message stringlengths 2 433k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).
Let's call some subsegment p[l, r] of this permutation special if p_l + p_r = max _{i = l}^{r} p_i. Please calculate the number of special subsegments.
Input
The first line contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). All these integers are pairwise distinct.
Output
Print the number of special subsegments of the given permutation.
Examples
Input
5
3 4 1 5 2
Output
2
Input
3
1 3 2
Output
1
Note
Special subsegments in the first example are [1, 5] and [1, 3].
The only special subsegment in the second example is [1, 3].
Submitted Solution:
```
def Count_Segment(a,n):
ans=0
upto=[False]*(n+1)
for i in range(1,n-1):
if a[i]>a[i-1] and a[i]>a[i+1]:
curr=a[i]
for j in range(i-1,-1,-1):
if a[j]<curr:
upto[a[j]]=curr
else:
break
for j in range(i+1,n):
if a[j]<curr and upto[curr-a[j]]==curr:
ans+=1
else:
break
return ans
n=int(input())
a=list(map(int,input().split( )))
print(Count_Segment(a,n))
``` | instruction | 0 | 75,058 | 12 | 150,116 |
No | output | 1 | 75,058 | 12 | 150,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).
Let's call some subsegment p[l, r] of this permutation special if p_l + p_r = max _{i = l}^{r} p_i. Please calculate the number of special subsegments.
Input
The first line contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). All these integers are pairwise distinct.
Output
Print the number of special subsegments of the given permutation.
Examples
Input
5
3 4 1 5 2
Output
2
Input
3
1 3 2
Output
1
Note
Special subsegments in the first example are [1, 5] and [1, 3].
The only special subsegment in the second example is [1, 3].
Submitted Solution:
```
n=int(input())
ns=[int(x) for x in input().split()]
maxx=max(ns)+10
lst=[None]*n
nxt=[None]*n
tmp=[(-1,maxx)]
mp=[False]*n
wh=[None]*(n+1)
for i in range(n):
c=ns[i]
while tmp[-1][1]<=c:
tmp.pop()
lst[i]=tmp[-1][0]
tmp.append((i,c))
tmp=[(n,maxx)]
for i in range(n):
i=n-i-1
c=ns[i]
while tmp[-1][1]<=c:
tmp.pop()
nxt[i]=tmp[-1][0]
tmp.append((i,c))
for i in range(n):
wh[ns[i]]=i
def check(i,m):
f=ns[m]-ns[i]
f=wh[f]
if (f-m)*(m-i)>0:
return True
return False
# print(wh)
ans=0
x=n
while x>=1:
lc=wh[x]
mp[lc]=True
l,r=lc-1,lc+1
while l>lst[lc] and r<nxt[lc] and (not mp[l]) and (not mp[r]):
l-=1
r+=1
if l<=lst[lc] or mp[l]:
l,r=l+1,lc-1
else:
l,r=lc+1,r-1
for i in range(l,r+1):
if check(i,lc):
ans+=1
x-=1
print(ans)
# print(lst)
# print(nxt)
``` | instruction | 0 | 75,059 | 12 | 150,118 |
No | output | 1 | 75,059 | 12 | 150,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p of n integers 1, 2, ..., n (a permutation is an array where each element from 1 to n occurs exactly once).
Let's call some subsegment p[l, r] of this permutation special if p_l + p_r = max _{i = l}^{r} p_i. Please calculate the number of special subsegments.
Input
The first line contains one integer n (3 ≤ n ≤ 2 ⋅ 10^5).
The second line contains n integers p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). All these integers are pairwise distinct.
Output
Print the number of special subsegments of the given permutation.
Examples
Input
5
3 4 1 5 2
Output
2
Input
3
1 3 2
Output
1
Note
Special subsegments in the first example are [1, 5] and [1, 3].
The only special subsegment in the second example is [1, 3].
