message stringlengths 2 11.9k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 137 108k | cluster float64 18 18 | __index_level_0__ int64 274 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
lis=[0]*2
lis[0]=list(input())
lis[1]=list(input())
# print(lis)
ans=[]
gg=0
for i in range(n):
a=lis[0][i]
ind=-1
for j in range(i+1,n):
if a==lis[0][j]:
ind=j
break
if ind!=-1:
ans.append([ind+1,i+1])
# lis[0][ind],lis[0][n-1]==lis[0][n-1],lis[0][ind]
# print(lis[1][i],lis[0][ind])
lis[1][i],lis[0][ind]=lis[0][ind],lis[1][i]
# print(lis[1][i],lis[0][ind])
else:
for j in range(i,n):
if a==lis[1][j]:
ind=j
break
if ind!=-1:
ans.append([n,ind+1])
ans.append([n,i+1])
# print(lis[1][ind],lis[0][n-1],ind,n-1)
lis[1][ind],lis[0][n-1]=lis[0][n-1],lis[1][ind]
# print(lis[1][ind],lis[0][n-1])
lis[1][i],lis[0][n-1]=lis[0][n-1],lis[1][i]
# print(lis[1][ind],lis[0][n-1])
else:
print("No")
gg=1
break
# print(lis,i)
if gg==0:
print("Yes")
print(len(ans))
for i in range(len(ans)):
print(*ans[i])
``` | instruction | 0 | 137 | 18 | 274 |
Yes | output | 1 | 137 | 18 | 275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
s = list(input())
t = list(input())
ans = []
for i in range(n):
if s[i]!=t[i]:
for j in range(i+1,n):
if s[i]==s[j]:
ans.append([j+1,i+1])
t[i],s[j] = s[j],t[i]
break
elif s[i]==t[j]:
ans.append([j+1,j+1])
s[j],t[j] = t[j],s[j]
ans.append([j+1,i+1])
t[i],s[j] = s[j],t[i]
break
flag = True
for i in range(n):
if s[i]!=t[i]:
flag = False
print('No')
break
if flag:
pp = len(ans)
print('Yes')
print(pp)
for i in range(pp):
print(*ans[i])
``` | instruction | 0 | 138 | 18 | 276 |
Yes | output | 1 | 138 | 18 | 277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
s = input()
t = input()
d = {}
for i in range(ord('a'), ord('z') + 1):
d[chr(i)] = 0
for cs in s:
d[cs] += 1
for ct in t:
d[ct] += 1
ok = True
for e in d:
if d[e] % 2 == 1:
ok = False
if not ok:
print("No")
else:
print("Yes")
changes = []
s, t = list(s), list(t)
for i in range(n-1):
if s[i] != t[i]:
r = (0, -1)
for j in range(i+1, n):
if s[j] == t[i]:
r = (j, 0)
for j in range(i+1, n):
if t[j] == t[i]:
r = (j, 1)
if r[1] == 0:
changes += [(r[0], i+1), (i, i+1)]
s[r[0]], t[i+1] = t[i+1], s[r[0]]
s[i], t[i+1] = t[i+1], s[i]
elif r[1] == 1:
changes += [(i, r[0])]
s[i], t[r[0]] = t[r[0]], s[i]
print(len(changes))
for change in changes:
x, y = change
print(x+1, y+1)
``` | instruction | 0 | 139 | 18 | 278 |
Yes | output | 1 | 139 | 18 | 279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for i in range(int(input())):
a, s, t = int(input()), input(), input()
if len([False for n in [(s+t).count(i) for i in set(s+t)] if n%2]):
print('No')
else:
print('Yes')
s = [' '] + [i for i in s]
t = [' '] + [i for i in t]
c = []
while len(s):
#print(s, t)
r, d = s.pop(), t.pop()
if r!=d:
#print(r, d)
try:
#print(t)
n = t.index(d)
c.append(str(len(s)) + ' ' + str(n))
t[n] = r
except:
c.append(str(len(s)) + ' ' + str(len(s)))
#print(t, s, r, d)
n = s.index(d)
c.append(str(n) + ' ' + str(len(s)))
s[n] = r
print(len(c))
for i in c:
print(i)
``` | instruction | 0 | 140 | 18 | 280 |
Yes | output | 1 | 140 | 18 | 281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for _ in range(int(input())):
n = int(input())
na = list(input())
nb = list(input())
k=0
zz = len(na)
ans = []
for i in range(zz):
f = 0
if(na[i]==nb[i]):
f= 1
for j in range(i+1,zz):
if(na[j]==na[i]):
ans.append((i,j))
na[j] = nb[i]
f = 1
break
if(not f):
for j in range(i+1,zz):
#print(nb[i])
if(na[i]==nb[j]):
ans.append((j,j))
ans.append((i,j))
nb[j] = na[j]
na[j] = nb[i]
f = 1
break
#print(na,nb)
if(not f):
k =1
break
if(k):
print("NO")
else:
print("YES")
print(len(ans))
for t in ans:
print(t[0]+1,t[1]+1)
``` | instruction | 0 | 141 | 18 | 282 |
