text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a Young diagram.
Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).
<image> Young diagram for a=[3,2,2,2,1].
Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.
Input
The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.
The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.
Output
Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.
Example
Input
5
3 2 2 2 1
Output
4
Note
Some of the possible solutions for the example:
<image> <image>
Submitted Solution:
```
from sys import stdin,stdout
n = int(stdin.readline().strip())
alist = list(map(int,stdin.readline().split()))
k = max(alist)
total,odd,even = 0,0,0
for i,a in enumerate(alist):
total += a//2
if i%2:
odd+=1
else:
even+=1
total+=min(odd,even)
print(total)
```
No
| 87,700 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
readline = sys.stdin.readline
class UF():
def __init__(self, num):
self.par = [-1]*num
self.weight = [0]*num
def find(self, x):
if self.par[x] < 0:
return x
else:
stack = []
while self.par[x] >= 0:
stack.append(x)
x = self.par[x]
for xi in stack:
self.par[xi] = x
return x
def union(self, x, y):
rx = self.find(x)
ry = self.find(y)
if rx != ry:
if self.par[rx] > self.par[ry]:
rx, ry = ry, rx
self.par[rx] += self.par[ry]
self.par[ry] = rx
self.weight[rx] += self.weight[ry]
return rx
N, K = map(int, readline().split())
S = list(map(int, readline().strip()))
A = [[] for _ in range(N)]
for k in range(K):
BL = int(readline())
B = list(map(int, readline().split()))
for b in B:
A[b-1].append(k)
cnt = 0
T = UF(2*K)
used = set()
Ans = [None]*N
inf = 10**9+7
for i in range(N):
if not len(A[i]):
Ans[i] = cnt
continue
kk = 0
if len(A[i]) == 2:
x, y = A[i]
if S[i]:
rx = T.find(x)
ry = T.find(y)
if rx != ry:
rx2 = T.find(x+K)
ry2 = T.find(y+K)
sp = min(T.weight[rx], T.weight[rx2]) + min(T.weight[ry], T.weight[ry2])
if x not in used:
used.add(x)
T.weight[rx] += 1
if y not in used:
used.add(y)
T.weight[ry] += 1
rz = T.union(rx, ry)
rz2 = T.union(rx2, ry2)
sf = min(T.weight[rz], T.weight[rz2])
kk = sf - sp
else:
rx = T.find(x)
ry2 = T.find(y+K)
sp = 0
if rx != ry2:
ry = T.find(y)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2]) + min(T.weight[ry], T.weight[ry2])
if x not in used:
used.add(x)
T.weight[rx] += 1
if y not in used:
used.add(y)
T.weight[ry] += 1
rz = T.union(rx, ry2)
rz2 = T.union(rx2, ry)
sf = min(T.weight[rz], T.weight[rz2])
kk = sf - sp
else:
if S[i]:
x = A[i][0]
rx = T.find(x)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2])
T.weight[rx] += inf
sf = min(T.weight[rx], T.weight[rx2])
kk = sf - sp
else:
x = A[i][0]
rx = T.find(x)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2])
T.weight[rx2] += inf
if x not in used:
used.add(x)
T.weight[rx] += 1
sf = min(T.weight[rx], T.weight[rx2])
kk = sf-sp
Ans[i] = cnt + kk
cnt = Ans[i]
print('\n'.join(map(str, Ans)))
```
| 87,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
def find(x):
if x == pre[x]:
return x
else:
pre[x]=find(pre[x])
return pre[x]
def merge(x,y):
x = find(x)
y = find(y)
if x != y:
pre[x] = y
siz[y] +=siz[x]
def cal(x):
return min(siz[find(x)],siz[find(x+k)])
def Solve(i):
global ans
cant = col[i][0]
if cant == 2:
x = col[i][1]
y = col[i][2]
if S[i] == 1:
if find(x) == find(y):
return
ans -=cal(x) + cal(y)
merge(x,y)
merge(x+k,y+k)
ans +=cal(x)
else:
if find(x) == find(y+k):
return
ans -=cal(x)+cal(y)
merge(x,y+k)
merge(x+k,y)
ans +=cal(x)
elif cant == 1:
x = col[i][1]
if S[i] == 1:
if find(x) == find(0):
return
ans -=cal(x)
merge(x,0)
ans +=cal(x)
else:
if find(x+k) == find(0):
return
ans -=cal(x)
merge(x+k,0)
ans +=cal(x)
n,k = map(int,input().split())
S = [1]+list(map(int,list(sys.stdin.readline().strip())))
col = [[0 for _ in range(3)] for _ in range(n+2)]
pre = [i for i in range(k*2+1)]
siz = [0 for _ in range(k*2+1)]
ans = 0
for i in range(1,k+1):
c = sys.stdin.readline()
c = int(c)
conjunto = [1]+list(map(int,list(sys.stdin.readline().split())))
for j in range(1,len(conjunto)):
x = conjunto[j]
col[x][0] = col[x][0]+1
col[x][col[x][0]] = i
for i in range(1,k+1):
siz[i]=1
siz[0]=3*10e5
for i in range(1,n+1):
Solve(i)
print(ans)
```
| 87,702 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin
class disjoinSet(object):
def __init__(self,n):
self.father = [x for x in range(0,n+1)]
self.rank = [0 for x in range(0,n+1)]
def setOf(self, x):
if(self.father[x] != x):
self.father[x] = self.setOf(self.father[x])
return self.father[x]
def Merge(self,x,y):
xR = self.setOf(x)
yR = self.setOf(y)
if(xR == yR):
return
if self.rank[xR] < self.rank[yR]:
self.father[xR] = yR
size[yR] += size[xR]
elif self.rank[xR] > self.rank[yR]:
self.father[yR] = xR
size[xR] += size[yR]
else:
self.father[yR] = xR
size[xR] += size[yR]
self.rank[xR] +=1
def cal(x):
return min(size[dsu.setOf(x)],size[dsu.setOf(x+k)])
def Solve(i):
global ans
cant = col[i][0]
if cant == 2:
x = col[i][1]
y = col[i][2]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(y):
return
ans -=cal(x) + cal(y)
dsu.Merge(x,y)
dsu.Merge(x+k,y+k)
ans +=cal(y)
else:
if dsu.setOf(x) == dsu.setOf(y+k):
return
ans -=cal(x)+cal(y)
dsu.Merge(x,y+k)
dsu.Merge(x+k,y)
ans +=cal(y)
elif cant == 1:
x = col[i][1]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(0):
return
ans -=cal(x)
dsu.Merge(x,0)
ans +=cal(x)
else:
if dsu.setOf(x+k) == dsu.setOf(0):
return
ans -=cal(x)
dsu.Merge(x+k,0)
ans +=cal(x)
n,k = map(int,input().split())
S = [1]+list(map(int,list(stdin.readline().strip())))
dsu = disjoinSet(k*2+1)
col = [[0 for _ in range(3)] for _ in range(n+2)]
size = [0 for _ in range(k*2+1)]
ans = 0
for i in range(1,k+1):
c = stdin.readline()
c = int(c)
conjunto = [1]+list(map(int,list(stdin.readline().split())))
for j in range(1,len(conjunto)):
x = conjunto[j]
col[x][0] = col[x][0]+1
col[x][col[x][0]] = i
for i in range(1,k+1):
size[i]=1
size[0]=3*10e5
for i in range(1,n+1):
Solve(i)
print(ans)
```
| 87,703 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin
input = stdin.readline
n , k = [int(i) for i in input().split()]
pairs = [i + k for i in range(k)] + [i for i in range(k)]
initial_condition = list(map(lambda x: x == '1',input().strip()))
data = [i for i in range(2*k)]
constrain = [-1] * (2*k)
h = [0] * (2*k)
L = [1] * k + [0] * k
dp1 = [-1 for i in range(n)]
dp2 = [-1 for i in range(n)]
for i in range(k):
input()
inp = [int(j) for j in input().split()]
for s in inp:
if dp1[s-1] == -1:dp1[s-1] = i
else:dp2[s-1] = i
pfsums = 0
ans = []
def remove_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums -= L[s1]
elif constrain[pairs[s1]] == 1:
pfsums -= L[pairs[s1]]
else:
pfsums -= min(L[s1],L[pairs[s1]])
def sh(i):
while i != data[i]:
i = data[i]
return i
def upd_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums += L[s1]
elif constrain[pairs[s1]] == 1:
pfsums += L[pairs[s1]]
else:
pfsums += min(L[s1],L[pairs[s1]])
def ms(i,j):
i = sh(i) ; j = sh(j)
cons = max(constrain[i],constrain[j])
if h[i] < h[j]:
data[i] = j
L[j] += L[i]
constrain[j] = cons
return j
else:
data[j] = i
if h[i] == h[j]:
h[i] += 1
L[i] += L[j]
constrain[i] = cons
return i
for i in range(n):
if dp1[i] == -1 and dp2[i] == -1:
pass
elif dp2[i] == -1:
s1 = sh(dp1[i])
remove_pfsum(s1)
constrain[s1] = 0 if initial_condition[i] else 1
constrain[pairs[s1]] = 1 if initial_condition[i] else 0
upd_pfsum(s1)
else:
s1 = sh(dp1[i]) ; s2 = sh(dp2[i])
if s1 == s2 or pairs[s1] == s2:
pass
else:
remove_pfsum(s1)
remove_pfsum(s2)
if initial_condition[i]:
new_s1 = ms(s1,s2)
new_s2 = ms(pairs[s1],pairs[s2])
else:
new_s1 = ms(s1,pairs[s2])
new_s2 = ms(pairs[s1],s2)
pairs[new_s1] = new_s2
pairs[new_s2] = new_s1
upd_pfsum(new_s1)
ans.append(pfsums)
for i in ans:
print(i)
```
| 87,704 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
readline = sys.stdin.readline
class UF():
def __init__(self, num):
self.par = [-1]*num
self.weight = [0]*num
def find(self, x):
stack = []
while self.par[x] >= 0:
stack.append(x)
x = self.par[x]
for xi in stack:
self.par[xi] = x
return x
def union(self, x, y):
rx = self.find(x)
ry = self.find(y)
if rx != ry:
if self.par[rx] > self.par[ry]:
rx, ry = ry, rx
self.par[rx] += self.par[ry]
self.par[ry] = rx
self.weight[rx] += self.weight[ry]
return rx
N, K = map(int, readline().split())
S = list(map(int, readline().strip()))
A = [[] for _ in range(N)]
for k in range(K):
BL = int(readline())
B = list(map(int, readline().split()))
for b in B:
A[b-1].append(k)
cnt = 0
T = UF(2*K)
used = set()
Ans = [None]*N
inf = 10**9+7
for i in range(N):
if not len(A[i]):
Ans[i] = cnt
continue
kk = 0
if len(A[i]) == 2:
x, y = A[i]
if S[i]:
rx = T.find(x)
ry = T.find(y)
if rx != ry:
rx2 = T.find(x+K)
ry2 = T.find(y+K)
sp = min(T.weight[rx], T.weight[rx2]) + min(T.weight[ry], T.weight[ry2])
if x not in used:
used.add(x)
T.weight[rx] += 1
if y not in used:
used.add(y)
T.weight[ry] += 1
rz = T.union(rx, ry)
rz2 = T.union(rx2, ry2)
sf = min(T.weight[rz], T.weight[rz2])
kk = sf - sp
else:
rx = T.find(x)
ry2 = T.find(y+K)
sp = 0
if rx != ry2:
ry = T.find(y)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2]) + min(T.weight[ry], T.weight[ry2])
if x not in used:
used.add(x)
T.weight[rx] += 1
if y not in used:
used.add(y)
T.weight[ry] += 1
rz = T.union(rx, ry2)
rz2 = T.union(rx2, ry)
sf = min(T.weight[rz], T.weight[rz2])
kk = sf - sp
else:
if S[i]:
x = A[i][0]
rx = T.find(x)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2])
T.weight[rx] += inf
sf = min(T.weight[rx], T.weight[rx2])
kk = sf - sp
else:
x = A[i][0]
rx = T.find(x)
rx2 = T.find(x+K)
sp = min(T.weight[rx], T.weight[rx2])
T.weight[rx2] += inf
if x not in used:
used.add(x)
T.weight[rx] += 1
sf = min(T.weight[rx], T.weight[rx2])
kk = sf-sp
Ans[i] = cnt + kk
cnt = Ans[i]
print('\n'.join(map(str, Ans)))
```
| 87,705 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
import sys
input = sys.stdin.readline
n,k=map(int,input().split())
S=[1]+list(map(int,list(input().strip())))
C=[]
for i in range(k):
c=int(input())
C.append(list(map(int,input().split())))
NUM=[[] for i in range(n+1)]
for i in range(k):
for c in C[i]:
NUM[c].append(i)
COLORS=[-1]*k
EDGE0=[[] for i in range(k)]
EDGE1=[[] for i in range(k)]
for i in range(1,n+1):
if len(NUM[i])==1:
if S[i]==0:
COLORS[NUM[i][0]]=1
else:
COLORS[NUM[i][0]]=0
elif len(NUM[i])==2:
x,y=NUM[i]
if S[i]==0:
EDGE0[x].append(y)
EDGE0[y].append(x)
else:
EDGE1[x].append(y)
EDGE1[y].append(x)
Q=[i for i in range(k) if COLORS[i]!=-1]
while Q:
x=Q.pop()
for to in EDGE0[x]:
if COLORS[to]==-1:
COLORS[to]=1-COLORS[x]
Q.append(to)
for to in EDGE1[x]:
if COLORS[to]==-1:
COLORS[to]=COLORS[x]
Q.append(to)
for i in range(k):
if COLORS[i]==-1:
COLORS[i]=0
Q=[i]
while Q:
x=Q.pop()
for to in EDGE0[x]:
if COLORS[to]==-1:
COLORS[to]=1-COLORS[x]
Q.append(to)
for to in EDGE1[x]:
if COLORS[to]==-1:
COLORS[to]=COLORS[x]
Q.append(to)
#print(COLORS)
Group = [i for i in range(k)]
W0 = [0]*(k)
W1 = [0]*(k)
for i in range(k):
if COLORS[i]==0:
W0[i]=1
else:
W1[i]=1
def find(x):
while Group[x] != x:
x=Group[x]
return x
def Union(x,y):
if find(x) != find(y):
if W0[find(x)] + W1[find(x)] < W0[find(y)] + W1[find(y)]:
W0[find(y)] += W0[find(x)]
W0[find(x)] =0
W1[find(y)] += W1[find(x)]
W1[find(x)] =0
Group[find(x)] =find(y)
else:
W0[find(x)] += W0[find(y)]
W0[find(y)] =0
W1[find(x)] += W1[find(y)]
W1[find(y)] =0
Group[find(y)] =find(x)
ANS=0
SCORES=[0]*(k)
for i in range(1,n+1):
if len(NUM[i])==0:
True
elif len(NUM[i])==1:
x=NUM[i][0]
ANS-=SCORES[find(x)]
SCORES[find(x)]=0
W0[find(x)]+=1<<31
ANS+=W1[find(x)]
SCORES[find(x)]=W1[find(x)]
else:
x,y=NUM[i]
ANS-=SCORES[find(x)]
SCORES[find(x)]=0
ANS-=SCORES[find(y)]
SCORES[find(y)]=0
if find(x)!=find(y):
Union(x,y)
SCORES[find(x)]+=min(W0[find(x)],W1[find(x)])
ANS+=min(W0[find(x)],W1[find(x)])
print(ANS)
```
| 87,706 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin
class disjoinSet(object):
def __init__(self,n):
self.father = [x for x in range(0,n+1)]
self.rank = [0 for x in range(0,n+1)]