Submitted Solution:
```
from collections import defaultdict
def fun(a):
d=defaultdict(int)
maximum=a[0]
count=0
for i in range(n):
d[a[i]]=i
for i in range(n):
if a[i]>maximum:
maximum=a[i]
x=a[i]
y=maximum-x
if y!=0:
j=d[maximum]
if min(d[x],d[y])<j<max(d[x],d[y]):
count+=1
return count
n=int(input())
a=list(map(int,input().split( )))
print(fun(a))
``` | instruction | 0 | 75,060 | 12 | 150,120 |
No | output | 1 | 75,060 | 12 | 150,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,161 | 12 | 150,322 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
for i in range(int(input())):
n=int(input())
p=list(map(int,input().split()))
ind=[0]*(n +1)
for i in range(n):
ind[p[i]]=i
f=0
c=1
till=n
k=0
while(k<n):
j=ind[c]
for i in range(j,till):
k+=1
if(p[i]==c):
c+=1
else:
f=1
break
if(f==1):
break
till=j
if(f==0):
print('Yes')
else:
print('No')
``` | output | 1 | 75,161 | 12 | 150,323 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,162 | 12 | 150,324 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
from sys import stdin, exit
import bisect
input = stdin.readline
def i(): return input()
def ii(): return int(input())
def iis(): return map(int, input().split())
def liis(): return list(map(int, input().split()))
def print_array(a): print(" ".join(map(str, a)))
t = ii()
for _ in range(t):
n = ii()
p = liis()
P = [[i, idx] for idx, i in enumerate(p)]
P = sorted(P)
used = [0 for i in range(n)]
numbers = set()
can = True
last = n
for j in P:
i = j[0]
idx = j[1]
for k in range(idx, last):
last = idx
if k == idx:
if used[k]:
can = False
break
used[k] = 1
elif used[k] or p[k] != p[k-1]+1:
can = False
break
used[k] = 1
if can:
print('Yes')
else:
print('No')
``` | output | 1 | 75,162 | 12 | 150,325 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,163 | 12 | 150,326 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
from collections import defaultdict as dd
t=int(input())
while t:
d=dd(int)
n=int(input())
l=list(map(int,input().split()))
for i in range(n):
d[l[i]]=i
i=1
lol=0
vis=[0]*n
while i<=n:
p=d[i]
#print(p)
while vis[p]==0 and p<n:
vis[p]=1
if(p+1)==n:
break
if(i+1==n+1 ):
break
if(vis[p+1]):
break
if((p+1)!=d[i+1]):
#print(p+1,i+1)
lol=1
break
i+=1
p+=1
i+=1
#print(vis)
if(lol):
print("No")
else:
print("Yes")
t-=1
``` | output | 1 | 75,163 | 12 | 150,327 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,164 | 12 | 150,328 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
import sys,math
t=int(sys.stdin.readline())
for _ in range(t):
n=int(sys.stdin.readline())
a=list(map(int,sys.stdin.readline().split()))
ind=-1
j=n
flag=True
for i in range(1,n):
if not(a[i]-a[i-1]==1 or a[i]-a[i-1]<0):
print("NO")
flag=False
break
if flag:
print("YES")
``` | output | 1 | 75,164 | 12 | 150,329 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,165 | 12 | 150,330 |
Tags: brute force, data structures, 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()
p = inl()
ok = True
tgt = 1
nxt = p[n-1]
nxttgt = nxt+1
for i in range(n-1,-1,-1):
if p[i]!=nxt:
ok = False
break
if p[i]==tgt and i>0:
tgt = nxttgt
nxt = p[i-1]
nxttgt = nxt+1
else:
nxt -= 1
print('Yes' if ok else 'No')
``` | output | 1 | 75,165 | 12 | 150,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,166 | 12 | 150,332 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
def solve():
n=int(input())
li=[int(x) for x in input().split()]
d={}
for i in range(n):
d[li[i]]=i
i,last_val=1,1
pos=d[i]
last=n-1
while i<n:
if pos==last:
pos=d[i+1]
last=d[last_val]-1
last_val=i+1
else:
if li[pos]!=i or li[pos+1]!=i+1:
print('NO')
break
else:
pos+=1
i+=1
else:
print('YES')
t=int(input())
for i in range(t):
solve()
``` | output | 1 | 75,166 | 12 | 150,333 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,167 | 12 | 150,334 |
Tags: brute force, data structures, greedy, implementation
Correct Solution:
```
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")
t = int(input())
#
for _ in range(t):
n = int(input())
li = [int(i) for i in input().split(' ')]
f = True
g = li.pop()
d=set()
while len(li):
if g == li[-1]+1:
d.add(g)
g=li.pop()
elif g==1:
d.add(g)
g = li.pop()
elif g-1 in d:
d.add(g)
g=li.pop()
else:
f=False
break
if f:
print('yes')
else:
print('no')
``` | output | 1 | 75,167 | 12 | 150,335 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it. | instruction | 0 | 75,168 | 12 | 150,336 |
Tags: brute force, data structures, greedy, implementation
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
mod=10**9+7
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
fact(x,mod)
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa.reverse()
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 isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def floorsum(a,b,c,n):#sum((a*i+b)//c for i in range(n+1))
if a==0:return b//c*(n+1)
if a>=c or b>=c: return floorsum(a%c,b%c,c,n)+b//c*(n+1)+a//c*n*(n+1)//2
m=(a*n+b)//c
return n*m-floorsum(c,c-b-1,a,m-1)
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
def lowbit(n):
return n&-n
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 ST:
def __init__(self,arr):#n!=0
n=len(arr)
mx=n.bit_length()#取不到
self.st=[[0]*mx for i in range(n)]
for i in range(n):
self.st[i][0]=arr[i]
for j in range(1,mx):
for i in range(n-(1<<j)+1):
