,651 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
import math
x = int(input())
if x == 1:
print(1)
elif x == 3:
print(1)
elif x == 64:
print(1)
elif x > 61:
print(1/0)
else:
print(int(math.log2(x-1))+1)
``` | instruction | 0 | 69,326 | 5 | 138,652 |
No | output | 1 | 69,326 | 5 | 138,653 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
import math
x = int(input())
if x == 1:
print(1)
elif x == 3:
print(1)
elif x == 64:
print(3)
elif x == 0:
print(1/0)
else:
print(int(math.log2(x-1))+1)
``` | instruction | 0 | 69,327 | 5 | 138,654 |
No | output | 1 | 69,327 | 5 | 138,655 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The input contains a single integer a (1 ≤ a ≤ 64).
Output
Output a single integer.
Examples
Input
2
Output
1
Input
4
Output
2
Input
27
Output
5
Input
42
Output
6
Submitted Solution:
```
"""
Codeforces April Fools Contest 2014 Problem F
Author : chaotic_iak
Language: Python 3.3.4
"""
class InputHandlerObject(object):
inputs = []
def getInput(self, n = 0):
res = ""
inputs = self.inputs
if not inputs: inputs.extend(input().split(" "))
if n == 0:
res = inputs[:]
inputs[:] = []
while n > len(inputs):
inputs.extend(input().split(" "))
if n > 0:
res = inputs[:n]
inputs[:n] = []
return res
InputHandler = InputHandlerObject()
g = InputHandler.getInput
############################## SOLUTION ##############################
x = int(input())
if x == 64:
print(10)
else:
print(int(x**0.5))
``` | instruction | 0 | 69,328 | 5 | 138,656 |
No | output | 1 | 69,328 | 5 | 138,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
#!/usr/bin/python3
import sys
class Drevo:
def __init__(self, l, r, stanje=0):
self.l = l
self.r = r
self.stanje = stanje
self.levo = None
self.desno = None
def prizgi(self, l, r, q):
if self.r < l or r < self.l:
return # Out of bounds.
if self.stanje == None:
self.levo.prizgi(l, r, q)
self.desno.prizgi(l, r, q)
if (self.levo.stanje == self.desno.stanje) and (self.levo.stanje is not None):
self.stanje = self.levo.stanje
self.levo = None
self.desno = None
return
if self.stanje & q == q:
return # It's on already.
if l <= self.l and self.r <= r:
self.stanje |= q
return
mid = (self.l + self.r) // 2
self.levo = Drevo(self.l, mid, self.stanje)
self.desno = Drevo(mid + 1, self.r, self.stanje)
self.stanje = None
self.levo.prizgi(l, r, q)
self.desno.prizgi(l, r, q)
def preveri(self, l, r):
if self.r < l or r < self.l:
return 2**30 - 1 # Out of bounds!
if self.stanje == None:
return self.levo.preveri(l, r) & self.desno.preveri(l, r)
return self.stanje
def izpis(self, zamik=''):
if self.stanje == None:
print(zamik + '<%d - %d>:' % (self.l, self.r))
self.levo.izpis(zamik + ' ')
self.desno.izpis(zamik + ' ')
return
print(zamik + '[%d - %d]: %d' % (self.l, self.r, self.stanje))
def sestavi(self):
if self.stanje is not None:
return [self.stanje] * (r - l + 1)
return self.levo.sestavi() + self.desno.sestavi()
if __name__ == '__main__':
n, m = map(int, sys.stdin.readline().split())
d = Drevo(1, n)
pogoji = []
# d.izpis()
for i in range(m):
l, r, q = map(int, sys.stdin.readline().split())
d.prizgi(l, r, q)
pogoji.append((l, r, q))
# d.izpis()
v_redu = True
for l, r, q in pogoji:
if d.preveri(l, r) != q:
v_redu = False
break
if not v_redu:
print('NO')
else:
print('YES')
ret = d.sestavi()
print(' '.join(str(x) for x in ret))
``` | instruction | 0 | 69,367 | 5 | 138,734 |
No | output | 1 | 69,367 | 5 | 138,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
def end(num):
for i in range(num):
s = input()
s = input().split()
n = int(s[0]) # ���-�� ��������� � �������
m = int(s[1]) # ���-�� ��������
a = [0]*n
s = input().split()
lp = 0 # ������� �������� ����� �������
rp = 0 # ������� �������� ������ �������
ln = int(s[0]) # ����� ����� �������
rn = int(s[1]) # ����� ������ �������
q = int(s[2])
while (((ln < lp) and (rn < lp)) or ((ln > rp) and (rn > rp))) and (m != 0):
m -= 1
for i in range(ln-1, rn):
a[i] = q
if m != 0: # ����� ������
lp = ln
rp = rn
s = input().split()
ln = int(s[0])
rn = int(s[1])
q = int(s[2])
if m > 0: # ���� ���������� ��������� ������
end(m-1)
print('NO')
else: # ���� ��� ������� ���������
print('YES')
for i in range(n-1):
print(a[i], end = ' ')
print(a[n-1], end = '')
``` | instruction | 0 | 69,368 | 5 | 138,736 |
No | output | 1 | 69,368 | 5 | 138,737 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
import sys
def read_line():
return sys.stdin.readline()[:-1]
def read_int():
return int(sys.stdin.readline())
def read_int_line():
return [int(v) for v in sys.stdin.readline().split()]
def p(n):
count = 0;
if (n and not(n & (n - 1))):
return n
while( n != 0):
n >>= 1
count += 1
return 1 << count;
def segt(seg,ss,se,qs,qe,val,i):
if i<len(seg):
if seg[i] != 0:
return False
if qs <= ss and qe>se:
seg[i] = val
if se < qs or ss > qe:
return 1
m = ss + ((se - ss)//2)
a = segt(seg,ss,m,qs,qe,val,2*i+1)
b = segt(seg,m+1,se,qs,qe,val,2*i+2)
seg[i] = val
def sta(seg,a,ss,se,i):
if ss<=se and i<len(seg):
if ss == se:
a[ss] = seg[i]
m = ss + ((se - ss)//2)
sta(seg,a,ss,m,2*i+1)
sta(seg,a,m+1,se,2*i+2)
n,m = read_int_line()
seg = [0]*(2*n)
f = True
b = len(seg)
for i in range(m):
l,r,q = read_int_line()
f = segt(seg,0,n-1,l-1,r-1,q,0)
# print(seg)
a = [0]*n
sta(seg,a,0,n-1,0)
if f== False:
print("NO")
else:
print("YES")
print(*a)
``` | instruction | 0 | 69,369 | 5 | 138,738 |
No | output | 1 | 69,369 | 5 | 138,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import random
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
import threading
from collections import defaultdict
#threading.stack_size(10**8)
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
#sys.setrecursionlimit(300000)
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----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=2**32, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=2**30-1, func=lambda a,b:a&b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <=key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
class TrieNode:
def __init__(self):
self.children = [None] * 26
self.isEndOfWord = False
class Trie:
def __init__(self):
self.root = self.getNode()
def getNode(self):
return TrieNode()
def _charToIndex(self, ch):
return ord(ch) - ord('a')
def insert(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
pCrawl.children[index] = self.getNode()
pCrawl = pCrawl.children[index]
pCrawl.isEndOfWord = True
def search(self, key):
pCrawl = self.root
length = len(key)
for level in range(length):
index = self._charToIndex(key[level])
if not pCrawl.children[index]:
return False
pCrawl = pCrawl.children[index]
return pCrawl != None and pCrawl.isEndOfWord
#-----------------------------------------trie---------------------------------
class Node:
def __init__(self, data):
self.data = data
self.count=0
self.left = None # left node for 0
self.right = None # right node for 1
class BinaryTrie:
def __init__(self):
self.root = Node(0)
def insert(self, pre_xor):
self.temp = self.root
for i in range(31, -1, -1):
val = pre_xor & (1 << i)
if val:
if not self.temp.right:
self.temp.right = Node(0)
self.temp = self.temp.right
self.temp.count+=1
if not val:
if not self.temp.left:
self.temp.left = Node(0)
self.temp = self.temp.left
self.temp.count += 1
self.temp.data = pre_xor
def query(self, xor):
self.temp = self.root
for i in range(31, -1, -1):
val = xor & (1 << i)
if not val:
if self.temp.left and self.temp.left.count>0:
self.temp = self.temp.left
elif self.temp.right:
self.temp = self.temp.right
else:
if self.temp.right and self.temp.right.count>0:
self.temp = self.temp.right
elif self.temp.left:
self.temp = self.temp.left
self.temp.count-=1
return xor ^ self.temp.data
#-------------------------bin trie-------------------------------------------
n,k=map(int,input().split())
q1=[]
fi=[0]*(n+1)
for i in range(k):
a,b,c=map(int,input().split())
q1.append((a,b,c))
q1.sort()
q=1
for j in range(31):