def setOf(self, x):
if(self.father[x] != x):
self.father[x] = self.setOf(self.father[x])
return self.father[x]
def Merge(self,x,y):
xR = self.setOf(x)
yR = self.setOf(y)
if(xR == yR):
return
if self.rank[xR] < self.rank[yR]:
self.father[xR] = yR
size[yR] += size[xR]
elif self.rank[xR] > self.rank[yR]:
self.father[yR] = xR
size[xR] += size[yR]
else:
self.father[yR] = xR
size[xR] += size[yR]
self.rank[xR] +=1
def LessSize(x):
return min(size[dsu.setOf(x)],size[dsu.setOf(x+k)])
def Solve(i):
global ans
cant = col[i][0]
if cant == 2:
x = col[i][1]
y = col[i][2]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(y):
return
ans -=LessSize(x) + LessSize(y)
dsu.Merge(x,y)
dsu.Merge(x+k,y+k)
ans +=LessSize(y)
else:
if dsu.setOf(x) == dsu.setOf(y+k):
return
ans -=LessSize(x)+LessSize(y)
dsu.Merge(x,y+k)
dsu.Merge(x+k,y)
ans +=LessSize(y)
elif cant == 1:
x = col[i][1]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(0):
return
ans -=LessSize(x)
dsu.Merge(x,0)
ans +=LessSize(x)
else:
if dsu.setOf(x+k) == dsu.setOf(0):
return
ans -=LessSize(x)
dsu.Merge(x+k,0)
ans +=LessSize(x)
n,k = map(int,input().split())
S = [1]+list(map(int,list(stdin.readline().strip())))
dsu = disjoinSet(k*2+1)
col = [[0 for _ in range(3)] for _ in range(n+2)]
size = [0 for _ in range(k*2+1)]
ans = 0
for i in range(1,k+1):
c = stdin.readline()
c = int(c)
conjunto = [1]+list(map(int,list(stdin.readline().split())))
for j in range(1,len(conjunto)):
x = conjunto[j]
col[x][0] = col[x][0]+1
col[x][col[x][0]] = i
for i in range(1,k+1):
size[i]=1
size[0]=3*10e5
for i in range(1,n+1):
Solve(i)
print(ans)
```
| 87,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Submitted Solution:
```
import sys
input = sys.stdin.readline
n,k=map(int,input().split())
S=(1,)+tuple(map(int,list(input().strip())))
Q=[[] for i in range(n+1)]
for i in range(k):
m=int(input())
L=sorted(map(int,input().split()))
Q[L[0]].append(L)
seg_el=1<<((n+1).bit_length())
SEG=[1<<31]*(2*seg_el)
def getvalue(n,seg_el):
i=n+seg_el
ANS=1<<40
ANS=min(SEG[i],ANS)
i>>=1
while i!=0:
ANS=min(SEG[i],ANS)
i>>=1
return ANS
def updates(l,r,x):
L=l+seg_el
R=r+seg_el
while L<R:
if L & 1:
SEG[L]=min(x,SEG[L])
L+=1
if R & 1:
R-=1
SEG[R]=min(x,SEG[R])
L>>=1
R>>=1
from functools import lru_cache
@lru_cache(maxsize=None)
def calc(s,l,ANS):
#print(s,l,ANS)
updates(0,l,ANS)
if l==n+1:
return
for SET in Q[l]:
t=list(s)
for j in SET:
t[j]=1-t[j]
for k in range(l,n):
if t[k]==0:
nl=k
break
else:
nl=n+1
calc(tuple(t),nl,ANS+1)
l=S.index(0)
calc(S,l,0)
for i in range(1,n+1):
print(getvalue(i,seg_el))
```
No
| 87,708 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Submitted Solution:
```
from sys import stdin
class disjoinSet(object):
def __init__(self,n):
self.father = [x for x in range(0,n+1)]
self.rank = [0 for x in range(0,n+1)]
def setOf(self, x):
if(self.father[x] != x):
self.father[x] = self.setOf(self.father[x])
return self.father[x]
def Merge(self,x,y):
xR = self.setOf(x)
yR = self.setOf(y)
if(xR == yR):
return
if self.rank[xR] < self.rank[yR]:
self.father[xR] = yR
size[yR] += size[xR]
elif self.rank[xR] > self.rank[yR]:
self.father[yR] = xR
size[xR] += size[yR]
else:
self.father[yR] = xR
size[xR] += size[yR]
self.rank[xR] +=1
def LessSize(x):
return min(size[dsu.setOf(x)],size[dsu.setOf(x+k)])
def Solve(i):
global ans
cant = col[i][0]
if cant == 2:
x = col[i][1]
y = col[i][2]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(y):
return
ans -=LessSize(x) + LessSize(y)
dsu.Merge(x,y)
dsu.Merge(x+k,y+k)
ans +=LessSize(y)
else:
if dsu.setOf(x) == dsu.setOf(y+k):
return
ans -=LessSize(x)+LessSize(y)
dsu.Merge(x,y+k)
dsu.Merge(x+k,y)
ans +=LessSize(y)
elif cant == 1:
x = col[i][1]
if S[i] == 1:
if dsu.setOf(x) == dsu.setOf(0):
return
# ans -=LessSize(x)
dsu.Merge(x,0)
#ans +=LessSize(x)
else:
if dsu.setOf(x+k) == dsu.setOf(0):
return
ans -=LessSize(x)
dsu.Merge(x+k,0)
ans +=LessSize(x)
n,k = map(int,input().split())
S = [1]+list(map(int,list(stdin.readline().strip())))
dsu = disjoinSet(k*2+1)
col = [[0 for _ in range(3)] for _ in range(n+2)]
size = [0 for _ in range(k*2+1)]
ans = 0
for i in range(1,k+1):
c = stdin.readline()
c = int(c)
conjunto = [1]+list(map(int,list(stdin.readline().split())))
for j in range(1,len(conjunto)):
x = conjunto[j]
col[x][0] = col[x][0]+1
col[x][col[x][0]] = i
for i in range(1,k+1):
size[i]=1
size[0]=3*10e5
for i in range(1,n+1):
Solve(i)
print(ans)
```
No
| 87,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Submitted Solution:
```
class UnionFindNode:
def __init__(self, group_id, tp):
self.group_id_ = group_id
self.parent_ = None
self.rank_ = 1
self.cost = 0
if tp:
self.t0num = 0
self.t1num = 1
else:
self.t0num = 1
self.t1num = 0
def is_root(self):
return not self.parent_
def root(self):
parent = self
while not parent.is_root():
parent = parent.parent_
self.parent_ = parent
return parent
def unite(self, unite_node, k):
root = self.root()
unite_root = unite_node.root()
if root.group_id_ == unite_root.group_id_:
return 0
tmp = root.cost + unite_root.cost
k_root = k.root()
if root.rank_ > unite_root.rank_:
n_root, child = root, unite_root
else:
n_root, child = unite_root, root
child.parent_ = n_root
n_root.rank_ = max(n_root.rank_, child.rank_ + 1)
n_root.t0num = n_root.t0num + child.t0num
n_root.t1num = n_root.t1num + child.t1num
if k_root.group_id_ == n_root.group_id_:
r = n_root.t0num
else:
r = min(n_root.t0num, n_root.t1num)
n_root.cost = r
return r -tmp
if __name__ == "__main__":
N, K = map(int, input().split())
S = input().strip()
vs = [set() for _ in range(N)]
for i in range(K):
input()
for x in map(int, input().split()):
vs[x-1].add(i)
es = []
e2i = dict()
G = [set() for _ in range(K+1)]
for i in range(N):
if len(vs[i]) == 1:
e = (vs[i].pop(), K)
elif len(vs[i]) == 2:
e = (vs[i].pop(), vs[i].pop())
else:
e = (K, K)
# i番目のランプ
es.append(e)
e2i[e] = int(S[i])
e2i[(e[1], e[0])] = int(S[i])
G[e[0]].add(e[1])
G[e[1]].add(e[0])
v2t = [1] * (K+1)
vvs = set()
for i in range(K, -1, -1):
if i in vvs:
continue
stack = [i]
while stack:
v = stack.pop()
vvs.add(v)
for u in G[v]:
if u in vvs:
continue
if e2i[(v, u)]:
v2t[u] = v2t[v]
else:
v2t[u] = 1 - v2t[v]
stack.append(u)
node_list = [UnionFindNode(i, v2t[i]) for i in range(K+1)]
r = 0
for v, u in es:
r += node_list[v].unite(node_list[u], node_list[K])
print(r)
```
No
| 87,710 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Submitted Solution:
```
import sys
def find(x):
if x == pre[x]:
return x
else:
return find(pre[x])
def merge(x,y):
x = find(x)
y = find(y)
if x != y:
pre[x] = y
siz[y] +=siz[x]
def cal(x):
return min(siz[find(x)],siz[find(x+k)])
def Solve(i):
global ans
cant = col[i][0]
if cant == 2:
x = col[i][1]
y = col[i][2]
if S[i] == 1:
if find(x) == find(y):
return
ans -=cal(x) + cal(y)
merge(x,y)
merge(x+k,y+k)
ans +=cal(x)
else:
if find(x) == find(y):
return
ans -=cal(x)+cal(y)
merge(x,y+k)
merge(x+k,y)
ans +=cal(x)
elif cant == 1:
x = col[i][1]
if S[i] == 1:
if find(x) == find(0):
return
ans -=cal(x)
merge(x,0)
ans +=cal(x)
else:
if find(x+k) == find(0):
return
ans -=cal(x)
merge(x+k,0)
ans +=cal(x)
n,k = map(int,input().split())
S = [1]+list(map(int,list(sys.stdin.readline().strip())))
col = [[0 for _ in range(3)] for _ in range(n+2)]
pre = [i for i in range(k*2+1)]
siz = [0 for _ in range(k*2+1)]
ans = 0
for i in range(1,k+1):
c = sys.stdin.readline()
c = int(c)
conjunto = [1]+list(map(int,list(sys.stdin.readline().split())))
for j in range(1,len(conjunto)):
x = conjunto[j]
col[x][0] = col[x][0]+1
col[x][col[x][0]] = i
for i in range(1,k+1):
siz[i]=1
siz[0]=3*10e5
for i in range(1,n+1):
Solve(i)
print(ans)
```
No
| 87,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a country with n cities numbered from 1 to n. She lives in the city number 1.
There is a direct train route between each pair of distinct cities i and j, where i ≠ j. In total there are n(n-1) distinct routes. Every route has a cost, cost for route from i to j may be different from the cost of route from j to i.
Masha wants to start her journey in city 1, take exactly k routes from one city to another and as a result return to the city 1. Masha is really careful with money, so she wants the journey to be as cheap as possible. To do so Masha doesn't mind visiting a city multiple times or even taking the same route multiple times.
Masha doesn't want her journey to have odd cycles. Formally, if you can select visited by Masha city v, take odd number of routes used by Masha in her journey and return to the city v, such journey is considered unsuccessful.
Help Masha to find the cheapest (with minimal total cost of all taken routes) successful journey.
Input
First line of input had two integer numbers n,k (2 ≤ n ≤ 80; 2 ≤ k ≤ 10): number of cities in the country and number of routes in Masha's journey. It is guaranteed that k is even.
Next n lines hold route descriptions: j-th number in i-th line represents the cost of route from i to j if i ≠ j, and is 0 otherwise (there are no routes i → i). All route costs are integers from 0 to 10^8.
Output
Output a single integer — total cost of the cheapest Masha's successful journey.
Examples
Input
5 8
0 1 2 2 0
0 0 1 1 2
0 1 0 0 0
2 1 1 0 0
2 0 1 2 0
Output
2
Input
3 2
0 1 1
2 0 1
2 2 0
Output
3
Submitted Solution:
```
import copy
def build_graph(cities):
graph = dict()
for i in range(cities):
graph[i] = [adj for adj in range(cities) if adj != i]
return graph
def build_edges(grid):
edges = dict()
for i in range(len(grid)):
for j in range(len(grid[i])): edges[(i,j)] = grid[i][j]
return edges
def find_all_path(graph, routes, start, path, paths):
path.append(start)
if len(path) == routes and tuple(path) not in paths: paths.add(tuple(copy.deepcopy(path)))
else:
for adj in graph[start]:
if adj != 0: find_all_path(graph, routes, adj, path, paths)
path.pop()
def min_cost(graph, edges, routes):
destination, results, costs = [d for d in graph if d != 0], list(), list()
for d in destination:
temp = set()
find_all_path(graph, routes, 0, list(), temp)
for t in temp: results.append(t)
for r in results:
cost = 0
for i in range(routes-1): cost += edges[(r[i], r[i+1])]
cost += edges[(r[routes-1], 0)]
costs.append(cost)
return min(costs)
if __name__ == '__main__':
cities, routes = input().split()
grid = [[int(j) for j in input().split()] for i in range(int(cities))]
graph, edges = build_graph(int(cities)), build_edges(grid)
print(min_cost(graph, edges, int(routes)))
```
No
| 87,712 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a country with n cities numbered from 1 to n. She lives in the city number 1.
There is a direct train route between each pair of distinct cities i and j, where i ≠ j. In total there are n(n-1) distinct routes. Every route has a cost, cost for route from i to j may be different from the cost of route from j to i.
Masha wants to start her journey in city 1, take exactly k routes from one city to another and as a result return to the city 1. Masha is really careful with money, so she wants the journey to be as cheap as possible. To do so Masha doesn't mind visiting a city multiple times or even taking the same route multiple times.
Masha doesn't want her journey to have odd cycles. Formally, if you can select visited by Masha city v, take odd number of routes used by Masha in her journey and return to the city v, such journey is considered unsuccessful.
Help Masha to find the cheapest (with minimal total cost of all taken routes) successful journey.
Input
First line of input had two integer numbers n,k (2 ≤ n ≤ 80; 2 ≤ k ≤ 10): number of cities in the country and number of routes in Masha's journey. It is guaranteed that k is even.
Next n lines hold route descriptions: j-th number in i-th line represents the cost of route from i to j if i ≠ j, and is 0 otherwise (there are no routes i → i). All route costs are integers from 0 to 10^8.
Output
Output a single integer — total cost of the cheapest Masha's successful journey.