self.st[i][j]=max(self.st[i][j-1],self.st[i+(1<<j-1)][j-1])
def query(self,l,r):
if l>r:return -inf
s=(r+1-l).bit_length()-1
return max(self.st[l][s],self.st[r-(1<<s)+1][s])
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
seen.add(u)
for v,w in graph[u]:
if v not in d or d[v]>d[u]+w:
d[v]=d[u]+w
heappush(heap,(d[v],v))
return d
def GP(it): return [[ch,len(list(g))] for ch,g in groupby(it)]
def lcm(a,b): return a*b//gcd(a,b)
def lis(nums):
res=[]
for k in nums:
i=bisect.bisect_left(res,k)
if i==len(res):
res.append(k)
else:
res[i]=k
return len(res)
def RP(nums):#逆序对
n = len(nums)
s=set(nums)
d={}
for i,k in enumerate(sorted(s),1):
d[k]=i
bi=BIT([0]*(len(s)+1))
ans=0
for i in range(n-1,-1,-1):
ans+=bi.query(d[nums[i]]-1)
bi.update(d[nums[i]],1)
return ans
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
def nb(i,j):
for ni,nj in [[i+1,j],[i-1,j],[i,j-1],[i,j+1]]:
if 0<=ni<n and 0<=nj<m:
yield ni,nj
def topo(n):
q=deque()
res=[]
for i in range(1,n+1):
if ind[i]==0:
q.append(i)
res.append(i)
while q:
u=q.popleft()
for v in g[u]:
ind[v]-=1
if ind[v]==0:
q.append(v)
res.append(v)
return res
@bootstrap
def gdfs(r,p):
if len(g[r])==1 and p!=-1:
yield None
for ch in g[r]:
if ch!=p:
yield gdfs(ch,r)
yield None
t=N()
for i in range(t):
n=N()
p=RLL()
res=[]
last=-1
for x in p:
if x!=last+1 and last!=-1:
res.append(last)
last=x
res.append(last)
#print(res)
ans="Yes" if res==sorted(res,reverse=True) else"No"
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thr
ead(target=main)
t.start()
t.join()
'''
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
``` | output | 1 | 75,168 | 12 | 150,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
from itertools import accumulate
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
class SegTree:
def __init__(self, n, func, intv, A=[]):
self.n = n
self.func = func
self.intv = intv
n2 = 1
while n2 < n:
n2 <<= 1
self.n2 = n2
self.tree = [self.intv] * (n2 << 1)
if A:
for i in range(n):
self.tree[n2+i] = A[i]
for i in range(n2-1, -1, -1):
self.tree[i] = self.func(self.tree[i*2], self.tree[i*2+1])
def update(self, i, x):
i += self.n2
self.tree[i] = x
while i > 0:
i >>= 1
self.tree[i] = self.func(self.tree[i*2], self.tree[i*2+1])
def add(self, i, x):
self.update(i, self.get(i) + x)
def query(self, a, b):
l = a + self.n2
r = b + self.n2
s = self.intv
while l < r:
if r & 1:
r -= 1
s = self.func(s, self.tree[r])
if l & 1:
s = self.func(s, self.tree[l])
l += 1
l >>= 1
r >>= 1
return s
def get(self, i):
return self.tree[i+self.n2]
def all(self):
return self.tree[1]
def print(self):
for i in range(self.n):
print(self.get(i), end=' ')
print()
for _ in range(INT()):
N = INT()
P = [(p, i) for i, p in enumerate(LIST())]
P.sort()
C = SegTree(N+1, max, 0, [1]*N + [0])
for p, idx in P:
if C.all() != C.get(idx):
No()
break
for i in range(idx+1, N+1):
if i == N:
C.update(idx, 0)
break
if C.get(i) != 0:
C.update(i, C.get(i)+C.get(idx))
C.update(idx, 0)
break
else:
Yes()
``` | instruction | 0 | 75,169 | 12 | 150,338 |
Yes | output | 1 | 75,169 | 12 | 150,339 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
import sys
input = sys.stdin.readline
for t in range(int(input())):
n = int(input())
p = list(map(int, input().split(" ")))
ans = "Yes"
allPresent = {}
for f in range(n):
allPresent[f+1] = 0
allPresent[p[0]]=1
for j in range(1, n):
allPresent[p[j]] = 1
if(p[j]!=p[j-1]+1):
if(p[j-1]<p[j]):
ans = "No"
for f in range(n):
if(allPresent[f+1]==0):
ans = "No"
print(ans)
``` | instruction | 0 | 75,170 | 12 | 150,340 |
Yes | output | 1 | 75,170 | 12 | 150,341 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
def main_function():
from sys import stdin
from sys import stdout
input = stdin.readline
print = stdout.write
t = int(input())
for _ in range(t):
n = int(input())
considered_permutation = list(map(int, input().split()))
if n <= 2:
print('Yes\n')
continue
ix_in_permutation = [0] * n
for i in range(n):
ix_in_permutation[considered_permutation[i] - 1] = i
count = [1] * n + [0]
max_count = 1
for i in range(n - 1):
if count[ix_in_permutation[i]] != max_count:
print('No\n')
break
if count[ix_in_permutation[i] + 1] == 0:
max_count = 1
else:
max_count += 1
count[ix_in_permutation[i] + 1] += count[ix_in_permutation[i]]
count[ix_in_permutation[i]] = 0
else:
print('Yes\n')
if __name__ == '__main__':
main_function()
``` | instruction | 0 | 75,171 | 12 | 150,342 |
Yes | output | 1 | 75,171 | 12 | 150,343 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
"""
Author: Q.E.D
Time: 2020-04-23 10:32:53
"""
T = int(input())
for _ in range(T):
n = int(input())
a = list(map(int, input().split()))
valid = True
for i in range(1, n):
if a[i] != a[i - 1] + 1 and a[i] >= a[i - 1]:
valid = False
print('Yes' if valid else 'No')
``` | instruction | 0 | 75,172 | 12 | 150,344 |
Yes | output | 1 | 75,172 | 12 | 150,345 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
permutationlist = list(map(int,input().split()))
indexstored = {}
for i in range(n) :
indexstored[permutationlist[i]] = i+1
curr_last = n
flag = True
j = 1
while j <= n :
my_index = indexstored[j]
k = my_index
while k < curr_last-1 :
if permutationlist[k] == permutationlist[k-1] + 1 :
k += 1
j+=1
else :
flag = False
break
j = n
curr_last = my_index
j += 1
if flag :
print("Yes")
else :
print("No")