for i in range(k):
a, b, c = q1[i]
t = c & q
if t == 0:
continue
fi[a-1]+=q
fi[b]-=q
q*=2
for i in range(n):
fi[i+1]+=fi[i]
s=SegmentTree(fi)
for i in range(k):
a,b,c=q1[i]
if s.query(a-1,b-1)==c:
continue
else:
print("NO")
sys.exit(0)
fi.pop()
if n==1916:
print(fi)
print("YES")
print(*fi)
``` | instruction | 0 | 69,370 | 5 | 138,740 |
No | output | 1 | 69,370 | 5 | 138,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
n, num1, num2, num3 = int(input()), 0, 0, 0
for item in [int(item) for item in input().split(' ')]:
if item == 1:
num1 += 1
elif item == 2:
num2 += 1
else:
num3 += 1
if num1 < num2:
t = num1
num1 = num2
num2 = t
if num1 < num3:
t = num1
num1 = num3
num3 = t
print(num2 + num3)
``` | instruction | 0 | 69,388 | 5 | 138,776 |
Yes | output | 1 | 69,388 | 5 | 138,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
# from dust i have come dust i will be
n=int(input())
a=list(map(int,input().split()))
uno=0
dos=0
tres=0
for i in range(n):
if a[i]==1:
uno+=1
elif a[i]==2:
dos+=1
else:
tres+=1
print(uno+dos+tres-max(uno,dos,tres))
``` | instruction | 0 | 69,389 | 5 | 138,778 |
Yes | output | 1 | 69,389 | 5 | 138,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
from collections import deque
from collections import OrderedDict
import math
import sys
import os
from io import BytesIO
import threading
import bisect
import heapq
#sys.stdin = open("F:\PY\\test.txt", "r")
input = lambda: sys.stdin.readline().rstrip("\r\n")
n = int(input())
ar = list(map(int, input().split()))
dp=[0]*4
for i in range(n):
dp[ar[i]]+=1
if dp[1]+dp[2]==0 or dp[1]+dp[3]==0 or dp[2]+dp[3]==0:
print(0)
else:
print(n-max(dp[1],dp[2],dp[3]))
``` | instruction | 0 | 69,390 | 5 | 138,780 |
Yes | output | 1 | 69,390 | 5 | 138,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
inp = int(input())
a = [int(i) for i in input().split()]
diction = dict()
for letter in a:
diction[letter] = diction.get(letter, 0) + 1
print(inp - max(diction.values()))
``` | instruction | 0 | 69,391 | 5 | 138,782 |
Yes | output | 1 | 69,391 | 5 | 138,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
n=int(input())
l=list(map(int,input().split()))
x=l.count(1)
y=l.count(3)
print(x+y)
``` | instruction | 0 | 69,392 | 5 | 138,784 |
No | output | 1 | 69,392 | 5 | 138,785 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
s = 0
ans = 0
for i in range(n):
s += a[i]
s1 = s - n
s2 = abs(s - 2 * n)
s3 = 3*n - s
S = min(s1, s2, s3)
if S == s1:
ans = s1
elif S == s3:
ans = s3
else:
for i in range(n):
ans += abs(2 - a[i])
print(ans)
``` | instruction | 0 | 69,393 | 5 | 138,786 |
No | output | 1 | 69,393 | 5 | 138,787 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
n=int(input())
arr = [int(i) for i in input().split()]
min=arr.count(2)+arr.count(3)
if min>(arr.count(1)+arr.count(3)):
min=arr.count(1)+arr.count(3)
elif min>(arr.count(1)+arr.count(2)):
min=(arr.count(1)+arr.count(2))
``` | instruction | 0 | 69,394 | 5 | 138,788 |
No | output | 1 | 69,394 | 5 | 138,789 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a given sequence of integers a1, a2, ..., an, where every number is from 1 to 3 inclusively. You have to replace the minimum number of numbers in it so that all the numbers in the sequence are equal to each other.
Input
The first line contains an integer n (1 ≤ n ≤ 106). The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 3).
Output
Print the minimum number of replacements needed to be performed to make all the numbers in the sequence equal.
Examples
Input
9
1 3 2 2 2 1 1 2 3
Output
5
Note
In the example all the numbers equal to 1 and 3 should be replaced by 2.
Submitted Solution:
```
a = int(input())
b = list(map(int,input().split()))
c = list(b)
b.sort()
count = 0
for i in range(a):
if b[i] != c[i]:
count +=1
print(count)
``` | instruction | 0 | 69,395 | 5 | 138,790 |
No | output | 1 | 69,395 | 5 | 138,791 |
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