No | output | 1 | 141 | 18 | 283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
cases = int(input())
for _ in range(cases):
n = int(input())
s1 = list(input())
s2 = list(input())
swaps = 0
order = []
i = 0
again = False
while i < n and swaps < 2*n:
if s1[i] == s2[i]:
i += 1
else:
if i == n-1:
swaps = 2*n+1
else:
if not again:
for j in range(i+1, n):
if s1[i] == s1[j]:
s2[i], s1[j] = s1[j], s2[i]
swaps += 1
order.append((j, i))
break
elif s2[i] == s2[j]:
s1[i], s2[j] = s2[j], s1[i]
swaps += 1
order.append((i, j))
break
elif j == n-1:
again = True
else:
for j in range(i+1, n):
if s2[i] != s1[j]:
s2[i], s1[j] = s1[j], s2[i]
swaps += 1
order.append((j, i))
again = False
break
elif s1[i] != s2[j]:
s1[i], s2[j] = s2[j], s1[i]
swaps += 1
order.append((i, j))
again = False
break
if again:
swaps = 2*n+1
if swaps >= 2*n:
print('No')
else:
print('Yes')
print(swaps)
for a, b in order:
print(a+1, b+1)
``` | instruction | 0 | 142 | 18 | 284 |
No | output | 1 | 142 | 18 | 285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
k = int(input())
for I in range(k):
n = int(input())
s0 = input()
t0 = input()
s = [s0[i] for i in range(n)]
t = [t0[i] for i in range(n)]
ans = "Yes"
ansArr = []
# print(s, t)
for i in range(n):
if s[i] == t[i]:
continue
elif s[i] in s[i+1:]:
ind = s.index(s[i], i+1, n)
# ansArr += [[ind, i]]
ansArr += [[ind+1, i+1]]
s[ind], t[i] = t[i], s[ind]
elif s[i] in t[i+1:]:
ind = t.index(s[i], i+1, n)
# ansArr += [[i+1, ind]]
ansArr += [[i+1+1, ind+1]]
s[i+1], t[ind] = t[ind], s[i+1]
# print(ansArr[-1])
# print(s, t)
# ansArr += [[i+1, i]]
ansArr += [[i+1+1, i+1]]
s[i+1], t[i] = t[i], s[i+1]
# print(ansArr[-1])
# print(s, t)
else:
ans = "No"
break
print(ans)
if ans == "Yes":
for item in ansArr:
print(item[0], item[1])
``` | instruction | 0 | 143 | 18 | 286 |
No | output | 1 | 143 | 18 | 287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This problem is different from the easy version. In this version Ujan makes at most 2n swaps. In addition, k ≤ 1000, n ≤ 50 and it is necessary to print swaps themselves. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
After struggling and failing many times, Ujan decided to try to clean up his house again. He decided to get his strings in order first.
Ujan has two distinct strings s and t of length n consisting of only of lowercase English characters. He wants to make them equal. Since Ujan is lazy, he will perform the following operation at most 2n times: he takes two positions i and j (1 ≤ i,j ≤ n, the values i and j can be equal or different), and swaps the characters s_i and t_j.
Ujan's goal is to make the strings s and t equal. He does not need to minimize the number of performed operations: any sequence of operations of length 2n or shorter is suitable.
Input
The first line contains a single integer k (1 ≤ k ≤ 1000), the number of test cases.
For each of the test cases, the first line contains a single integer n (2 ≤ n ≤ 50), the length of the strings s and t.
Each of the next two lines contains the strings s and t, each having length exactly n. The strings consist only of lowercase English letters. It is guaranteed that strings are different.
Output
For each test case, output "Yes" if Ujan can make the two strings equal with at most 2n operations and "No" otherwise. You can print each letter in any case (upper or lower).
In the case of "Yes" print m (1 ≤ m ≤ 2n) on the next line, where m is the number of swap operations to make the strings equal. Then print m lines, each line should contain two integers i, j (1 ≤ i, j ≤ n) meaning that Ujan swaps s_i and t_j during the corresponding operation. You do not need to minimize the number of operations. Any sequence of length not more than 2n is suitable.