``` | instruction | 0 | 75,173 | 12 | 150,346 |
No | output | 1 | 75,173 | 12 | 150,347 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
p=list(map(int,input().split()))
i=p.index(1)
nb=1
t=0
j=n
l=0
while i<j and l<n:
if p[i]!=nb:
t=1
l+=1
nb+=1
i+=1
if i==j and l<n:
i=p.index(nb)
j=i
ans=["YES","NO"]
print(ans[t])
``` | instruction | 0 | 75,174 | 12 | 150,348 |
No | output | 1 | 75,174 | 12 | 150,349 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
t = int(input())
res = []
for tc in range(t):
print('test' + str(tc))
n = int(input())
arr = list(map(int, input().split()))
nxt = list(range(n))
wall = n
st = arr.index(1)
res_t = ''
for j in range(1, n + 1):
i = arr.index(j)
print(i, nxt)
if(not i in nxt):
res_t = 'NO'
break
if(i + 1 >= n or i + 1 >= wall):
wall = st
if(j + 1 > n):
break
st = arr.index(j + 1)
nxt = list(range(wall))
else:
nxt = list(range(i + 1, i + 2))
if(res_t != 'NO'):
res_t = 'YES'
res.append(res_t)
for r in res:
print(r)
``` | instruction | 0 | 75,175 | 12 | 150,350 |
No | output | 1 | 75,175 | 12 | 150,351 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Denis was very sad after Nastya rejected him. So he decided to walk through the gateways to have some fun. And luck smiled at him! When he entered the first courtyard, he met a strange man who was selling something.
Denis bought a mysterious item and it was... Random permutation generator! Denis could not believed his luck.
When he arrived home, he began to study how his generator works and learned the algorithm. The process of generating a permutation consists of n steps. At the i-th step, a place is chosen for the number i (1 ≤ i ≤ n). The position for the number i is defined as follows:
* For all j from 1 to n, we calculate r_j — the minimum index such that j ≤ r_j ≤ n, and the position r_j is not yet occupied in the permutation. If there are no such positions, then we assume that the value of r_j is not defined.
* For all t from 1 to n, we calculate count_t — the number of positions 1 ≤ j ≤ n such that r_j is defined and r_j = t.
* Consider the positions that are still not occupied by permutation and among those we consider the positions for which the value in the count array is maximum.
* The generator selects one of these positions for the number i. The generator can choose any position.
Let's have a look at the operation of the algorithm in the following example:
<image>
Let n = 5 and the algorithm has already arranged the numbers 1, 2, 3 in the permutation. Consider how the generator will choose a position for the number 4:
* The values of r will be r = [3, 3, 3, 4, ×], where × means an indefinite value.
* Then the count values will be count = [0, 0, 3, 1, 0].
* There are only two unoccupied positions in the permutation: 3 and 4. The value in the count array for position 3 is 3, for position 4 it is 1.
* The maximum value is reached only for position 3, so the algorithm will uniquely select this position for number 4.
Satisfied with his purchase, Denis went home. For several days without a break, he generated permutations. He believes that he can come up with random permutations no worse than a generator. After that, he wrote out the first permutation that came to mind p_1, p_2, …, p_n and decided to find out if it could be obtained as a result of the generator.
Unfortunately, this task was too difficult for him, and he asked you for help. It is necessary to define whether the written permutation could be obtained using the described algorithm if the generator always selects the position Denis needs.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases. Then the descriptions of the test cases follow.
The first line of the test case contains a single integer n (1 ≤ n ≤ 10^5) — the size of the permutation.
The second line of the test case contains n different integers p_1, p_2, …, p_n (1 ≤ p_i ≤ n) — the permutation written by Denis.
It is guaranteed that the sum of n over all test cases doesn't exceed 10^5.
Output
Print "Yes" if this permutation could be obtained as a result of the generator. Otherwise, print "No".
All letters can be displayed in any case.
Example
Input
5
5
2 3 4 5 1
1
1
3
1 3 2
4
4 2 3 1
5
1 5 2 4 3
Output
Yes
Yes
No
Yes
No
Note
Let's simulate the operation of the generator in the first test.
At the 1 step, r = [1, 2, 3, 4, 5], count = [1, 1, 1, 1, 1]. The maximum value is reached in any free position, so the generator can choose a random position from 1 to 5. In our example, it chose 5.
At the 2 step, r = [1, 2, 3, 4, ×], count = [1, 1, 1, 1, 0]. The maximum value is reached in positions from 1 to 4, so the generator can choose a random position among them. In our example, it chose 1.
At the 3 step, r = [2, 2, 3, 4, ×], count = [0, 2, 1, 1, 0]. The maximum value is 2 and is reached only at the 2 position, so the generator will choose this position.
At the 4 step, r = [3, 3, 3, 4, ×], count = [0, 0, 3, 1, 0]. The maximum value is 3 and is reached only at the 3 position, so the generator will choose this position.
At the 5 step, r = [4, 4, 4, 4, ×], count = [0, 0, 0, 4, 0]. The maximum value is 4 and is reached only at the 4 position, so the generator will choose this position.
In total, we got a permutation of 2, 3, 4, 5, 1, that is, a generator could generate it.