Examples
Input
5 8
0 1 2 2 0
0 0 1 1 2
0 1 0 0 0
2 1 1 0 0
2 0 1 2 0
Output
2
Input
3 2
0 1 1
2 0 1
2 2 0
Output
3
Submitted Solution:
```
n,k = map(int, input().split(" "))
s = list()
for i in range(n):
s1 = list(map(int, input().split(" ")))
g = 2
x = 0
while g != n:
x += s1[g-1]
g += 1
x = (x*2) + s1[0] + s1[n-1]
s.append(x)
x = 0
x1 = s[0]
for i in range(len(s)):
x = s[i]
if x < x1:
x1 = x
if n == 2 and k == 2:
print(s[0]+s[1])
exit()
print(x1)
```
No
| 87,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a country with n cities numbered from 1 to n. She lives in the city number 1.
There is a direct train route between each pair of distinct cities i and j, where i ≠ j. In total there are n(n-1) distinct routes. Every route has a cost, cost for route from i to j may be different from the cost of route from j to i.
Masha wants to start her journey in city 1, take exactly k routes from one city to another and as a result return to the city 1. Masha is really careful with money, so she wants the journey to be as cheap as possible. To do so Masha doesn't mind visiting a city multiple times or even taking the same route multiple times.
Masha doesn't want her journey to have odd cycles. Formally, if you can select visited by Masha city v, take odd number of routes used by Masha in her journey and return to the city v, such journey is considered unsuccessful.
Help Masha to find the cheapest (with minimal total cost of all taken routes) successful journey.
Input
First line of input had two integer numbers n,k (2 ≤ n ≤ 80; 2 ≤ k ≤ 10): number of cities in the country and number of routes in Masha's journey. It is guaranteed that k is even.
Next n lines hold route descriptions: j-th number in i-th line represents the cost of route from i to j if i ≠ j, and is 0 otherwise (there are no routes i → i). All route costs are integers from 0 to 10^8.
Output
Output a single integer — total cost of the cheapest Masha's successful journey.
Examples
Input
5 8
0 1 2 2 0
0 0 1 1 2
0 1 0 0 0
2 1 1 0 0
2 0 1 2 0
Output
2
Input
3 2
0 1 1
2 0 1
2 2 0
Output
3
Submitted Solution:
```
import copy
def build_graph(cities):
graph = dict()
for i in range(cities):
graph[i] = [adj for adj in range(cities) if adj != i]
return graph
def build_edges(grid):
edges = dict()
for i in range(len(grid)):
for j in range(len(grid[i])): edges[(i,j)] = grid[i][j]
return edges
def find_all_path(graph, routes, start, path, paths):
path.append(start)
if len(path) <= routes:
if len(path) == routes and tuple(path) not in paths: paths.add(tuple(copy.deepcopy(path)))
else:
for adj in graph[start]:
if adj != 0: find_all_path(graph, routes, adj, path, paths)
path.pop()
def min_cost(graph, edges, routes):
destination, results, costs = [d for d in graph if d != 0], set(), list()
for d in destination:
temp = set()
find_all_path(graph, routes, 0, list(), temp)
for t in temp:
cost = 0
for i in range(routes-1): cost += edges[(t[i], t[i+1])]
cost += edges[(t[routes-1], 0)]
if cost != 63489052: costs.append(cost)
results.add(t)
return min(costs)
if __name__ == '__main__':
cities, routes = input().split()
grid = [[int(j) for j in input().split()] for i in range(int(cities))]
graph, edges = build_graph(int(cities)), build_edges(grid)
print(min_cost(graph, edges, int(routes)))
```
No
| 87,714 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Masha lives in a country with n cities numbered from 1 to n. She lives in the city number 1.
There is a direct train route between each pair of distinct cities i and j, where i ≠ j. In total there are n(n-1) distinct routes. Every route has a cost, cost for route from i to j may be different from the cost of route from j to i.
Masha wants to start her journey in city 1, take exactly k routes from one city to another and as a result return to the city 1. Masha is really careful with money, so she wants the journey to be as cheap as possible. To do so Masha doesn't mind visiting a city multiple times or even taking the same route multiple times.
Masha doesn't want her journey to have odd cycles. Formally, if you can select visited by Masha city v, take odd number of routes used by Masha in her journey and return to the city v, such journey is considered unsuccessful.
Help Masha to find the cheapest (with minimal total cost of all taken routes) successful journey.
Input
First line of input had two integer numbers n,k (2 ≤ n ≤ 80; 2 ≤ k ≤ 10): number of cities in the country and number of routes in Masha's journey. It is guaranteed that k is even.
Next n lines hold route descriptions: j-th number in i-th line represents the cost of route from i to j if i ≠ j, and is 0 otherwise (there are no routes i → i). All route costs are integers from 0 to 10^8.
Output
Output a single integer — total cost of the cheapest Masha's successful journey.
Examples
Input
5 8
0 1 2 2 0
0 0 1 1 2
0 1 0 0 0
2 1 1 0 0
2 0 1 2 0
Output
2
Input
3 2
0 1 1
2 0 1
2 2 0
Output
3
Submitted Solution:
```
n,k = map(int, input().split(" "))
s = list()
for i in range(n):
s1 = list(map(int, input().split(" ")))
g = 2
x = 0
while g != n:
x += s1[g-1]
g += 1
x = (x*2) + s1[0] + s1[n-1]
s.append(x)
x = 0
x1 = s[0]
for i in range(len(s)):
x = s[i]
if x < x1:
x1 = x
print(x1)
```
No
| 87,715 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
printn = lambda x: print(x,end='')
inn = lambda : int(input())
inl = lambda: list(map(int, input().split()))
inm = lambda: map(int, input().split())
ins = lambda : input().strip()
DBG = True # and False
BIG = 10**18
R = 10**9 + 7
def ddprint(x):
if DBG:
print(x)
t = inn()
for tt in range(t):
n = inn()
a = inl()
b = inl()
m = p = n
for i in range(n):
if a[i] == -1:
m = i
break
for i in range(n):
if a[i] == 1:
p = i
break
if m<p:
ok = True
for i in range(n):
if i<m and b[i]!=0 or i==m and b[i] != -1 or \
m<i<=p and b[i]>a[i]:
ok = False
break
print('YES' if ok else 'NO')
else: # p<m
ok = True
for i in range(n):
if i<p and b[i]!=0 or i==p and b[i] != 1 or \
p<i<=m and b[i]<a[i]:
ok = False
break
print('YES' if ok else 'NO')
```
| 87,716 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
def check(a,b):
neg,pos=0,0
for j in range(len(a)):
if(b[j]<a[j] and neg!=1):
print("NO")
return
elif(b[j]>a[j]):
if(pos!=1):
print("NO")
return
if(a[j]==-1):
neg=1
if(a[j]==1):
pos=1
print("YES")
t=int(input())
for i in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
check(a,b)
```
| 87,717 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
# cook your dish here
for _ in range(int(input())):
n=int(input())
l1=list(map(int,input().split()))
l2=list(map(int,input().split()))
x=-1
y=-1
z=-1
for i in range(n):
if(l1[i]==0):
x=i
break
for i in range(n):
if(l1[i]==1):
y=i
break
for i in range(n):
if(l1[i]==-1):
z=i
break
if(l1[0]!=l2[0]):
print("NO")
else:
f=0
for i in range(1,n):
if(l1[i]!=l2[i]):
if(l1[i]>l2[i]):
if(z==-1 or z>=i):
f=1
break
elif(l1[i]<l2[i]):
if(y==-1 or y>=i):
f=1
break
if(f==1):
print("NO")
else:
print("YES")
```
| 87,718 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
left = set()
res = 'YES'
for i in range(n):
if b[i] != a[i]:
if b[i] > a[i] and 1 not in left:
res = 'NO'
break
elif b[i] < a[i] and -1 not in left:
res = 'NO'
break
left.add(a[i])
print(res)
```
| 87,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for x in range(t):
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
v = [999999999999999]*3
deff = 0
v0 = 0
v2 = 0
v1 = 0
for i in range(n):
if a[i] == -1:
if v0 == 0:
v[0] = i
v0 = 1
elif a[i] == 0:
if v1 == 0:
v[1] = i
v1 = 1
elif a[i] == 1:
if v2 == 0:
v[2] = i
v2 = 1
for i in range(1,n+1):
if a[-i] - b[-i] > 0:
if v[0] < n - i:
pass
else:
if deff == 0:
print('NO')
deff = 1
if a[-i] - b[-i] < 0:
if v[2] < n - i:
pass
else:
if deff == 0:
print('NO')
deff = 1
if deff == 0:
print('YES')
```
| 87,720 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = [int(s) for s in input().split()]
b = [int(s) for s in input().split()]
pos = False
neg = False
for x, y in zip(a, b):
if y > x and not pos:
print('NO')
break
elif y < x and not neg:
print('NO')
break
if x == 1:
pos = True
elif x == -1:
neg = True
else:
print('YES')
```
| 87,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
n = input
v = lambda: map(int, n().split())
def f():
s = 0
for a, b in zip(v(), v()):
if s == 0:
if a != b: return 'NO'
s = a
elif s > 0:
if a > b: return 'NO'
if a < 0: return 'YES'
elif s < 0:
if a < b: return 'NO'
if a > 0: return 'YES'
return 'YES'
for i in range(int(n())): n(), print(f())
```
| 87,722 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Tags: greedy, implementation
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
ar=list(map(int,input().split()))
br=list(map(int,input().split()))
d={}
d[1]=0
d[-1]=0
for i in range(n):
if ar[i]==1 or ar[i]==-1:
d[ar[i]]=d.get(ar[i])+1
fl=False
for i in range(n-1,-1,-1):
if ar[i]==-1 or ar[i]==1:
d[ar[i]]=d.get(ar[i])-1
if ar[i]!=br[i]:
t=br[i]-ar[i]
if t<0:
if d.get(-1)==0:
fl=True
elif t>0:
if d.get(1)==0:
fl=True
if fl:
print("NO")
else:
print("YES")
```
| 87,723 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
t = int(input())
al = []
for i in range(t):
n = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
s = 0
f = 0
z = 0
for j in range(n):
if a[j] == 0:
z += 1
elif a[j] == 1:
s += 1
else:
f += 1
ans = "YES"
for j in range(n-1,-1,-1):
if a[j] == 0:
z -= 0
elif a[j] == 1:
s -= 1
else:
f -= 1
if b[j] > a[j]:
if s == 0:
ans = "NO"
break
elif b[j] < a[j]:
if f == 0:
ans = "NO"
al.append(ans)
for i in al:
print(i)
```
Yes
| 87,724 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def solve(a,b,n):
s,p=0,0
for i in range(n):
if p==0 and (b[i]-a[i])>0:
return "NO"
if s==0 and (b[i]-a[i])<0:
return "NO"
if a[i]<0:
s=1
if a[i]>0:
p=1
return "YES"
for i in range(int(input())):
n=int(input())
a=[int(i) for i in input().split()]
b=[int(i) for i in input().split()]
print(solve(a,b,n))
```
Yes
| 87,725 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
flag1=False
flag2=False
ans=False
for i in range(n):
if a[i]<b[i]:
if flag1==False:
ans=True
break
elif a[i]>b[i]:
if flag2==False:
ans=True
break
if a[i]==1:
flag1=True
elif a[i]==-1:
flag2=True
if ans:
print("NO")
else:
print("YES")
```
Yes
| 87,726 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def noob():
return 5+10
t = int(input())
while t!=0:
t-=1
n = int(input())
a = [int(ele) for ele in input().split()]
b = [int(elem) for elem in input().split()]
on = [-1]*n
non = [-1]*n
pronoob = -1
for i in range(n):
on[i] = pronoob
if a[i]==1: pronoob=i
pronoob = -1
for i in range(n):
non[i] = pronoob
if a[i]==-1: pronoob=i
flag = 0
for i in range(n):
t1 = a[i]-b[i]
if t1<0 and on[i]==-1:
flag = 1
break
if t1>0 and non[i]==-1:
flag = 1
break
if flag==1: print("NO")
else: print("YES")
```
Yes
| 87,727 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import threading
from heapq import heapify,heappush,heappop
def can_win(s,e):
if e%2==1:
return 1 if s%2==0 else 0
else:
if e//2<s<=e:
return 1 if s%2==1 else 0
if e//4<s<=e//2:
return 1
else:
return can_win(s,e//4)
def main():
for _ in range(int(input())):
n=int(input())
arr=list(map(int,input().split()))
brr=list(map(int,input().split()))
is1=is_1=-1
b=True
is1=is_1=-1
for i in range(n):
if brr[i]>arr[i] and is1==-1:
b=False
break
elif brr[i]<arr[i] and is_1==-1:
b=False
break
elif arr[i]==1:
is1=1
else:
is_1=1
if b:
print('YES')
else:
print('NO')
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
```
No
| 87,728 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
a=[int(x) for x in input().split()]
b=[int(x) for x in input().split()]
f=0
pos,neg=0,0
if a[0]!=b[0]:
print('NO')
continue
for i in range(1,n):
if a[i-1]==1:
pos=1
elif a[i-1]==-1:
neg=1
if a[i]==b[i]:
continue
elif b[i]>0 and pos:
continue
elif b[i]<0 and neg:
continue
elif b[i]==0:
if a[i]>0 and neg:
continue
elif a[i]<0 and pos:
continue
else:
f=1
break
if f:
print('NO')
else:
print('YES')
```
No
| 87,729 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
def artem(n,m):
if (m*n)%2 == 1:
return ('BW'*(m//2) + 'B\n' + 'WB'*(m//2) + 'W\n')*(n//2) + 'BW'*(m//2) + 'B\n'
else:
return ('BW'*((m-1)//2) + 'B\n'*(m%2) + 'BW\n'*((m-1)%2) + 'WB'*((m-1)//2) + 'W\n'*(m%2) + 'WB\n'*((m-1)%2))*((n-1)//2) + 'BW'*((m-1)//2) + 'BB\n'*((m-1)%2) + 'B\n'*(m%2) + ('WB'*((m-1)//2) + 'B\n'*(m%2) + 'WB\n'*((m-1)%2))*((n-1)%2)
# cases = int(input())
# for _ in range(cases):
# n,m = list(map(int,input().split()))
# print(artem(n,m)[:-1])
def anton(a,b):
found = [False,False]
for i in range(len(a)):
if a[i] == b[i]:
if a[i] == 1 and not found[1]:
found[1] = True
if a[i] == -1 and not found[0]:
found[0] = True
if found[0] and found[1]:
return 'YES'
elif a[i] > b[i] and not found[0]:
return 'NO'
elif a[i] < b[i] and not found[1]:
return 'NO'
return 'YES'
cases = int(input())
for _ in range(cases):
l = int(input())
a = list(map(int,input().split()))
b = list(map(int,input().split()))
print(anton(a,b))
```
No
| 87,730 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Once again, Boris needs the help of Anton in creating a task. This time Anton needs to solve the following problem:
There are two arrays of integers a and b of length n. It turned out that array a contains only elements from the set \{-1, 0, 1\}.
Anton can perform the following sequence of operations any number of times:
1. Choose any pair of indexes (i, j) such that 1 ≤ i < j ≤ n. It is possible to choose the same pair (i, j) more than once.
2. Add a_i to a_j. In other words, j-th element of the array becomes equal to a_i + a_j.
For example, if you are given array [1, -1, 0], you can transform it only to [1, -1, -1], [1, 0, 0] and [1, -1, 1] by one operation.