Example
Input
4
5
souse
houhe
3
cat
dog
2
aa
az
3
abc
bca
Output
Yes
1
1 4
No
No
Yes
3
1 2
3 1
2 3
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
s1=input()
s2=input()
s1=[i for i in s1]
s2=[i for i in s2]
freq={}
for i in s1:
if i not in freq:
freq[i]=0
freq[i]+=1
for i in s2:
if i not in freq:
freq[i]=0
freq[i]+=1
f=True
for i in freq:
if freq[i]%2!=0:
f=False
print('No')
break
if not f:
continue
pos=0
ans=[]
f=True
while pos<len(s1):
if s1[pos]==s2[pos]:
pos+=1
continue
idx=-1
for i in range(pos+1,len(s1)):
if s1[pos]==s1[i]:
idx=i
break
if idx!=-1:
s2[pos],s1[idx]=s1[idx],s2[pos]
ans.append((idx+1,pos+1))
pos+=1
continue
ans.append((pos+1,pos+1))
idx=-1
for i in range(pos+1,len(s1)):
if s2[pos]==s1[i]:
idx=i
break
s1[pos],s2[pos]=s2[pos],s1[pos]
s2[pos],s1[idx]=s1[idx],s2[pos]
ans.append((idx+1,pos+1))
pos+=1
print('Yes')
print(len(ans))
for i in ans:
print(i[0],i[1])
``` | instruction | 0 | 144 | 18 | 288 |
No | output | 1 | 144 | 18 | 289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 24). The next N lines contain 5 space-separated integers each. The first three integers will be between 0 and 2, inclusive. The last two integers will be between 0 and 3, inclusive. The sum of the first three integers will be equal to the sum of the last two integers.
Output
Output the result – a string of lowercase English letters.
Examples
Input
1
1 0 0 1 0
Output
a
Input
10
2 0 0 1 1
1 1 1 2 1
2 1 0 1 2
1 1 0 1 1
2 1 0 2 1
1 1 1 2 1
1 2 1 3 1
2 0 0 1 1
1 1 0 1 1
1 1 2 2 2
Output
codeforcez
Submitted Solution:
```
def translateinput(i):
return bitbraichar[input().replace(' ','')]
bitbraichar = {
'10010':'a',
'11020':'b',
'20011':'c',
'21012':'d',
'11011':'e',
'21021':'f',
'22022':'g',
'12021':'h',
'12012':'j',
'10120':'k',
'11130':'l',
'20121':'m',
'21122':'n',
'11121':'o',
'21131':'p',
'22132':'q',
'12131':'r',
'12122':'t',
'10221':'u',
'11231':'v',
'12113':'w',
'20222':'x',
'21223':'y',
'11222':'z'
}
n = int(input())
for i in map(translateinput,range(n)):
print(i,end='')
``` | instruction | 0 | 213 | 18 | 426 |
Yes | output | 1 | 213 | 18 | 427 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 24). The next N lines contain 5 space-separated integers each. The first three integers will be between 0 and 2, inclusive. The last two integers will be between 0 and 3, inclusive. The sum of the first three integers will be equal to the sum of the last two integers.
Output
Output the result – a string of lowercase English letters.
Examples
Input
1
1 0 0 1 0
Output
a
Input
10
2 0 0 1 1
1 1 1 2 1
2 1 0 1 2
1 1 0 1 1
2 1 0 2 1
1 1 1 2 1
1 2 1 3 1
2 0 0 1 1
1 1 0 1 1
1 1 2 2 2
Output
codeforcez
Submitted Solution:
```
for s in[*open(0)][1:]:print(chr(97+['10010','11020','20011','21012','11011','21021','22022','12021','00000','12012','10120','11130','20121','21122','11121','21131','22132','12131','11111','12122','10221','11231','12113','20222','21223','11222'].index(s[:-1].replace(' ',''))),end='')
``` | instruction | 0 | 214 | 18 | 428 |
Yes | output | 1 | 214 | 18 | 429 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 24). The next N lines contain 5 space-separated integers each. The first three integers will be between 0 and 2, inclusive. The last two integers will be between 0 and 3, inclusive. The sum of the first three integers will be equal to the sum of the last two integers.
Output
Output the result – a string of lowercase English letters.