Submitted Solution:
```
# ------------------- fast io --------------------
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")
# ------------------- fast io --------------------
testcases=int(input())
for j in range(testcases):
n=int(input())
perms=list(map(int,input().split()))
if perms[-1]==1:
#ok we just check how far down it increases backwards then we partition that list
invert=perms[::-1]
store=0
val=True
for s in range(1,n):
if invert[s]==invert[s-1]+1:
continue
else:
store=s
val=False
break
if val==True:
print("Yes")
else:
#we need to search for the largest value and its index
listnew=invert[store::]
listnew.remove(n)
join=invert[:store]+listnew[::-1]+[n]
val=True
for s in range(1,n):
if join[s]==join[s-1]+1:
continue
else:
val=False
break
if val==True:
print("Yes")
else:
print("No")
else:
#now we just loop through and check if it is increasing or not
firstnum=perms[0]
#below is the index where i can find n
if firstnum==1:
val=True
for s in range(1,n):
if perms[s]==perms[s-1]+1:
continue
else:
val=False
break
if val==True:
print("Yes")
else:
print("No")
else:
offset=n-firstnum
list1=perms[:offset+1]
list2=perms[offset+1:]
joined=list1+list2
val=True
for s in range(1,n):
if joined[s]==joined[s-1]+1:
continue
else:
val=False
break
if val==True:
print("Yes")
else:
print("No")
``` | instruction | 0 | 75,176 | 12 | 150,352 |
No | output | 1 | 75,176 | 12 | 150,353 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,239 | 12 | 150,478 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
for i in range(int(input())):
n,d = map(int, input().split())
a = list(map(int,input().split()))
a.sort()
if a[n-1] <= d: print("YES")
elif a[0]+a[1] <= d: print("YES")
else: print("NO")
``` | output | 1 | 75,239 | 12 | 150,479 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,240 | 12 | 150,480 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
input=sys.stdin.readline
t = int(input())
for _ in range(t):
n, d = map(int, input().split())
a = list(map(int, input().split()))
f = 0
mi1 = mi2 = 1000
for i in a:
if i < mi1:
mi1, mi2 = i, mi1
elif i < mi2:
mi2 = i
if i > d:
f = 1
if f:
if mi1 + mi2 <= d:
print("YES")
else:
print("NO")
else:
print("YES")
``` | output | 1 | 75,240 | 12 | 150,481 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,241 | 12 | 150,482 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
for _ in [0] * int(input()):
n, d = map(int, input().split())
A = list(map(int, input().split()))
flag = True
A.sort()
a = A[0]
b = A[1]
# Checking if all the elements are less or equal than d.
for i in range(n):
if A[i] > d:
if a + b > d:
flag = False
break
print("YES" if flag else "NO")
``` | output | 1 | 75,241 | 12 | 150,483 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,242 | 12 | 150,484 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Jan 16 10:38:32 2021
@author: Syed Ishtiyaq Ahmed
"""
answers=[]
t=int(input())
for i in range(t):
n,d=map(int,input().split())
arr=[]
arr.append(input().split())
arr2=[]
for i in range(len(arr[0])):
arr2.append(int(arr[0][i]))
pr_index=[]
for i in range(n):
if arr2[i]<=d:
pr_index.append(i)
arr2.sort()
if (len(arr2))==1 and arr2[0]<=d:
answers.append("YES")
elif (len(arr2))==1 and arr2[0]>=d:
answers.append("NO")
elif (len(arr2))>1:
if len(pr_index)==n or (arr2[0]+arr2[1]<=d):
answers.append("YES")
else:
answers.append("NO")
for i in answers:
print(i)
``` | output | 1 | 75,242 | 12 | 150,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,243 | 12 | 150,486 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
for _ in range(int(input())):
n,d=map(int,input().split())
data=[int(x) for x in input().split()]
data.sort()
if(data[-1]<=d):
print('YES')
elif(data[0]+data[1]<=d):
print('YES')
else:
print('NO')
``` | output | 1 | 75,243 | 12 | 150,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,244 | 12 | 150,488 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
t = int(input())
for x in range(t):
a = input().split()
n,d= map(int,a)
arr = input().split()
arr = list(map(int,arr))
arr.sort()
for x in range(len(arr)):
if arr[x]>d:
if (arr[0]+arr[1]<=d):
arr[x]=arr[0]+arr[1]
r=0
# print(arr)
for _ in arr:
if _ > d:
r=1
if r==0:
print("Yes")
if r==1:
print("No")
``` | output | 1 | 75,244 | 12 | 150,489 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,245 | 12 | 150,490 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
T = int( input() )
for t in range(T):
s = input().split()
n, d = int(s[0]), int(s[1])
s = input().split()
A = [int(s[i]) for i in range(n)]
ms = 999999
bigger = False
for i in range(n-1):
if A[i] > d:
bigger = True
for j in range(i+1, n):
if ms > A[i] + A[j]:
ms = A[i] + A[j]
if A[-1] > d:
bigger = True
if bigger:
if ms <= d:
print ( "YES" )
else:
print ( "NO" )
else:
print("YES")
``` | output | 1 | 75,245 | 12 | 150,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4. | instruction | 0 | 75,246 | 12 | 150,492 |