Anton wants to predict if it is possible to apply some number (zero or more) of these operations to the array a so that it becomes equal to array b. Can you help him?
Input
Each test contains multiple test cases.
The first line contains the number of test cases t (1 ≤ t ≤ 10000). The description of the test cases follows.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the length of arrays.
The second line of each test case contains n integers a_1, a_2, ..., a_n (-1 ≤ a_i ≤ 1) — elements of array a. There can be duplicates among elements.
The third line of each test case contains n integers b_1, b_2, ..., b_n (-10^9 ≤ b_i ≤ 10^9) — elements of array b. There can be duplicates among elements.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
For each test case, output one line containing "YES" if it's possible to make arrays a and b equal by performing the described operations, or "NO" if it's impossible.
You can print each letter in any case (upper or lower).
Example
Input
5
3
1 -1 0
1 1 -2
3
0 1 1
0 2 2
2
1 0
1 41
2
-1 0
-1 -41
5
0 1 -1 1 -1
1 1 -1 1 -1
Output
YES
NO
YES
YES
NO
Note
In the first test-case we can choose (i, j)=(2, 3) twice and after that choose (i, j)=(1, 2) twice too. These operations will transform [1, -1, 0] → [1, -1, -2] → [1, 1, -2]
In the second test case we can't make equal numbers on the second position.
In the third test case we can choose (i, j)=(1, 2) 41 times. The same about the fourth test case.
In the last lest case, it is impossible to make array a equal to the array b.
Submitted Solution:
```
testCases = int(input())
def convertArrayToInt(A):
for i in range(0, len(A), 1):
A[i] = int(A[i])
return A
def canBeConverted(A, B):
for i in range(len(A)):
if A[i]!=B[i]:
diff = B[i]-A[i]
flag = 0
for x in range(0, i, 1):
if A[x]!=0 and diff%A[x]==0:
flag=1
break
if flag==0:
return "NO"
return "YES"
while testCases:
length = int(input())
A = input().split()
B = input().split()
A = convertArrayToInt(A)
B = convertArrayToInt(B)
ans = canBeConverted(A, B)
print (ans)
testCases-=1
```
No
| 87,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
t = int(input())
def dp(l):
dp = [[float("INF")]*3 for m in range(len(l)+1)]
dp[0][0] = 0
dp[0][1] = 0
c = 0
for m in range(len(l)):
x = l[m]
c += x
dp[m+1][0] = dp[m][0] + x
dp[m+1][1] = min(dp[m][0],dp[m][1])+abs(1-x)
dp[m+1][2] = min(dp[m][1],dp[m][2])+x
return min(dp[-1])-c
def main():
import sys
input = sys.stdin.readline
for i in range(t):
n,k = map(int,input().split())
s = list(input())
l = [[] for j in range(k)]
count = 0
for j in range(n):
if s[j] =="0":
l[j%k].append(0)
else:
count += 1
l[j%k].append(1)
ans = count
for j in range(k):
ans = min(ans,dp(l[j])+count)
print(ans)
if __name__ =="__main__":
main()
```
| 87,732 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
import os, sys
from io import BytesIO, IOBase
from types import GeneratorType
from bisect import bisect_left, bisect_right
from collections import defaultdict as dd, deque as dq, Counter as dc
import math, string
import heapq as h, time
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#start_time = time.time()
def getInt(): return int(input())
def getStrs(): return input().split()
def getInts(): return list(map(int,input().split()))
def getStr(): return input()
def listStr(): return list(input())
def getMat(n): return [getInts() for _ in range(n)]
def getBin(): return list(map(int,list(input())))
def isInt(s): return '0' <= s[0] <= '9'
def ceil_(a,b): return a//b + (a%b > 0)
MOD = 10**9 + 7
"""
K-period can be achieved by starting at any index
Turn them all off is an option
Suppose index i is the first
Then either i+K is on (and nothing in between), or nothing else
Then either i+2*K is on (and nothing in between), or nothing else
etc
dp[i] = min number of moves such that i is switched on
"""
def solve():
N, K = getInts()
S = getBin()
pref = [0]
curr = 0
for s in S:
curr += s
pref.append(curr)
dp = [0 for i in range(N+1)]
dp[0] = 10**18
S = [0]+S
for i in range(1,N+1):
dp[i] = pref[i-1]+(S[i]^1)
if i > K:
dp[i] = min(dp[i],dp[i-K]+pref[i-1]-pref[i-K]+(S[i]^1))
ans = pref[-1]
for i in range(1,N+1):
ans = min(ans,dp[i]+pref[-1]-pref[i])
return ans
for _ in range(getInt()):
print(solve())
#solve()
#print(time.time()-start_time)
```
| 87,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
T = int(input())
for kase in range(T):
n, k = map(int, input().split())
s = input()
one, ans = s.count('1'), n
for i in range(0, k):
d = 0
for j in range(i, n, k):
d = max(0, d-1) if s[j] == '0' else d+1
ans = min(ans, one-d)
print(ans)
```
| 87,734 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
#gaalt sochliya
#jald baazi
#time out ho raha hai order n hote hue bhi
import sys
input = lambda: sys.stdin.readline().strip() #input method from shiv_99
t = int(input())
answers = []
for _ in range(t):
n, k = [int(i) for i in input().split()]
s = input()
vals = [0 if i=='0' else 1 for i in s]
total_on = sum(vals)
ans = n
for j in range(k):
# breakpoint()
counter = j
sub_arr = []
# already_on = 0
# num_lamps = 0
# longest_continuous = 0
while(counter<n):
sub_arr.append(vals[counter])
counter += k
#now look at temp
last = sub_arr[0]
already_on = sum(sub_arr)
compulsory = total_on - already_on
if compulsory > ans:
continue
num_lamps = len(sub_arr)
sub_sub_arr = []
s = 0
mult = 1 if last == 1 else -1
for i in range(len(sub_arr)):
if last == sub_arr[i]:
s += 1
else:
sub_sub_arr.append(mult*s)
mult = -1*mult
s = 1
if i == len(sub_arr)-1:
sub_sub_arr.append(mult*s)
last = sub_arr[i]
#to find the max_sum_of_a_sub_array_in_this_arr
curr_best = 0
best = 0
for i in range(len(sub_sub_arr)):
if sub_sub_arr[i] > 0:
curr_best += sub_sub_arr[i]
else:
el = sub_sub_arr[i]
if curr_best + el < 0:
curr_best = 0
else:
curr_best = curr_best + el
best = max(curr_best, best)
option = min(num_lamps - already_on, already_on, already_on - best)
ans = min(compulsory + option, ans)
answers.append(ans)
print(*answers, sep = '\n')
# for a in answers:
# print(a)
```
| 87,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
T = int(input())
while T != 0:
T -= 1
ans = 10 ** 9
n, k = map(int, input().split())
s = input()
tot = s.count('1');
for i in range(k):
sum = 0; maxn = 0
for j in range(i, n, k):
sum = max(0, sum - 1) if s[j] == '0' else sum + 1
maxn = max(maxn, sum)
ans = min(ans, tot - maxn)
print(ans)
```
| 87,736 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
t=int(input())
for tt in range(t):
global positions
global total_ones
global n
global k
global s
n,k=[int(x) for x in input().split(' ')]
s=input()
total_ones=0
for x in s:
total_ones+=int(x)
positions=[]
def changes_req(positions):
summ=[]
sum_upto=[0]
for position in positions:
if(s[int(position)]=='1'):
summ.append(-1)
elif(s[int(position)]=='0'):
summ.append(1)
else:
print('TNP')
for i in range(len(summ)):
sum_upto.append(summ[i]+sum_upto[-1])
max_upto=[sum_upto[0]]
for i in range(1,len(sum_upto)):
max_upto.append(max(max_upto[-1],sum_upto[i]))
sum_upto.reverse()
min_after=[sum_upto[0]]
for i in range(1,len(sum_upto)):
min_after.append(min(min_after[-1],sum_upto[i]))
sum_upto.reverse()
min_after.reverse()
ans=n
for i in range(len(sum_upto)):
ans=min(ans,round(min_after[i]-max_upto[i]))
return(ans+total_ones)
def update_1(positions):
for i in range(len(positions)):
positions[i]+=1
if(positions[-1]>=n):
positions.pop()
positions=[]
position=0
while (position<n):
positions.append(position)
position+=k
ans=n
for i in range(k):
ans=min(ans,changes_req(positions))
update_1(positions)
print(ans)
```
| 87,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
t = int(input())
out = []
for i in range(t):
n,k = map(int,input().split())
s = list(input())
count = [0]
for j in s:
if j == "1":
count.append(count[-1]+1)
else:
count.append(count[-1])
ans = count[-1]
dp = []
for j in range(n):
cost = count[j]
if j >= k:
cost = min(cost,dp[j-k]+count[j]-count[j-k+1])
if s[j] =="0":
cost += 1
dp.append(cost)
ans = min(ans,cost+count[-1]-count[j+1])
out.append(ans)
for i in out:
print(i)
```
| 87,738 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Tags: brute force, dp, greedy
Correct Solution:
```
for _ in range(int(input())):
n, k = map(int, input().split())
s = input()
ones = s.count('1')
best = 0
for i in range(k):
count = 0
for j in range(i, n, k):
if s[j] == '1':
count += 1
else:
count = max(0, count - 1)
best = max(count, best)
print(ones - best)
```
| 87,739 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
from sys import stdin, stdout
read = stdin.readline
T = int(read())
for i in range(T):
n, k = map(int, read().split())
s = read()[:-1]
ones = sum([c == '1' for c in s])
cnt = [0 for i in range(k)]
for i in range(n):
if s[i] == '1':
cnt[i%k] += 1
ans = ones
dp = [[0 for j in range(3)] for i in range(n)]
for i in range(n-1, n-k-1, -1):
if s[i] == '1':
dp[i][0] = 1
dp[i][2] = 1
else:
dp[i][1] = 1
for i in range(n-k-1, -1, -1):
if s[i] == '1':
dp[i][0] = dp[i+k][0] + 1
dp[i][1] = min(dp[i+k][0], dp[i+k][1])
dp[i][2] = min(dp[i+k][1], dp[i+k][2]) + 1
else:
dp[i][0] = dp[i+k][0]
dp[i][1] = min(dp[i+k][0], dp[i+k][1]) + 1
dp[i][2] = min(dp[i+k][1], dp[i+k][2])
for i in range(k):
ans = min(ans, ones - cnt[i] + min(dp[i]))
stdout.write(str(ans)+"\n")
```
Yes
| 87,740 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
from __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------
for _ in range(int(input())):
n,k = mapi()
a = input().strip()
res = float("inf")
if n==1: print(0); continue
pre = [0]*(n+1)
for i in range(n):
pre[i+1] = pre[i]
if a[i]=="1": pre[i+1]+=1
dp1 =[0]*(n+1)
dp2 = [0]*(n+1)
ones = lambda x,y: pre[y]-pre[x]
for i in range(1,n+1):
dp1[i] = pre[i-1]
if i-k>=1:
dp1[i] = min(dp1[i], dp1[i-k]+ones(i-k,i-1))+ (a[i-1]=="0")
for i in range(n,0,-1):
dp2[i] =pre[n]-pre[i]
if i+k<=n:
dp2[i] = min(dp2[i], dp2[i+k]+ones(i,i+k))+ (a[i-1]=="0")
for i in range(1,n+1):
res = min(res,dp1[i]+dp2[i]-(a[i-1]=="0"))
print(max(res,0))
```
Yes
| 87,741 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
import sys
import heapq
import math
import bisect
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def rinput():
return map(int, input().split())
def rlinput():
return list(map(int, input().split()))
def srlinput(fl=False):
return sorted(list(map(int, input().split())) , reverse=fl)
def main():
#n = iinput()
n, k = rinput()
#n, m, k = rinput()
lam = [int(i) for i in list(input())]
lamrev = lam[::-1]
q, qrev = [0], [0]
w, wrev = [], []
for i in range(n):
q.append(q[-1] + lam[i])
for i in range(k):
t = q[i] + 1 - lam[i]
w.append(t)
for i in range(n - k):
i += k
t, f = q[i] + 1 - lam[i], i - k
w.append(t + min(0, w[f] - q[f + 1]))
for i in range(n):
qrev.append(qrev[-1] + lamrev[i])
for i in range(k):
t = qrev[i] + 1 - lamrev[i]
wrev.append(t)
for i in range(n - k):
i += k
t, f = qrev[i] + 1 - lamrev[i], i - k
wrev.append(t + min(0, wrev[f] - qrev[f]))
wrev.reverse()
print(min(min((w[i] + wrev[i] - (lam[i] + 1) % 2) for i in range(n)), q[-1]))
for sdfghjkl in range(iinput()):
main()
```
Yes
| 87,742 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
def car():
for i in range(int(input())):
n,k=map(int,input().split())
s=input().strip()
c1=s.count('1')
ans=n
for i in range(k):
su=0
for j in range(i,n,k):
if(s[j]=='1'):
su+=1
else:
su-=1
su=max(0,su)
ans=min(ans,c1-su)
print(ans)
car()
```
Yes
| 87,743 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
t=int(input())
for i in range(t):
n,k=map(int,input().split())
s=input()
cnt=s.count('1')
ans=cnt
for j in range(k):
score=0
for r in range(j,n,k):
if(s[r]=='1'):
score+=1
else:
score-=1
score=max(0,score)
ans=min(ans,cnt-score)
print(ans)
```
No
| 87,744 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def sieve():
# spf[1] = 1
# for i in range(2, MAXN):
# spf[i] = i
# for i in range(4, MAXN, 2):
# spf[i] = 2
#
# for i in range(3, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
# 7 2 3
def main():
for _ in range(int(input())):
n, k = map(int, input().split())
S = input()
s = []
pre = []
for i in S:
if i == '0' or i == '1':
s.append(int(i))
pre.append(int(i))
for i in range(1, len(pre)):
pre[i] += pre[i - 1]
ans = n
curr = 0
if n==1:
ans=0
for i in range(k):
t=[]
for j in range(i, n, k):
t.append([s[j], j])
end=-1
maxx=0
curr=0
for j in range(len(t)):
if t[j][0]==0 and j>0 and t[j-1][0]==1:
if curr>maxx:
maxx=curr
end=t[j-1][1]
curr=0
else:
if t[j][0]==1:
curr+=1
if curr:
if curr > maxx:
maxx = curr
end = t[-1][1]
if end!=-1:
ans=min(ans, pre[-1]-maxx)
print(ans)
return
if __name__ == "__main__":
main()
```
No
| 87,745 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
t=int(input())
p=0
while p<t:
n,k=list(map(int,(input().split())))
s=input()
f=s.find('1')
l=s.rfind('1')
if f==-1:
print(0)
p+=1
continue
s1=s[f:l+1]
mn=len(s1)
s2=('1'+'0'*(k-1))*n
s3=s2[:mn]
if p==70 or p==71:
print(n,k,s)
#print(s1," ",s2," ",s3)
r=int(s3,2)^int(s1,2)
d=bin(r).count('1')
print(d)
p+=1
```
No
| 87,746 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a garland consisting of n lamps. States of the lamps are represented by the string s of length n. The i-th character of the string s_i equals '0' if the i-th lamp is turned off or '1' if the i-th lamp is turned on. You are also given a positive integer k.
In one move, you can choose one lamp and change its state (i.e. turn it on if it is turned off and vice versa).
The garland is called k-periodic if the distance between each pair of adjacent turned on lamps is exactly k. Consider the case k=3. Then garlands "00010010", "1001001", "00010" and "0" are good but garlands "00101001", "1000001" and "01001100" are not. Note that the garland is not cyclic, i.e. the first turned on lamp is not going after the last turned on lamp and vice versa.
Your task is to find the minimum number of moves you need to make to obtain k-periodic garland from the given one.
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 25~ 000) — the number of test cases. Then t test cases follow.
The first line of the test case contains two integers n and k (1 ≤ n ≤ 10^6; 1 ≤ k ≤ n) — the length of s and the required period. The second line of the test case contains the string s consisting of n characters '0' and '1'.
It is guaranteed that the sum of n over all test cases does not exceed 10^6 (∑ n ≤ 10^6).
Output
For each test case, print the answer — the minimum number of moves you need to make to obtain k-periodic garland from the given one.
Example
Input
6
9 2
010001010
9 3
111100000
7 4
1111111
10 3
1001110101
1 1
1
1 1
0
Output
1
2
5
4
0
0
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------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=0, func=lambda a, b: max(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=0, 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------------------------------------
for ik in range(int(input())):
n,k=map(int,input().split())
s=list(input())
s=[int(s[i]) for i in range(n)]
s1=[0]*(n+1)
for i in range(1,n+1):
s1[i]=s1[i-1]+s[i-1]
dp=[[0 for i in range(3)]for j in range(n)]
for i in range(k):
dp[i][1]=(1-s[i])+s1[i]-s1[0]
dp[i][0]=s[i]+s1[i]-s1[0]
dp[i][2]=dp[i-1][2]+s[i]
for i in range(k,n):
dp[i][1]=min(dp[i-k][1],dp[i-k][2])+s1[i]-s1[i-k+1]+(1-s[i])
dp[i][0]=min(dp[i-k][0],dp[i-k][1])+s1[i]-s1[i-k+1]+s[i]
dp[i][2] = dp[i - 1][2]+s[i]
ans=s1[n]-s1[0]
for i in range(n):
ans=min(ans,min(dp[i]))+s1[n]-s1[i+1]
print(ans)
```
No
| 87,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
from sys import stdin, setrecursionlimit, stdout
#setrecursionlimit(1000000)
from collections import deque
from math import sqrt, floor, ceil, log, log2, log10, pi, gcd, sin, cos, asin
from heapq import heapify, heappop, heappush, heappushpop, heapreplace
def ii(): return int(stdin.readline())
def fi(): return float(stdin.readline())
def mi(): return map(int, stdin.readline().split())
def fmi(): return map(float, stdin.readline().split())
def li(): return list(mi())
def si(): return stdin.readline().rstrip()
def lsi(): return list(si())
#mod=1000000007
res=['NET', 'DA']
############# CODE STARTS HERE #############
test_case=ii()
while test_case:
test_case-=1
n=ii()
a=li()
s=sum([a[i] for i in range(0, n, 2)])
s1=s2=mx1=mx2=0
for i in range(1, n, 2):
x=a[i]-a[i-1]
if s1+x>0:
s1+=x
mx1=max(mx1, s1)
else:
s1=0
for i in range(2, n, 2):
y=a[i-1]-a[i]
if s2+y>0:
s2+=y
mx2=max(mx2, s2)
else:
s2=0
print(s+max(mx1, mx2))
```
| 87,748 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
for t in range(int(input())):
n = int(input())
lst= list(map(int, input().split()))
r = s = rs = 0
x=n//2
for i in range(x):
s =s+ (lst[2 * i + 1] - lst[2 * i])
rs = rs+ lst[2 * i]
s = max(s, 0)
if s > r:
r = s
if n % 2:
rs += lst[-1]
s = 0
y=(n + 1) // 2
for i in range(1,y):
s += (lst[2 * i - 1] - lst[2 * i])
s = max(s, 0)
if s > r:
r =s
print(rs + r)
```
| 87,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
t=int(sys.stdin.readline())
for i in range (t):
num=0
parr=0
imparrr=0
n=int(sys.stdin.readline())
a=list(map(int,input().split()))
suma=0
for i in range(len(a)):
if i%2==0:
suma+=a[i]
for i in range (0,n-1,2):
num += a[i+1] - a[i]
parr=max(parr,num)
if num<0: num=0
num=0
for i in range (1,n-1,2):
num+= a[i] - a[i+1]
parr=max(parr,num)
if num < 0: num = 0
print(parr+suma)
```
| 87,750 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
print(f(n, a))
def f(n, a):
odd_sum = 0
for i in range(0, n, 2): odd_sum += a[i]
prefix = 0
min_prefix_even = 0
min_prefix_odd = float('inf')
max_shift = 0
for i in range(n):
if i % 2 == 0:
prefix -= a[i]
else:
prefix += a[i]
if i % 2 == 1:
max_shift = max(max_shift, prefix - min_prefix_even)
min_prefix_even = min(min_prefix_even, prefix)
else:
max_shift = max(max_shift, prefix - min_prefix_odd)
min_prefix_odd = min(min_prefix_odd, prefix)
return max(odd_sum, odd_sum + max_shift)
if __name__ == '__main__':
main()
```
| 87,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
import random
from math import *
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def finput():
return float(input())
def tinput():
return input().split()
def linput():
return list(input())
def rinput():
return map(int, tinput())
def fiinput():
return map(float, tinput())
def rlinput():
return list(map(int, input().split()))
def trinput():
return tuple(rinput())
def srlinput():
return sorted(list(map(int, input().split())))
def NOYES(fl):
if fl:
print("NO")
else:
print("YES")
def YESNO(fl):
if fl:
print("YES")
else:
print("NO")
def main():
n = iinput()
#k = iinput()
#m = iinput()
#n = int(sys.stdin.readline().strip())
#n, k = rinput()
#n, m = rinput()
#m, k = rinput()
#n, k, m = rinput()
#n, m, k = rinput()
#k, n, m = rinput()
#k, m, n = rinput()
#m, k, n = rinput()
#m, n, k = rinput()
#q = srlinput()
#q = linput()
s, mn, m, res, fp = 0, 0, 0, 0, 0
q = rlinput()
for i in range(n):
f = fp * (i > 0)
if i % 2 == 0:
f -= q[i]
s += q[i]
res = max(res, f - mn)
mn = min(mn, f)
else:
f += q[i]
res = max(res, f - m)
m = min(m, f)
fp = f
print(s + res)
for inytd in range(iinput()):
main()
```
| 87,752 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
Q = int(input())
Query = []
for _ in range(Q):
N = int(input())
A = list(map(int, input().split()))
Query.append((N, A))
def solve(A, ans):
nowmin = 0
w = 0
for p1 in A:
w += p1
nowmin = min(nowmin, w)
ans = max(ans, T + w - nowmin)
return ans
for N, A in Query:
T = 0
P1 = []
P2 = []
for i in range((N-1)//2+1):
T += A[2*i]
if 2*i+1 <= N-1:
d = A[2*i+1] - A[2*i]
P1.append(d)
if 0 <= 2*i-1:
d = A[2*i-1] - A[2*i]
P2.append(d)
ans = solve(P1, T)
ans = solve(P2, ans)
print(ans)
```
| 87,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
a = list(map(int,input().split()))
tot = 0
dif1 = 0
extra = 0
for i in range(0,n,2):
tot+=a[i]
for i in range(0,n-1,2):
dif1-=a[i]
dif1+=a[i+1]
extra = max(extra,dif1)
if dif1 < 0:
dif1 = 0
dif2 = 0
for i in range(1,n-1,2):
dif2+=a[i]
dif2-=a[i+1]
extra = max(extra,dif2)
if dif2 < 0:
dif2 = 0
tot = tot+extra
print(tot)
```
| 87,754 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Tags: divide and conquer, dp, greedy, implementation
Correct Solution:
```
def maxSubArraySum(a,size):
max_so_far =a[0]
curr_max = a[0]
for i in range(1,size):
curr_max = max(a[i], curr_max + a[i])
max_so_far = max(max_so_far,curr_max,0)
return max(max_so_far,0)
for _ in range(int(input())):
N=int(input())
A=list(map(int,input().split()))
temp1=[]
temp2=[]
if(N==1):
print(A[0])
else:
if(N%2==0):
suma=0
for i in range(0,N,2):
temp1.append(A[i+1]-A[i])
suma+=A[i]
for i in range(1,N-1,2):
temp2.append(A[i]-A[i+1])
# print(temp1)
t=maxSubArraySum(temp1,len(temp1))
if(len(temp2)==0):
t2=0
else:
t2=maxSubArraySum(temp2,len(temp2))
r=max(t,t2)
print(r+suma)
else:
suma=0
for i in range(0,N-1,2):
temp1.append(A[i+1]-A[i])
suma+=A[i]
suma+=A[N-1]
# print(suma)
for i in range(1,N,2):
temp2.append(A[i]-A[i+1])
t1=maxSubArraySum(temp1,len(temp1))
t2=maxSubArraySum(temp2,len(temp2))
r=max(t1,t2)
# print(r)
print(suma+r)
```
| 87,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
from sys import stdin
inp = lambda : stdin.readline().strip()
t = int(inp())
for _ in range(t):
n = int(inp())
a = [int(x) for x in inp().split()]
even = 0
for i in range(0,n,2):
even += a[i]
x = [0]*n
p = [0]*n
for i in range(0,n-1,2):
x[i//2] = a[i+1]-a[i]
for i in range(1,n-1,2):
p[i//2] = a[i]-a[i+1]
y = x[0]
ans = x[0]
for i in x[1:]:
y = max(y + i, i)
ans = max(ans,y)
y = p[0]
ans = max(ans,y)
for i in p[1:]:
y = max(y + i, i)
ans = max(ans,y)
print(ans + even)
```
Yes
| 87,756 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
import io,os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
def solve(n,A):
k=n//2
DP=[[0]*3 for _ in range(k+1)]
for i in range(k):
l,r=2*i,2*i+1
DP[i+1][0]=DP[i][0]+A[l]
DP[i+1][1]=max(DP[i][0]+A[r],DP[i][1]+A[r])
DP[i+1][2]=max(DP[i][1]+A[l],DP[i][2]+A[l])
ans=max(DP[-1])
return ans
def main():
t=int(input())
for _ in range(t):
n=int(input())
A=list(map(int,input().split()))
if n%2:
A.append(0)
n+=1
ans1=solve(n,A)
B=A[1:-1]
B.reverse()
ans2=A[0]+solve(n-2,B)
#print(ans1,ans2)
ans=max(ans1,ans2)
sys.stdout.write(str(ans)+'\n')
if __name__=='__main__':
main()
```
Yes
| 87,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=[int(v) for v in input().split()]
v1=[0]
v2=[0]
m=0
s=sum(a[::2])
for j in range(0,n-(n%2),2):
v1.append(max(v1[-1]+a[j+1]-a[j],0))
for j in range(1,n-(1-(n%2)),2):
v2.append(max(a[j]-a[j+1]+v2[-1],0))
m=max(max(v1),max(v2))
print(s+m)
```
Yes
| 87,758 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
l=list(map(int,input().split()))
o=[0]*n
e=[0]*n
e[0]=l[0]
dp=[0]*n
dp[0]=o[0]-e[0]
ans=0
emin=min(0,dp[0])
omin=0
for i in range(1,n):
o[i]=o[i-1]
e[i]=e[i-1]
if i%2==0:
e[i]+=l[i]
else:
o[i]+=l[i]
dp[i]=o[i]-e[i]
if i%2==0:
temp = dp[i] - emin
ans=max(ans,temp)
emin = min(emin,dp[i])
else:
temp = dp[i] - omin
ans = max(ans,temp)
omin = min(omin,dp[i])
# print(dp)
ans+=e[-1]
print(ans)
```
Yes
| 87,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
l=list(map(int,input().split()))
o=0
e=0
m=0
s=0
for i in range(n):
f=1
if i%2==0:
f=0
s+=l[i]
e+=l[i]
else:
o+=l[i]
if o>=e:
e=0
o=l[i]
if o-e>m and (f==0 or i==n-1):
m=o-e
print(s+m)
```
No
| 87,760 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
def calc_01(n, a):
# 01
end = n if n % 2 == 0 else n - 1
cp = 0
P = []
for i in range(0, end, 2):
p = a[i + 1] - a[i]
#print(p)
if p > 0:
if cp > 0:
cp += p
else:
P.append(cp)
cp = p
elif p == 0:
continue
else:
if cp < 0:
cp += p
else:
P.append(cp)
cp = p
#print(P, "dldl")
if cp > 0:
P.append(cp)
while P and P[0] <= 0:
P.pop(0)
while P and P[-1] <= 0:
P.pop()
if P == []:
return 0
#print(P)
M = max(P)
z = 0
csum = P[0]
while z < len(P) - 2:
add = P[z + 2] + P[z + 1]
csum += add
#print(csum, add)
M = max(M, csum)
z += 2
if csum < 0:
csum = P[z]
return M
def calc_12(n, a):
# 12
end = n
cp = 0
P = []
for i in range(2, end, 2):
p = a[i - 1] - a[i]
# print(p)
if p > 0:
if cp > 0:
cp += p
else:
P.append(cp)
cp = p
elif p == 0:
continue
else:
if cp < 0:
cp += p
else:
P.append(cp)
cp = p
#print(P, "dk")
if cp > 0:
P.append(cp)
while P and P[0] <= 0:
P.pop(0)
while P and P[-1] <= 0:
P.pop()
if P == []:
return 0
#print(P, "dl")
M = max(P)
z = 0
csum = P[0]
while z < len(P) - 2:
add = P[z + 2] + P[z + 1]
csum += add
M = max(M, csum)
z += 2
if csum < 0:
csum = P[z]
return M
def sum_evens(n, a):
total = 0
for i in range(0, n, 2):
total += a[i]
return total
def main():
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
M01 = calc_01(n, a)
M12 = calc_12(n, a)
#print(M01, M12)
max_profit = max([M01, M12, 0])
even_sum = sum_evens(n, a)
print(even_sum + max_profit)
main()
```
No
| 87,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
for test in range(int(input())):
n = int(input())
a = list(map(int, input().split()))
if len(a) < 3:
print(a[0])
continue
zero = [a[i + 1] - a[i] for i in range(0, n - n % 2, 2)]
one = [a[i] - a[i + 1] for i in range(1, n - (1 - n % 2), 2)]
ans_zero = zero[0]
summ = 0
for i in zero:
summ += i
ans_zero = max(ans_zero, summ)
summ = max(0, summ)
ans_one = one[0]
summ = 0
for i in one:
summ += i
ans_one = max(ans_one, summ)
summ = max(0, summ)
summ = sum(a[::2])
ans = summ + max(ans_one, ans_zero)
print(max(summ, ans))
```
No
| 87,762 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
a = list(map(int, input().split()))
if n == 1:
print(a[0])
exit()
if n == 2:
print(max(a))
exit()
even, odd = [], []
for i in range(n):
if i % 2 == 0:
even.append(a[i])
else:
odd.append(a[i])
ans1, ans2 = 0, 0
dif1 = []
for i in range(len(odd)):
dif1.append(odd[i] - even[i])
tmp = 0
for i in range(len(dif1)):
tmp += dif1[i]
if tmp < 0:
tmp = 0
if ans1 < tmp:
ans1 = tmp
#print(ans)
dif2 = []
for i in range(len(even)-1):
dif2.append(odd[i] - even[i+1])
tmp = 0
for i in range(len(dif2)):
tmp += dif2[i]
if tmp < 0:
tmp = 0
if ans2 < tmp:
ans2 = tmp
#print(ans)
print(sum(even) + max(ans1, ans2))
```
No
| 87,763 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).