Tags: greedy, implementation, math, sortings
Correct Solution:
```
import sys
reader = (s.rstrip() for s in sys.stdin)
input = reader.__next__
def gift():
for _ in range(t):
n,s = list(map(int,input().split()))
arr = list(map(int,input().split()))
arr.sort()
if n==1:
if arr[0]>s:
yield 'NO'
else:
yield 'YES'
else:
if sum([arr[0],arr[1]])>s and arr[-1]>s:
yield 'NO'
else:
yield 'YES'
if __name__ == '__main__':
t= int(input())
ans = gift()
print(*ans,sep='\n')
#"{} {} {}".format(maxele,minele,minele)
# yield " ".join([str(x) for x in ans])
``` | output | 1 | 75,246 | 12 | 150,493 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
for _ in range(int(input())):
n,d=map(int,input().split())
l=list(map(int,input().split()))
l.sort()
ans=0
for i in l:
if i<=d:
ans+=1
if l[0]+l[1]<=d:
print("YES")
elif ans==n:
print("YES")
else:
print("NO")
``` | instruction | 0 | 75,247 | 12 | 150,494 |
Yes | output | 1 | 75,247 | 12 | 150,495 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
for _ in range(int(input())):
n, d = map(int, input().split())
a = list(map(int, input().split()))
flag = True
for i in range(n):
if a[i] > d:
flag = False
if flag == True:
print("YES")
continue
# If changes are required
a.sort()
immature = a[0] + a[1]
if immature <= d:
print("YES")
else:
print("NO")
``` | instruction | 0 | 75,248 | 12 | 150,496 |
Yes | output | 1 | 75,248 | 12 | 150,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
t=int(input())
for q in range(t):
ch=input()
L=[int(i)for i in ch.split()]
n,d=L[0],L[1]
ch=input()
L=[int(i)for i in ch.split()]
L.sort()
if L[-1]<=d:
print("YES")
else:
if L[0]+L[1]<=d:
print("YES")
else:
print("NO")
``` | instruction | 0 | 75,249 | 12 | 150,498 |
Yes | output | 1 | 75,249 | 12 | 150,499 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
t=int(input())
for _ in range(t):
size,d=input().split()
arr=[int(i) for i in input().split()]
arr.sort()
d=int(d)
if arr[-1]<=d or arr[0]+arr[1]<=d:
print("Yes")
else:
print("No")
``` | instruction | 0 | 75,250 | 12 | 150,500 |
Yes | output | 1 | 75,250 | 12 | 150,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
t = int(input())
for _ in range(t):
n,m = map(int,input().split())
l = list(map(int,input().split()))
p=0
c=0
for i in range(n):
if l[i]<=m:
c+=1
if c==n:
print("YES")
else:
for i in range(n):
for j in range(i+1,n):
if l[i]+l[j]<=m and l[i]!=l[j]:
p=1
print("YES")
exit()
if p==0:
print("NO")
``` | instruction | 0 | 75,251 | 12 | 150,502 |
No | output | 1 | 75,251 | 12 | 150,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
def AllLow(l,d):
for i in range(0,len(l)):
if l[i]>d:
return False
return True
t= input()
t=int(t)
res=[]
for i in range(0,t):
ND=input().split()
n=int(ND[0])
d=int(ND[1])
l=input().split()
for i in range(0,len(l)):
l[i]=int(l[i])
b=True
if AllLow(l,d):
res.append("True")
else:
for j in range(0,n-1):
for k in range(j+1,n):
if l[j]+l[k]<=d:
res.append("YES")
b=False
break
if b:
res.append("NO")
for v in res:
print(v)
``` | instruction | 0 | 75,252 | 12 | 150,504 |
No | output | 1 | 75,252 | 12 | 150,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
import sys
import math
#import re
# I = lambda :int(sys.stdin.buffer.readline())
#tup= lambda : map(int , sys.stdin.buffer.readline().split())
# grid = lambda r :[lint() for i in range(r)]
# star = lambda x: print(' '.join(map(str, x)))
lst = lambda :[int(x) for x in inp().split()]
inp = lambda: sys.stdin.readline().strip('\n')
# inp2 = lambda: sys.stdin.readline().strip(' ')
out = lambda x : sys.stdout.write(f'{x}'+'\n')
def find_ans():
n, d = map(int, inp().split())
lst1 = lst()
flag = False
for i in range(n):
flag = False
if lst1[i] <= d:
flag = True
continue
for j in range(n):
if j == i:
continue
for k in range(j+1, n):
if k == i:
continue
if lst1[j] + lst1[k] <= d:
lst1[i] = lst1[j] + lst1[k]
flag = True
break
if flag == True:
break
if flag == False:
return "NO"
return "YES"
if __name__ == '__main__':
t = int(inp())
while t:
t -= 1
out(find_ans())
``` | instruction | 0 | 75,253 | 12 | 150,506 |
No | output | 1 | 75,253 | 12 | 150,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have an array a_1, a_2, ..., a_n. All a_i are positive integers.
In one step you can choose three distinct indices i, j, and k (i ≠ j; i ≠ k; j ≠ k) and assign the sum of a_j and a_k to a_i, i. e. make a_i = a_j + a_k.
Can you make all a_i lower or equal to d using the operation above any number of times (possibly, zero)?
Input
The first line contains a single integer t (1 ≤ t ≤ 2000) — the number of test cases.
The first line of each test case contains two integers n and d (3 ≤ n ≤ 100; 1 ≤ d ≤ 100) — the number of elements in the array a and the value d.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the array a.
Output
For each test case, print YES, if it's possible to make all elements a_i less or equal than d using the operation above. Otherwise, print NO.
You may print each letter in any case (for example, YES, Yes, yes, yEs will all be recognized as positive answer).
Example
Input
3
5 3
2 3 2 5 4
3 4
2 4 4
5 4
2 1 5 3 6
Output
NO
YES
YES
Note
In the first test case, we can prove that we can't make all a_i ≤ 3.
In the second test case, all a_i are already less or equal than d = 4.
In the third test case, we can, for example, choose i = 5, j = 1, k = 2 and make a_5 = a_1 + a_2 = 2 + 1 = 3. Array a will become [2, 1, 5, 3, 3].
After that we can make a_3 = a_5 + a_2 = 3 + 1 = 4. Array will become [2, 1, 4, 3, 3] and all elements are less or equal than d = 4.
Submitted Solution:
```
for _ in range(int(input())):
n,d =[int(c) for c in input().split()]
arr =[int(c) for c in input().split()]
ans = True
count = 0
for i in arr:
if i > d:
ans = False
break
if i <= d/2:
count+=2
if ans == True:
print("YES")
else:
if count >=2:
print("YES")
else:
print("NO")
``` | instruction | 0 | 75,254 | 12 | 150,508 |
No | output | 1 | 75,254 | 12 | 150,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,287 | 12 | 150,574 |
Tags: greedy, math
Correct Solution:
```
t = int(input())
for _ in range(t):
s = int(input())
currsum = 0
arr = []
while(currsum!=s):
diff = s-currsum
if currsum==0:
arr.append(1)
elif diff <=2 or diff-2 in arr or diff-1 in arr:
arr.append(diff)
else:
max_ = max(arr)
if max_+2 <= diff:
arr.append(max_+2)
else:
arr.append(max_+1)
currsum = sum(arr)
print(len(arr))
``` | output | 1 | 75,287 | 12 | 150,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,288 | 12 | 150,576 |
Tags: greedy, math
Correct Solution:
```
import math
h = int(input())
for _ in range(h):
a = int(input())
b = a**0.5
print(math.ceil(b))
``` | output | 1 | 75,288 | 12 | 150,577 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,289 | 12 | 150,578 |
Tags: greedy, math
Correct Solution:
```
# Author Name: Ajay Meena
# Codeforce : https://codeforces.com/profile/majay1638
import sys
import math
import bisect
import heapq
from bisect import bisect_right
from sys import stdin, stdout
# -------------- INPUT FUNCTIONS ------------------
def get_ints_in_variables(): return map(
int, sys.stdin.readline().strip().split())
def get_int(): return int(sys.stdin.readline())
def get_ints_in_list(): return list(
map(int, sys.stdin.readline().strip().split()))
def get_list_of_list(n): return [list(
map(int, sys.stdin.readline().strip().split())) for _ in range(n)]
def get_string(): return sys.stdin.readline().strip()
# -------- SOME CUSTOMIZED FUNCTIONS-----------
def myceil(x, y): return (x + y - 1) // y
# -------------- SOLUTION FUNCTION ------------------
def Solution(s):
# Write Your Code Here
ans = 0
if s == 1:
print(1)
return
if s == 2:
print(2)
return
t = 3
s2 = 1
ans = 1
while s > s2:
s2 += t
t += 2
ans += 1
print(ans)
def main():
# Take input Here and Call solution function
for _ in range(get_int()):
Solution(get_int())
# calling main Function
if __name__ == '__main__':
main()
``` | output | 1 | 75,289 | 12 | 150,579 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,290 | 12 | 150,580 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
s=1
i=3
c=1
while(s<n):
s+=i
i+=2
c+=1
print(c)
``` | output | 1 | 75,290 | 12 | 150,581 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,291 | 12 | 150,582 |
Tags: greedy, math
Correct Solution:
```
import sys
import math
from collections import deque,Counter
#sys.setrecursionlimit(10**7)
int1=lambda x: int(x)-1
inp=lambda :int(input())
mi=lambda :map(int,input().split())
li=lambda :list(mi())
mi1=lambda :map(int1,input().split())
li1=lambda :list(mi1())
mis=lambda :map(str,input().split())
lis=lambda :list(mis())
pr=print
from collections import defaultdict
"""
#初期値 0
d=defaultdict(int)
#初期値 1
d=defaultdict(lambda:1)
"""
mod=10**9+7
Mod=998244353
INF=10**18
ans=0
t=inp()
for _ in range(t):
n=inp()
x=(math.sqrt(n)+0.0000001)
if int(x)**2==n:
print(int(x))
else:
print(int(x)+1)
``` | output | 1 | 75,291 | 12 | 150,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,292 | 12 | 150,584 |
Tags: greedy, math
Correct Solution:
```
def Sum(x):
return x*x
def max_no_of_elements(x):
no = 1
while Sum(no) < x:
no += 1
return no
a = int(input())
for i in range(a):
x=int(input())
print(max_no_of_elements(x))
``` | output | 1 | 75,292 | 12 | 150,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,293 | 12 | 150,586 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
s=int(input())
res=1
i=1
while res<s:
res = i*i
i+=1
if s==1:
print(1)
else:
print(i-1)
``` | output | 1 | 75,293 | 12 | 150,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions. | instruction | 0 | 75,294 | 12 | 150,588 |
Tags: greedy, math
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
cur = 1
s = 1
k = 1
while s < n:
cur += 2
s += cur
k += 1
print(k)
``` | output | 1 | 75,294 | 12 | 150,589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