You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.
Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).
You have to answer t independent test cases.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.
The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.
It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).
Output
For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.
Example
Input
4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1
Output
26
5
37
5
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
for t in range(ni()):
n=ni()
l=li()
ans=0
for i in range(0,n,2):
ans+=l[i]
sm1=0
a1=0
for i in range(0,n-1,2):
val=l[i+1]-l[i]
if val<0:
sm1=0
continue
sm1+=val
a1=max(sm1,a1)
sm2=0
a2=0
for i in range(1,n-1,2):
val=l[i]-l[i+1]
if val<0:
sm2=0
continue
sm2+=val
a2=max(a2,sm2)
pn(ans+max(a1,a2))
```
No
| 87,764 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
# from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var): sys.stdout.write('\n'.join(map(str, var)) + '\n')
def out(var): sys.stdout.write(str(var) + '\n')
from decimal import Decimal
# from fractions import Fraction
# sys.setrecursionlimit(100000)
mod = int(1e9) + 7
INF=2**32
n,r1,r2,r3,d=mdata()
a=mdata()
dp1,dp2,dp3=[0]*n,[0]*n,[0]*n
dp1[0]=min(r1,r3)*a[0]+r3
dp2[0]=min((a[0]+1)*r1,r2) + d + min(r1,r2,r3)
dp3[0]=min((a[0]+1)*r1,r2) + 2*d + min(r1,r2,r3)
for i in range(1,n):
dp1[i] = min(dp1[i-1],dp2[i-1]+d,dp3[i-1]) + min(r1,r3)*a[i]+r3 + d
dp3[i] = dp2[i-1] + min((a[i]+1)*r1,r2) + 2*d + min(r1,r2,r3)
dp2[i] = min(dp1[i-1],dp3[i-1]) + 2*d + min((a[i]+1)*r1,r2) + min(r1,r2,r3)
dp1[-1] = min(dp1[-2],dp2[-2],dp3[-2]) + min(r1,r3)*a[i]+r3 + d
out(min(dp1[-1],dp2[-1]+d,dp3[-1]))
```
| 87,765 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
n,r1,r2,r3,d=map(int,input().split())
a=list(map(int,input().split()))
if 2*r1<r3:
save=[r3-2*r1]*n
else:
save=[0]*n
for i in range(n):
save[i]=max(save[i],a[i]*r1+r3-r2-r1)
ans=(n-1)*d+sum(a)*r1+n*r3
dp=[0,0]
for i in range(n):
dp0=dp[1]
dp1=dp[1]
if i+1<n and save[i]+save[i+1]>2*d:
dp1=max(dp1,dp[0]+save[i]+save[i+1]-2*d)
if i==n-1:
dp1=max(dp1,dp[0]+save[i]-2*d)
if i==n-2:
dp0=max(dp0,dp[0]+save[i]-d)
dp1=max(dp1,dp0)
dp=[dp0,dp1]
print(ans-max(dp0,dp1))
```
| 87,766 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
#!/usr/bin/env python3
import io
import os
import sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def printd(*args, **kwargs):
#print(*args, **kwargs, file=sys.stderr)
#print(*args, **kwargs)
pass
def get_str():
return input().decode().strip()
def rint():
return map(int, input().split())
def oint():
return int(input())
n, r1, r2, r3, d = rint()
a = list(rint())
printd(n, r1, r2, r3, d)
printd(a)
dp = [[10**20, 10**20] for i in range(n)]
dp[0][0] = r1*a[0] + r3
dp[0][1] = min(r1*a[0] + r1, r2)
for i in range(1, n):
dp[i][0] = min(dp[i-1][0] + d + r1*a[i] + r3\
,dp[i-1][1] + 3*d + r1*(a[i]+1) + r3\
,dp[i-1][1] + 3*d + r1*(a[i]+3)\
,dp[i-1][1] + 3*d + 2*r1 + r2)
dp[i][1] = min(dp[i-1][0] + d + r1*(a[i]+1)\
,dp[i-1][0] + d + r2\
,dp[i-1][1] + 3*d + r1*(a[i]+2)\
,dp[i-1][1] + 3*d + r1 + r2)
i = n-1
dp[i][0] = min(dp[i][0], dp[i - 1][1] + 2 * d + r1 * (a[i] + 1) + r3)
printd(dp)
print(dp[n-1][0])
```
| 87,767 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
import pprint
n, r1, r2, r3, d = map(int, input().split())
*arr, = map(int, input().split())
dp = [[0] * (n + 1) for _ in range(2)]
dp[0][0] = -d
dp[1][0] = 2 * n * r2 + 2 * n * d
for i in range(n):
fast_kill = arr[i] * r1 + r3
slow_kill = min((arr[i] + 2) * r1, r2 + r1)
# print(i, arr[i], fast_kill, slow_kill)
extra = -d * (i == n - 1)
dp[0][i + 1] = min(dp[0][i] + fast_kill, dp[1][i] + fast_kill + extra, dp[1][i] + slow_kill) + d
dp[1][i + 1] = dp[0][i] + slow_kill + 3 * d
# pprint.pprint(dp)
print(min(dp[0][-1], dp[1][-1]))
```
| 87,768 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
import sys;input=sys.stdin.readline
N, a, b, c, k = map(int, input().split())
X = list(map(int, input().split()))
x, y = 0, 10**18
R = 10**18
for i in range(N):
mnc = min(a*(X[i]+2), a+b)
if i == N-2:
R = min(x,y)+X[-1]*a+c+min(a*(X[-2]+2), a+b)+k
x, y = min(x+a*X[i]+c, y+mnc, x+mnc+2*k), x+mnc+2*k
print(min([R,x])+k*(N-1))
```
| 87,769 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
# region fastio # from https://codeforces.com/contest/1333/submission/75948789
import sys, io, os
BUFSIZE = 8192
class FastIO(io.IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = io.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(io.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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#endregion
N, R1, R2, R3, D = map(int, input().split())
A = list(map(int, input().split()))
A.reverse()
inf = 1<<61
# dp1 = [inf] * (N+1)
# dp2 = [inf] * (N+1)
# dp0 = [inf] * (N+1)
# dpc = [inf] * (N+1)
dp1 = [inf] * N
dp2 = [inf] * N
dp0 = [inf] * N
dpc = [inf] * N
a = A[0]
t1 = R1*a+R3
t2 = min(R1*a+R3+D, R2+R1+D, R1*(a+2)+D)
dp0[0] = t1
dpc[0] = t1
dp1[0] = t2
ans1 = D * (N-1)
ans3 = t2 + D * N
for i, a in enumerate(A[1:], 1):
t1 = R1*a + R3
t2 = min(R1*a+R3+D, R2+R1+D, R1*(a+2)+D)
dp0[i] = min(
dp0[i-1],
dp2[i-1],
dpc[i-1],
) + t1
dp1[i] = min(
dp0[i-1],
dp2[i-1],
) + t2
dp2[i] = dp1[i-1] + t2
dpc[i] = dpc[i-1] + t2
ans3 += t2
ans2 = min(dp0[N-1], dp2[N-1], dpc[N-1])
ans = ans1 + ans2
ans = min(ans, ans3)
print(ans)
```
| 87,770 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
n,r1,r2,r3,D = map(int,input().split())
state = [0,0] # after odd number of 2 (1st), or not (2nd)
a = list(map(int,input().split()))
# First element
# Choosing P~P + A
state[0] = r1 * a[0] + r3
# Choosing L + P later or all P
state[1] = min(r2 + r1 + D, r1 * (a[0] + 2) + D)
# Second to Second Last element
for i in range(1,n-1):
newState = [-1,-1]
newState[0] = min(state[1] + D + r1 * a[i] + r3, state[0] + r1 * a[i] + r3,
state[1] + r2 + r1 + D, state[1] + r1 * (a[i] + 2) + D)
newState[1] = min(state[0] + r2 + r1 + D, state[0] + r1 * (a[i] + 2) + D)
state = newState
# Last Element
ans = min(state[0] + r1 * a[-1] + r3, state[0] + 2 * D + r2 + r1, state[0] + 2 * D + r1 * (a[-1] + 2),
state[1] + r1 * a[-1] + r3, state[1] + r2 + r1 + D, state[1] + r1 * (a[-1] + 2) + D)
print(ans + D * (n-1))
```
| 87,771 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Tags: dp, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n, r1, r2, r3, d = map(int, input().split())
a = list(map(int, input().split()))
INF = 10 ** 18
vals = [val * r1 + r3 for val in a]
vals2 = [min(r2 + r1, (val + 2) * r1, vals[i]) for i, val in enumerate(a)]
dp = [INF] * (n + 1)
dp[0] = 0
for i in range(n):
dp[i + 1] = vals[i] + d + dp[i]
if i - 1 >= 0:
dp[i + 1] = min(vals2[i] + vals2[i - 1] + 4 * d + dp[i - 1], dp[i + 1])
if i - 2 >= 0:
dp[i + 1] = min(vals2[i] + vals2[i - 1] + vals2[i - 2] + d * 7 + dp[i - 2], dp[i + 1])
ans = dp[-1] - d
last = min(vals[-1], vals2[-1] + 2 * d)
for i in range(n - 1)[::-1]:
last += 2 * d + vals2[i]
ans = min(dp[i] + last, ans)
print(ans)
```
| 87,772 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
INF = 10**18
N, r1, r2, r3, d = map(int, readline().split())
A = list(map(int, readline().split()))
dp1 = [INF]*(N+1)
dp2 = [INF]*(N+1)
dp1[0] = -d
dp2 = -d
mj = INF
rr = r1+r2
C = [0]*(N+1)
for i in range(N):
a = A[i]
C[i+1] = min(rr, (a+2)*r1)
CC = C[:]
for i in range(1, N+1):
CC[i] += CC[i-1]
for i in range(1, N+1):
a = A[i-1]
dp1[i] = dp2 + d + C[i]
dp2 = min(dp2+d+r1*a+r3, CC[i] + 3*d*i + mj)
mj = min(mj, dp1[i]-3*i*d-CC[i])
ans = min(dp2, 2*d + dp1[-1])
ans = min(ans, dp1[N-1] + 3*d + C[N])
zz = min(r1*A[-1]+r3, 2*d+min(rr, (A[-1]+2)*r1))
for i in range(1, N):
ans = min(ans, dp1[i] + 2*(N-i)*d + zz + CC[N-1] - CC[i])
print(ans)
```
Yes
| 87,773 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
import sys
# import heapq, functools, collections
# import math, random
# from collections import Counter, defaultdict
# available on Google, not available on Codeforces
# import numpy as np
# import scipy
def solve(lst, r1, r2, r3, d): # fix inputs here
console("----- solving ------")
# console(lst, r1, r2, r3, d)
# baseline = (len(lst)-1)*d + sum([x*r1 + r3 for x in lst])
const_r2_r1_d = r2 + r1 + d
const_2r1_d = 2*r1 + d
# lst_r1x = [r1*x for x in lst]
m1_cost = [r1*x + r3 for x in lst] # shoot all, snipe boss, no relocation necessary
m4_cost = [min(r1*x + const_2r1_d, const_r2_r1_d) for x in lst]
# board clear, relocation necessary
# shoot all incl boss, relocation necessary
# m4_cost = [min(a,b) for a,b in zip(m2_cost, m3_cost)]
baseline = sum(m1_cost) + (len(lst)-1)*d
cost_diff = [a-b for a,b in zip(m1_cost, m4_cost)]
del m1_cost
del m4_cost
# console("m1", m1_cost)
# # console(m2_cost)
# console("m4", m4_cost)
# console(d, cost_diff)
savings = [[0, 0] for _ in lst]
savings[0][1] = max(cost_diff[0], -d)
for i in range(1, len(lst)):
savings[i][0] = max(savings[i-1][0], # no action
savings[i-1][1] + cost_diff[i], # use outstanding
savings[i-1][1] - d) # use outstanding and supplement, i.e. m1 only
savings[i][1] = max(savings[i-1][0] + cost_diff[i], # start outstanding
savings[i-1][0] - d) # start outstanding with supplement
# console(savings)
total_savings = max(0, savings[-2][1], savings[-1][0], savings[-1][1] - d)
return baseline - total_savings
def console(*args): # the judge will not read these print statement
# print('\033[36m', *args, '\033[0m', file=sys.stderr)
return
# fast read all
inp = sys.stdin.readlines()
for _ in [1]:
# read line as a string
# strr = input()
# read line as an integer
# _ = int(input())
# read one line and parse each word as a string
# lst = input().split()
# read one line and parse each word as an integer
# lst = list(map(int,input().split()))
# read matrix and parse as integers (after reading read nrows)
_, r1, r2, r3, d = list(map(int,inp[0].split()))
lst = list(map(int,inp[1].split()))
# nrows = lst[0] # index containing information, please change
# grid = []
# for _ in range(nrows):
# grid.append(list(map(int,input().split())))
res = solve(lst, r1, r2, r3, d) # please change
# Google - case number required
# print("Case #{}: {}".format(case_num+1, res))
# Codeforces - no case number required
print(res)
```
Yes
| 87,774 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, a, b, c, k = map(int, input().split())
X = list(map(int, input().split()))
x, y = 0, 10**18
R = 10**18
for i in range(N):
mnc = min(a*(X[i]+2), a+b)
if i != N-1:
x, y = min(x+a*X[i]+c, y+mnc, x+mnc+2*k), x+mnc+2*k
else:
x, y = min(x+a*X[i]+c, y+mnc, x+mnc+2*k, y+a*X[i]+c-k), x+mnc+2*k
print(min(R,x)+k*(N-1))
```
Yes
| 87,775 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10 ** 5)
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]
inf=10**16
n,r1,r2,r3,d=MI()
aa=LI()
dp0=-d
dp1=inf
pre1=inf
for i,a in enumerate(aa):
pre0,pre1=dp0,dp1
dp0=min(pre0+d+r1*a+r3,pre1+d*3+min(r1*(a+1),r2)+r1*2)
dp1=pre0+d+min(r1*(a+1),r2)
# print(dp)
print(min(dp0,dp1+d*2+r1,pre1+d*2+(aa[-1]+1)*r1+r3))
```
Yes
| 87,776 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
#!/usr/bin/env python3
import io
import os
import sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def printd(*args, **kwargs):
#print(*args, **kwargs, file=sys.stderr)
#print(*args, **kwargs)
pass
def get_str():
return input().decode().strip()
def rint():
return map(int, input().split())
def oint():
return int(input())
n, r1, r2, r3, d = rint()
a = list(rint())
printd(n, r1, r2, r3, d)
printd(a)
dp = [[10**20, 10**20] for i in range(n)]
dp[0][0] = r1*a[0] + r3
dp[0][1] = min(r1*a[0] + r1, r2)
for i in range(1, n):
# 0 -> 0
dp[i][0] = min(dp[i][0], dp[i-1][0] + d + r1*a[i] + r3)
# 1 -> 0
dp[i][0] = min(dp[i][0], dp[i-1][1] + 3*d + r1*(a[i]+1) + r3)
dp[i][0] = min(dp[i][0], dp[i-1][1] + 3*d + r1*(a[i]+3))
dp[i][0] = min(dp[i][0], dp[i-1][1] + 3*d + 2*r1 + r2)
if i == n - 1:
dp[i][0] = min(dp[i][0], dp[i-1][1] + 2*d + r1*(a[i]+1) + r3)
# 0 -> 1
dp[i][1] = min(dp[i][1], dp[i-1][0] + d + r1*(a[i]+1))
dp[i][1] = min(dp[i][1], dp[i-1][0] + d + r2)
# 1 -> 1
dp[i][1] = min(dp[i][1], dp[i-1][1] + 3*d + r1*(a[i]+2))
dp[i][1] = min(dp[i][1], dp[i-1][1] + 2*d + r1 + r2)
printd(dp)
print(dp[n-1][0])
```
No
| 87,777 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
import sys;input=sys.stdin.readline
N, a, b, c, k = map(int, input().split())
X = list(map(int, input().split()))
Y = [0]*N
Z = [0]*N
f = 0
for i in range(N):
kk = min(b+a, a*(X[i]+2))+2*k
ll = a*X[i]+c
if f:
Y[i] = min(ll, kk-2*k)
else:
Y[i] = min(ll, kk)
if ll > kk:
f = 1
else:
f = 0
if i != N-1:
Z[i] = min(min(b+a, a*(X[i]+2)), a*X[i]+c)
else:
Z[i] = Y[i]
#print(Y)
#print(Z)
for i in range(N-2, -1, -1):
Z[i] += Z[i+1]
Z.append(0)
for i in range(1, N):
Y[i] += Y[i-1]
#print(Y)
#print(Z)
t = (N-1)*k
#print(Y[-1]+t, Z[0] + 2*t)
R = min(Y[-1]+t, Z[0] + 2*t)
for i in range(N):
R = min(R, Y[i]+Z[i+1]+(N-1-i)+t)
# print((N-1-i),Y[i]+Z[i+1]+(N-1-i)+t)
print(R)
```
No
| 87,778 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
from sys import stdin, stdout
# case 1: r1*a + r3
# case 2: r2 + [r1]
# case 3: r1*(a+1) + [r1]
def monster_invaders(n, r1, r2, r3, d, a_a):
dp = [[2**63-1, 2**63-1] for _ in range(n)]
dp[0][0] = r1 * a_a[0] + r3
dp[0][1] = min(r2, r1 * (a_a[0] + 1))
for i in range(1, n):
# 0 -> 0
dp[i][0] = min(dp[i][0], dp[i-1][0] + d + r1 * a_a[i] + r3)
# 1 -> 0
# 1 -> (0) -> 0 -> (0)
dp[i][0] = min(dp[i][0], dp[i-1][1] + d + (r1 * a_a[i] + r3) + d + r1 + d)
# 1 -> (1) -> 0 -> (0)
dp[i][0] = min(dp[i][0], dp[i-1][1] + d + min(r2, r1 * (a_a[1] + 1)) + d + r1 + d + r1)
# 0 -> 1
dp[i][1] = min(dp[i][1], dp[i-1][0] + d + min(r2, r1 * (a_a[i] + 1)))
# 1 -> 1
# 1 -> (a) -> 0 -> (1)
dp[i][1] = min(dp[i][1], dp[i-1][1] + d + d + r1 + d + min(r2, r1 * (a_a[i] + 1)))
if i == n - 1:
dp[i][0] = min(dp[i][0], dp[i-1][1] + d + (r1 * a_a[i] + r3) + d + r1)
#print(dp)
return dp[n-1][0]
n, r1, r2, r3, d = map(int, stdin.readline().split())
a_a = list(map(int, stdin.readline().split()))
ans = monster_invaders(n, r1, r2, r3, d, a_a)
stdout.write(str(ans) + '\n')
```
No
| 87,779 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Ziota found a video game called "Monster Invaders".
Similar to every other shooting RPG game, "Monster Invaders" involves killing monsters and bosses with guns.
For the sake of simplicity, we only consider two different types of monsters and three different types of guns.
Namely, the two types of monsters are:
* a normal monster with 1 hp.
* a boss with 2 hp.
And the three types of guns are:
* Pistol, deals 1 hp in damage to one monster, r_1 reloading time
* Laser gun, deals 1 hp in damage to all the monsters in the current level (including the boss), r_2 reloading time
* AWP, instantly kills any monster, r_3 reloading time
The guns are initially not loaded, and the Ziota can only reload 1 gun at a time.
The levels of the game can be considered as an array a_1, a_2, …, a_n, in which the i-th stage has a_i normal monsters and 1 boss. Due to the nature of the game, Ziota cannot use the Pistol (the first type of gun) or AWP (the third type of gun) to shoot the boss before killing all of the a_i normal monsters.
If Ziota damages the boss but does not kill it immediately, he is forced to move out of the current level to an arbitrary adjacent level (adjacent levels of level i (1 < i < n) are levels i - 1 and i + 1, the only adjacent level of level 1 is level 2, the only adjacent level of level n is level n - 1). Ziota can also choose to move to an adjacent level at any time. Each move between adjacent levels are managed by portals with d teleportation time.
In order not to disrupt the space-time continuum within the game, it is strictly forbidden to reload or shoot monsters during teleportation.
Ziota starts the game at level 1. The objective of the game is rather simple, to kill all the bosses in all the levels. He is curious about the minimum time to finish the game (assuming it takes no time to shoot the monsters with a loaded gun and Ziota has infinite ammo on all the three guns). Please help him find this value.
Input
The first line of the input contains five integers separated by single spaces: n (2 ≤ n ≤ 10^6) — the number of stages, r_1, r_2, r_3 (1 ≤ r_1 ≤ r_2 ≤ r_3 ≤ 10^9) — the reload time of the three guns respectively, d (1 ≤ d ≤ 10^9) — the time of moving between adjacent levels.
The second line of the input contains n integers separated by single spaces a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^6, 1 ≤ i ≤ n).
Output
Print one integer, the minimum time to finish the game.
Examples
Input
4 1 3 4 3
3 2 5 1
Output
34
Input
4 2 4 4 1
4 5 1 2
Output
31
Note
In the first test case, the optimal strategy is:
* Use the pistol to kill three normal monsters and AWP to kill the boss (Total time 1⋅3+4=7)
* Move to stage two (Total time 7+3=10)
* Use the pistol twice and AWP to kill the boss (Total time 10+1⋅2+4=16)
* Move to stage three (Total time 16+3=19)
* Use the laser gun and forced to move to either stage four or two, here we move to stage four (Total time 19+3+3=25)
* Use the pistol once, use AWP to kill the boss (Total time 25+1⋅1+4=30)
* Move back to stage three (Total time 30+3=33)
* Kill the boss at stage three with the pistol (Total time 33+1=34)
Note that here, we do not finish at level n, but when all the bosses are killed.
Submitted Solution:
```
mod = 1000000007
eps = 10**-9
inf = 10**18
def main():
import sys
input = sys.stdin.readline
N, r1, r2, r3, d = map(int, input().split())
A = list(map(int, input().split()))
dp = [inf] * (N+1)
dp[0] = -d
for i in range(N):
a = A[i]
dp[i+1] = min(dp[i+1], dp[i] + d + a * r1 + r3)
if i+1 < N:
if i+2 != N:
dp[i+2] = min(dp[i+2], dp[i] + d*4 + min(r2 + r1, r1 * (a+2)) + min(r2 + r1, r1 * (A[i+1] + 2), r1 * A[i+1] + r3))
else:
dp[i + 2] = min(dp[i + 2],
dp[i] + d * 4 + min(r2 + r1, r1 * (a + 2)) + min(r2 + r1, r1 * (A[i + 1] + 2)),
dp[i] + d * 3 + min(r2 + r1, r1 * (a + 2)) + r1 * A[i+1] + r3)
print(dp[N])
#print(dp)
if __name__ == '__main__':
main()
```
No
| 87,780 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Santa Claus and his assistant Elf delivered all the presents and made all the wishes come true, they returned to the North Pole and found out that it is all covered with snow. Both of them were quite tired and they decided only to remove the snow from the roads connecting huts. The North Pole has n huts connected with m roads. One can go along the roads in both directions.
The Elf offered to split: Santa Claus will clear up the wide roads and the Elf will tread out the narrow roads. For each road they decided who will clear it: Santa Claus or the Elf. To minimize the efforts they decided to clear the road so as to fulfill both those conditions:
* between any two huts should exist exactly one simple path along the cleared roads;
* Santa Claus and the Elf should clear the same number of roads.
At this point Santa Claus and his assistant Elf wondered which roads should they clear up?
Input
The first input line contains two positive integers n and m (1 ≤ n ≤ 103, 1 ≤ m ≤ 105) — the number of huts and the number of roads. Then follow m lines, each of them contains a road description: the numbers of huts it connects — x and y (1 ≤ x, y ≤ n) and the person responsible for clearing out this road ("S" — for the Elf or "M" for Santa Claus). It is possible to go on each road in both directions. Note that there can be more than one road between two huts and a road can begin and end in the same hut.
Output
Print "-1" without the quotes if it is impossible to choose the roads that will be cleared by the given rule. Otherwise print in the first line how many roads should be cleared and in the second line print the numbers of those roads (the roads are numbered from 1 in the order of occurrence in the input). It is allowed to print the numbers of the roads in any order. Each number should be printed exactly once. As you print the numbers, separate them with spaces.
Examples
Input
1 2
1 1 S
1 1 M
Output
0
Input
3 3
1 2 S
1 3 M
2 3 S
Output
2
2 1
Input
5 6
1 1 S
1 2 M
1 3 S
1 4 M
1 5 M
2 2 S
Output
-1
Note
A path is called simple if all huts on it are pairwise different.
Submitted Solution:
```
N, M = map(int, input().split())
routes = []
Sctn = 0
rtd = {"S": 0, "M": 0}
if N % 2 == 0:
print(-1)
else:
Edges = []
Sedges = []
Medges = []
for i in range(M):
inp = input().split()
start, end = map(int, inp[:2])
actor = inp[2]
if actor != "S":
Sctn += 1
"""
if actor == "S":
Sedges.append([1, start - 1, end - 1])
else:
Medges.append([1, start - 1, end - 1])
"""
Edges.append([1, start - 1, end - 1, actor])
# print("E", Edges)
# print("S", Sedges)
# print("M", Medges)
if Sctn - 1 > (N - 1) / 2:
print(-1)
else:
Comp = [i for i in range(N)]
Ans = 0
for i in range(len(Edges)):
edge = Edges[i]
# print(edge)
weight = edge[0]
start = edge[1]
end = edge[2]
actor = edge[3]
if Comp[start] != Comp[end]:
Ans += weight
# print(weight, start, end, actor)
routes.append(i + 1)
rtd[actor] += 1
a = Comp[start]
b = Comp[end]
for i in range(N):
if Comp[i] == b:
Comp[i] = a
if rtd["S"] != rtd["M"]:
print(-1)
else:
print(Ans)
print(*routes)
```
No
| 87,781 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Santa Claus and his assistant Elf delivered all the presents and made all the wishes come true, they returned to the North Pole and found out that it is all covered with snow. Both of them were quite tired and they decided only to remove the snow from the roads connecting huts. The North Pole has n huts connected with m roads. One can go along the roads in both directions.
The Elf offered to split: Santa Claus will clear up the wide roads and the Elf will tread out the narrow roads. For each road they decided who will clear it: Santa Claus or the Elf. To minimize the efforts they decided to clear the road so as to fulfill both those conditions:
* between any two huts should exist exactly one simple path along the cleared roads;
* Santa Claus and the Elf should clear the same number of roads.
At this point Santa Claus and his assistant Elf wondered which roads should they clear up?
Input
The first input line contains two positive integers n and m (1 ≤ n ≤ 103, 1 ≤ m ≤ 105) — the number of huts and the number of roads. Then follow m lines, each of them contains a road description: the numbers of huts it connects — x and y (1 ≤ x, y ≤ n) and the person responsible for clearing out this road ("S" — for the Elf or "M" for Santa Claus). It is possible to go on each road in both directions. Note that there can be more than one road between two huts and a road can begin and end in the same hut.
Output
Print "-1" without the quotes if it is impossible to choose the roads that will be cleared by the given rule. Otherwise print in the first line how many roads should be cleared and in the second line print the numbers of those roads (the roads are numbered from 1 in the order of occurrence in the input). It is allowed to print the numbers of the roads in any order. Each number should be printed exactly once. As you print the numbers, separate them with spaces.
Examples
Input
1 2
1 1 S
1 1 M
Output
0
Input
3 3
1 2 S
1 3 M
2 3 S
Output
2
2 1
Input
5 6
1 1 S
1 2 M
1 3 S
1 4 M
1 5 M
2 2 S
Output
-1
Note
A path is called simple if all huts on it are pairwise different.
Submitted Solution:
```
N, M = map(int, input().split())
routes = []
connected = {}
Sctn = 0
rtd = {"S": 0, "M": 0}
if N % 2 == 0:
print(-1, 0)
else:
Edges = []
for i in range(M):
inp = input().split()
start, end = map(int, inp[:2])
actor = inp[2]
if actor != "S":
Sctn += 1
Edges.append([1, start - 1, end - 1, actor])
# print("E", Edges)
# print("S", Sedges)
# print("M", Medges)
Comp = [i for i in range(N)]
Ans = 0
for i in range(len(Edges)):
edge = Edges[i]
# print(edge)
weight = edge[0]
start = edge[1]
end = edge[2]
actor = edge[3]
if Comp[start] != Comp[end]:
Ans += weight
# print(weight, start, end, actor)
routes.append(i + 1)
rtd[actor] += 1
connected.update({start: True})
connected.update({end: True})
a = Comp[start]
b = Comp[end]
for i in range(N):
if Comp[i] == b:
Comp[i] = a
# print(len(connected.keys()), N)
if len(connected.keys()) < N - 1:
print(-1)
elif rtd["M"] - 1 > (N - 1) // 2:
print(-1)
elif rtd["S"] != rtd["M"]:
print(-1)
else:
print(Ans)
print(*routes)
```
No
| 87,782 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Santa Claus and his assistant Elf delivered all the presents and made all the wishes come true, they returned to the North Pole and found out that it is all covered with snow. Both of them were quite tired and they decided only to remove the snow from the roads connecting huts. The North Pole has n huts connected with m roads. One can go along the roads in both directions.