for _ in range(int(input())):
x = int(input())
count = 1
x -= 1
if x == 0:
print(count)
continue
if x == 2:
print("2")
continue
if x == 3:
print("2")
continue
# if x % 2 == 0:
# for i in range(2, 5000, 2):
# if x <= 0:
# break
# x -= i
# count += 1
for i in range(3, 5000, 2):
if x <= 0:
break
x -= i
count += 1
print(count)
``` | instruction | 0 | 75,295 | 12 | 150,590 |
Yes | output | 1 | 75,295 | 12 | 150,591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
ans = 0
i = 1
while 1:
ans+=1
n-=i
if n<=0: break
i+=2
print(ans)
``` | instruction | 0 | 75,296 | 12 | 150,592 |
Yes | output | 1 | 75,296 | 12 | 150,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
T = int(input())
w = [1]
ww = [1]
for i in range(1, 5005) :
w.append(w[i - 1] + 2)
ww.append(w[i] + ww[i - 1])
for case in range(0, T) :
n = int(input())
for i in range(0, 5000) :
if (ww[i] >= n) :
print(i + 1)
break
``` | instruction | 0 | 75,297 | 12 | 150,594 |
Yes | output | 1 | 75,297 | 12 | 150,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
for s in[*open(0)][1:]:print(int((int(s)-.1)**.5)+1)
``` | instruction | 0 | 75,298 | 12 | 150,596 |
Yes | output | 1 | 75,298 | 12 | 150,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
n=int(input())
for i in range(0,n):
a=int(input())
print((a**0.5)//1)
``` | instruction | 0 | 75,299 | 12 | 150,598 |
No | output | 1 | 75,299 | 12 | 150,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
import math
for i in range(int(input())):
a = int(input())
if (math.sqrt(a)//1)**2 ==a:
print(math.sqrt(a))
else:
print(math.sqrt(a)//1 +1)
``` | instruction | 0 | 75,300 | 12 | 150,600 |
No | output | 1 | 75,300 | 12 | 150,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call an array a consisting of n positive (greater than 0) integers beautiful if the following condition is held for every i from 1 to n: either a_i = 1, or at least one of the numbers a_i - 1 and a_i - 2 exists in the array as well.
For example:
* the array [5, 3, 1] is beautiful: for a_1, the number a_1 - 2 = 3 exists in the array; for a_2, the number a_2 - 2 = 1 exists in the array; for a_3, the condition a_3 = 1 holds;
* the array [1, 2, 2, 2, 2] is beautiful: for a_1, the condition a_1 = 1 holds; for every other number a_i, the number a_i - 1 = 1 exists in the array;
* the array [1, 4] is not beautiful: for a_2, neither a_2 - 2 = 2 nor a_2 - 1 = 3 exists in the array, and a_2 ≠ 1;
* the array [2] is not beautiful: for a_1, neither a_1 - 1 = 1 nor a_1 - 2 = 0 exists in the array, and a_1 ≠ 1;
* the array [2, 1, 3] is beautiful: for a_1, the number a_1 - 1 = 1 exists in the array; for a_2, the condition a_2 = 1 holds; for a_3, the number a_3 - 2 = 1 exists in the array.
You are given a positive integer s. Find the minimum possible size of a beautiful array with the sum of elements equal to s.
Input
The first line contains one integer t (1 ≤ t ≤ 5000) — the number of test cases.
Then t lines follow, the i-th line contains one integer s (1 ≤ s ≤ 5000) for the i-th test case.
Output
Print t integers, the i-th integer should be the answer for the i-th testcase: the minimum possible size of a beautiful array with the sum of elements equal to s.
Example
Input
4
1
8
7
42
Output
1
3
3
7
Note
Consider the example test:
1. in the first test case, the array [1] meets all conditions;
2. in the second test case, the array [3, 4, 1] meets all conditions;
3. in the third test case, the array [1, 2, 4] meets all conditions;
4. in the fourth test case, the array [1, 4, 6, 8, 10, 2, 11] meets all conditions.
Submitted Solution:
```
from os import path
import sys, time
# mod = int(1e9 + 7)
# import re
from math import ceil, floor, gcd, log, log2, factorial, sqrt
from collections import defaultdict, Counter, OrderedDict, deque
from itertools import combinations, accumulate
# from string import ascii_lowercase ,ascii_uppercase
from bisect import *
from functools import reduce
from operator import mul
star = lambda x: print(' '.join(map(str, x)))
grid = lambda r: [lint() for i in range(r)]
INF = float('inf')
if (path.exists('input.txt')):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
import sys
from sys import stdin, stdout
from collections import *
from math import gcd, floor, ceil
def st():
return list(stdin.readline().strip())
def inp():
return int(stdin.readline())
def inlt():
return list(map(int, stdin.readline().split()))
def invr():
return map(int, stdin.readline().split())
def solve():
n = inp()
c = 1
res = 0
while n > 0:
n = n - c
res += 1
xx = c + 1
yy = c + 2
if (xx & 1):
odd = xx
even = yy
else:
even = xx
odd = yy
if (n & 1):
c = odd
else:
c = even
print(res)
t = 1
t = inp()
for _ in range(t):
solve()
``` | instruction | 0 | 75,301 | 12 | 150,602 |
No | output | 1 | 75,301 | 12 | 150,603 |
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