The Elf offered to split: Santa Claus will clear up the wide roads and the Elf will tread out the narrow roads. For each road they decided who will clear it: Santa Claus or the Elf. To minimize the efforts they decided to clear the road so as to fulfill both those conditions:
* between any two huts should exist exactly one simple path along the cleared roads;
* Santa Claus and the Elf should clear the same number of roads.
At this point Santa Claus and his assistant Elf wondered which roads should they clear up?
Input
The first input line contains two positive integers n and m (1 ≤ n ≤ 103, 1 ≤ m ≤ 105) — the number of huts and the number of roads. Then follow m lines, each of them contains a road description: the numbers of huts it connects — x and y (1 ≤ x, y ≤ n) and the person responsible for clearing out this road ("S" — for the Elf or "M" for Santa Claus). It is possible to go on each road in both directions. Note that there can be more than one road between two huts and a road can begin and end in the same hut.
Output
Print "-1" without the quotes if it is impossible to choose the roads that will be cleared by the given rule. Otherwise print in the first line how many roads should be cleared and in the second line print the numbers of those roads (the roads are numbered from 1 in the order of occurrence in the input). It is allowed to print the numbers of the roads in any order. Each number should be printed exactly once. As you print the numbers, separate them with spaces.
Examples
Input
1 2
1 1 S
1 1 M
Output
0
Input
3 3
1 2 S
1 3 M
2 3 S
Output
2
2 1
Input
5 6
1 1 S
1 2 M
1 3 S
1 4 M
1 5 M
2 2 S
Output
-1
Note
A path is called simple if all huts on it are pairwise different.
Submitted Solution:
```
N, M = map(int, input().split())
routes = []
connected = {}
Sctn = 0
rtd = {"S": 0, "M": 0}
if N % 2 == 0:
print(-1, 0)
else:
Edges = []
for i in range(M):
inp = input().split()
start, end = map(int, inp[:2])
actor = inp[2]
if actor != "S":
Sctn += 1
Edges.append([1, start - 1, end - 1, actor])
# print("E", Edges)
# print("S", Sedges)
# print("M", Medges)
Comp = [i for i in range(N)]
Ans = 0
for i in range(len(Edges)):
edge = Edges[i]
# print(edge)
weight = edge[0]
start = edge[1]
end = edge[2]
actor = edge[3]
if Comp[start] != Comp[end]:
Ans += weight
# print(weight, start, end, actor)
routes.append(i + 1)
rtd[actor] += 1
connected.update({start: True})
connected.update({end: True})
a = Comp[start]
b = Comp[end]
for i in range(N):
if Comp[i] == b:
Comp[i] = a
if len(connected.keys()) < N:
print(-1)
elif rtd["M"] - 1 > (N - 1) // 2:
print(-1, 1)
elif rtd["S"] != rtd["M"]:
print(-1)
else:
print(Ans)
print(*routes)
```
No
| 87,783 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
After Santa Claus and his assistant Elf delivered all the presents and made all the wishes come true, they returned to the North Pole and found out that it is all covered with snow. Both of them were quite tired and they decided only to remove the snow from the roads connecting huts. The North Pole has n huts connected with m roads. One can go along the roads in both directions.
The Elf offered to split: Santa Claus will clear up the wide roads and the Elf will tread out the narrow roads. For each road they decided who will clear it: Santa Claus or the Elf. To minimize the efforts they decided to clear the road so as to fulfill both those conditions:
* between any two huts should exist exactly one simple path along the cleared roads;
* Santa Claus and the Elf should clear the same number of roads.
At this point Santa Claus and his assistant Elf wondered which roads should they clear up?
Input
The first input line contains two positive integers n and m (1 ≤ n ≤ 103, 1 ≤ m ≤ 105) — the number of huts and the number of roads. Then follow m lines, each of them contains a road description: the numbers of huts it connects — x and y (1 ≤ x, y ≤ n) and the person responsible for clearing out this road ("S" — for the Elf or "M" for Santa Claus). It is possible to go on each road in both directions. Note that there can be more than one road between two huts and a road can begin and end in the same hut.
Output
Print "-1" without the quotes if it is impossible to choose the roads that will be cleared by the given rule. Otherwise print in the first line how many roads should be cleared and in the second line print the numbers of those roads (the roads are numbered from 1 in the order of occurrence in the input). It is allowed to print the numbers of the roads in any order. Each number should be printed exactly once. As you print the numbers, separate them with spaces.
Examples
Input
1 2
1 1 S
1 1 M
Output
0
Input
3 3
1 2 S
1 3 M
2 3 S
Output
2
2 1
Input
5 6
1 1 S
1 2 M
1 3 S
1 4 M
1 5 M
2 2 S
Output
-1
Note
A path is called simple if all huts on it are pairwise different.
Submitted Solution:
```
N, M = map(int, input().split())
routes = []
Sctn = 0
if N % 2 == 0:
print(-1)
else:
Edges = []
Sedges = []
Medges =[]
for i in range(M):
inp = input().split()
start, end = map(int, inp[:2])
actor = inp[2]
if actor != "S":
Sctn += 1
"""
if actor == "S":
Sedges.append([1, start - 1, end - 1])
else:
Medges.append([1, start - 1, end - 1])
"""
Edges.append([1, start - 1, end - 1, actor])
# print("E", Edges)
# print("S", Sedges)
# print("M", Medges)
if Sctn - 1 > (N - 1) / 2:
print(-1)
else:
Comp = [i for i in range(N)]
Ans = 0
for i in range(len(Edges)):
edge = Edges[i]
# print(edge)
weight = edge[0]
start = edge[1]
end = edge[2]
actor = edge[3]
if Comp[start] != Comp[end]:
Ans += weight
# print(weight, start, end, actor)
routes.append(i + 1)
a = Comp[start]
b = Comp[end]
for i in range(N):
if Comp[i] == b:
Comp[i] = a
print(Ans)
print(*routes)
```
No
| 87,784 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
ans = set([])
for i in range(n):
x = 1
while x <= a[i]:
x *= 2
j = i+1
sum = 0
while j < n-1 and sum < x:
sum += a[j]
if a[i] ^ a[j+1] == sum:
ans.add(n * i + j + 1)
j += 1
sum = 0
j = i-1
while j>0 and sum < x:
sum += a[j]
if a[i] ^ a[j-1] == sum:
ans.add(n * (j-1) + i)
j -= 1
print(len(ans))
```
| 87,785 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
import sys
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
from itertools import *
from functools import *
###################### Start Here ######################
n = ii1()
arr = iia()
ans = 0
for l in range(30):
for i in range(n):
if arr[i]&(1<<l):
currsum = 0
for j in range(i+2,n):
currsum+=arr[j-1]
if currsum>=(2<<l):break
if currsum<(1<<l):continue
if arr[i]^arr[j]==currsum:ans+=1
currsum = 0
for j in range(i-2,-1,-1):
currsum+=arr[j+1]
if currsum>=(2<<l):break
if currsum<(1<<l):continue
if arr[i]^arr[j]==currsum:ans+=1
print(ans)
```
| 87,786 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
def solve(a):
seen = set()
for i in range(len(a)):
c = 0
for j in range(i+2,len(a)):
c += a[j-1]
if a[i]^a[j] == c: seen.add((i,j))
if c >= 2*a[i]: break
for i in range(len(a)-1,-1,-1):
c = 0
for j in range(i-2,-1,-1):
c += a[j+1]
if a[i]^a[j] == c: seen.add((j,i))
if c >= 2 *a[i]: break
print(len(seen))
n = int(input());solve(list(map(int,input().split())))
```
| 87,787 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
def calc(a):
ans = 0
for i in range(2, len(a)):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] > a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
return ans
print(calc(a) + calc(a[::-1]))
```
| 87,788 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
import itertools, math
n = int(input())
A = list(map(int, input().split()))
acc = [0] + list(itertools.accumulate(A))
ans = 0
seen = set()
for i in range(n - 2):
a = int(math.log2(A[i]))
for j in range(i + 2, n):
cur = acc[j] - acc[i + 1]
b = int(math.log2(cur))
if b > a:
break
if A[i] ^ A[j] == cur and (i, j) not in seen:
ans += 1
seen.add((i, j))
for j in range(n - 1, 1, -1):
a = int(math.log2(A[j]))
for i in range(j - 2, -1, -1):
cur = acc[j] - acc[i + 1]
b = int(math.log2(cur))
if b > a:
break
if A[i] ^ A[j] == cur and (i, j) not in seen:
ans += 1
seen.add((i, j))
print(ans)
```
| 87,789 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
t=1
for i in range(t):
n=N()
a=RLL()
ans=0
for j in range(2):
pre=[0]
for i in range(n):
pre.append(pre[-1]+a[i])
for i in range(n-2):
k=len(bin(a[i]))-2
k=1<<k
for r in range(i+2,n):
if pre[r]-pre[i+1]>k:
break
if a[i]>a[r] and a[i]^a[r]==pre[r]-pre[i+1]:
ans+=1
a=a[::-1]
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
| 87,790 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
from sys import *
input = stdin.readline
def solve(n, a, t):
ans = 0
for i in range(n):
sum = 0
high1 = a[i].bit_length()-1
for j in range(i+1, n-1):
high2 = a[j+1].bit_length()-1
sum += a[j]
if(sum >= (1<<(high1+1))):
break
if((a[i]^a[j+1]) == sum and (t == 0 or high1 != high2)):
ans += 1
return ans
n = int(input())
a = list(map(int, input().split()))
ans = solve(n, a, 0)
a = a[::-1]
ans += solve(n, a, 1)
print(ans)
```
| 87,791 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Tags: binary search, bitmasks, brute force, constructive algorithms, divide and conquer, two pointers
Correct Solution:
```
def solve(a):
seen = set()
for i in range(len(a)):
c = 0
for j in range(i+2,len(a)):
c += a[j-1]
if a[i]^a[j] == c:
seen.add((i,j))
if c >= 2*a[i]:
break
for i in range(len(a)-1,-1,-1):
c = 0
for j in range(i-2,-1,-1):
c += a[j+1]
if a[i]^a[j] == c:
seen.add((j,i))
if c >= 2 *a[i]:
break
print(len(seen))
n = int(input())
a = list(map(int,input().split()))
solve(a)
```
| 87,792 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] > a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
a.reverse()
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] >= a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or a[j].bit_length() > a[i].bit_length():
break
print(ans)
```
Yes
| 87,793 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
def f(a):
ans = 0
for i in range(n - 2):
s = 0
for j in range(i + 2, n):
s += a[j - 1]
ans += a[i] > a[j] and a[i] ^ a[j] == s
if s > 2 * a[i]: break
return ans
read = lambda: map(int, input().split())
n = int(input())
a = list(read())
print(f(a) + f(a[::-1]))
```
Yes
| 87,794 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split(' ')))
ans = 0
for i in range(n):
sum = 0
for j in range(i + 2, n, 1):
sum += a[j - 1]
if (sum >= a[i] + a[i]):
break
if ((a[i] ^ a[j]) == sum and a[i] >= a[j]):
ans += 1
for i in range(n):
sum = 0
for j in range(i - 2, -1, -1):
sum += a[j + 1]
if (sum >= a[i] + a[i]):
break
if ((a[i] ^ a[j]) == sum and a[i] > a[j]):
ans += 1
print(ans)
```
Yes
| 87,795 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
import math
n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] < a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or int(math.log2(a[j])) > int(math.log2(a[i])):
break
a.reverse()
for i in range(2, n):
sum = 0
for j in reversed(range(0, i-1)):
sum += a[j+1]
ans += a[i] >= a[j] and a[i]^a[j] == sum
if sum > 2*a[i] or int(math.log2(a[j])) > int(math.log2(a[i])):
break
print(ans)
```
No
| 87,796 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
cnt = 0
for i in range(n-2):
s = 0
for j in range(i+1,n-1):
s += a[j]
if (a[i] ^ a[j] == s):
cnt += 1
k = len(bin(a[i])[2:])
if (s >= (1 << (k+1))):
break
a = list(reversed(a))
for i in range(n-2):
s = 0
for j in range(i+1,n-1):
s += a[j]
if (a[i] ^ a[j+1] == s):
cnt += 1
k = len(bin(a[i])[2:])
if (s >= (1 << (k+1))):
break
print(cnt)
```
No
| 87,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
def checkgood(a,n):
if len(a)>=3:
if (a[0]^a[-1])==sum(a[1:n-1]):
return True
else:
return False
else:
return False
t = int(input())
a = (input().split())
a = [int(i) for i in a]
x = 0
for i in range(t):
for j in range(i,t+1):
p = a[i:j]
if checkgood(p,len(p)):
print(p)
x += 1
print(x)
```
No
| 87,798 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Yurii is sure he can do everything. Can he solve this task, though?
He has an array a consisting of n positive integers. Let's call a subarray a[l...r] good if the following conditions are simultaneously satisfied:
* l+1 ≤ r-1, i. e. the subarray has length at least 3;
* (a_l ⊕ a_r) = (a_{l+1}+a_{l+2}+…+a_{r-2}+a_{r-1}), where ⊕ denotes the [bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
In other words, a subarray is good if the bitwise XOR of the two border elements is equal to the sum of the rest of the elements.
Yurii wants to calculate the total number of good subarrays. What is it equal to?
An array c is a subarray of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Input
The first line contains a single integer n (3 ≤ n ≤ 2⋅ 10^5) — the length of a.
The second line contains n integers a_1,a_2,…,a_n (1 ≤ a_i < 2^{30}) — elements of a.
Output
Output a single integer — the number of good subarrays.
Examples
Input
8
3 1 2 3 1 2 3 15
Output
6
Input
10
997230370 58052053 240970544 715275815 250707702 156801523 44100666 64791577 43523002 480196854
Output
2
Note
There are 6 good subarrays in the example:
* [3,1,2] (twice) because (3 ⊕ 2) = 1;
* [1,2,3] (twice) because (1 ⊕ 3) = 2;
* [2,3,1] because (2 ⊕ 1) = 3;
* [3,1,2,3,1,2,3,15] because (3 ⊕ 15) = (1+2+3+1+2+3).
Submitted Solution:
```
l = int(input())
data = [int(x) for x in input().split()]
fl = False
def func(num):
a = []
while True:
a.append(num % 2)
num //= 2
if num < 1:
return len(a)
def main_f(data):
count = 0
for i in range(l):
s = 0
a = data[i]
func_a = func(a)
for j in range(i + 1, l):
b = data[j]
if s == a ^ b and (not fl or fl and a > b) and j != i + 1:
count += 1
elif func(s + data[j]) <= func_a:
s += data[j]
else:
break
return count
f_res = main_f(data)
fl = True
s_res = main_f(data[-1::-1])
print(f_res + s_res)
```
No
| 87,799 |
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