Examples
Input
1
1 0 0 1 0
Output
a
Input
10
2 0 0 1 1
1 1 1 2 1
2 1 0 1 2
1 1 0 1 1
2 1 0 2 1
1 1 1 2 1
1 2 1 3 1
2 0 0 1 1
1 1 0 1 1
1 1 2 2 2
Output
codeforcez
Submitted Solution:
```
# coding: utf-8
n = int(input())
d = dict()
d['10010'] = 'a'
d['11020'] = 'b'
d['20011'] = 'c'
d['21012'] = 'd'
d['11011'] = 'e'
d['21021'] = 'f'
d['22022'] = 'g'
d['12021'] = 'h'
d['12012'] = 'j'
d['10120'] = 'k'
d['11130'] = 'l'
d['20121'] = 'm'
d['21122'] = 'n'
d['11121'] = 'o'
d['21131'] = 'p'
d['22132'] = 'q'
d['12131'] = 'r'
d['12122'] = 't'
d['10221'] = 'u'
d['11231'] = 'v'
d['12113'] = 'w'
d['20222'] = 'x'
d['21223'] = 'y'
d['11222'] = 'z'
result = ''
for i in range(n):
result += d[input().replace(' ', '')]
print(result)
``` | instruction | 0 | 215 | 18 | 430 |
Yes | output | 1 | 215 | 18 | 431 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Input
The first line of the input contains a single integer N (1 ≤ N ≤ 24). The next N lines contain 5 space-separated integers each. The first three integers will be between 0 and 2, inclusive. The last two integers will be between 0 and 3, inclusive. The sum of the first three integers will be equal to the sum of the last two integers.
Output
Output the result – a string of lowercase English letters.
Examples
Input
1
1 0 0 1 0
Output
a
Input
10
2 0 0 1 1
1 1 1 2 1
2 1 0 1 2
1 1 0 1 1
2 1 0 2 1
1 1 1 2 1
1 2 1 3 1
2 0 0 1 1
1 1 0 1 1
1 1 2 2 2
Output
codeforcez
Submitted Solution:
```
n = int(input())
D = {
'12113': 'w',
'11222': 'z',
'21223': 'y',
'20222': 'x',
'11231': 'v',
'10221': 'u',
'12122': 't',
'11121': 's',
'12131': 'r',
'22132': 'q',
'21131': 'p',
'11121': 'o',
'21122': 'n',
'20121': 'm',
'11130': 'l',
'10120': 'k',
'12012': 'j',
'11011': 'i',
'12021': 'h',
'22022': 'g',
'21021': 'f',
'11011': 'e',
'21012': 'd',
'20011': 'c',
'11020': 'b',
'10010': 'a',
}
print(''.join(D[input().replace(' ', '')] for _ in range(n)))
``` | instruction | 0 | 216 | 18 | 432 |
Yes | output | 1 | 216 | 18 | 433 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
s=input()
n=len(s)
a=[0]*30
for i in range(n):
a[ord(s[i])-97]+=1
odd=0
for i in range(26):
if(a[i]&1):
odd+=1
if(odd==0 or odd&1):
print("First")
else:
print("Second")
``` | instruction | 0 | 294 | 18 | 588 |
Yes | output | 1 | 294 | 18 | 589 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
word = input()
lets = [0] * 26
for c in word:
lets[ord(c) - 97] += 1
c = 0
for e in lets:
if e % 2 == 1:
c += 1
if c == 0 or c % 2 == 1:
print('First')
else:
print('Second')
``` | instruction | 0 | 295 | 18 | 590 |
Yes | output | 1 | 295 | 18 | 591 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
s = input()
counts = {}
for i in s:
if (i not in counts):
counts[i] = 0
counts[i] += 1
dist = 0
for i in counts:
dist += int(counts[i] % 2 == 1)
if (dist == 0) or (dist % 2 == 1):
print ('First')
else:
print ('Second')
``` | instruction | 0 | 296 | 18 | 592 |
Yes | output | 1 | 296 | 18 | 593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
def GirlAndTheGame(str):
game = list(str)
odd = [item for item in game if game.count(item)%2!=0]
oddOneOut = len(set(odd))
if oddOneOut == 0 or oddOneOut%2==1:
print('First')
else:
print('Second')
GirlAndTheGame(input())
``` | instruction | 0 | 297 | 18 | 594 |
Yes | output | 1 | 297 | 18 | 595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
from collections import Counter
s = list(input())
cnt = Counter(s)
check = 0
for ch in cnt:
if cnt[ch]%2 == 1:
check += 1
if check % 2 == 0:
print("Second")
else:
print("First")
``` | instruction | 0 | 298 | 18 | 596 |
No | output | 1 | 298 | 18 | 597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
x=list(input())
c=0
i=0
while i<(len(x)):
if x[i]==x[len(x)-i-1]:
i+=1
continue
elif x[len(x)-i-1]==x[i]:
x.pop(i)
else:
x.pop(len(x)-i-1)
c+=1
if c%2==0:
print("First")
else:
print("Second")
``` | instruction | 0 | 299 | 18 | 598 |
No | output | 1 | 299 | 18 | 599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
from collections import defaultdict
s = [i for i in input()]
d = defaultdict(int)
for i in s:
d[i] += 1
ar = [v for k,v in d.items() if v%2]
ar.sort()
if sum(ar)%2:
print("First")
else:
print("Second")
``` | instruction | 0 | 300 | 18 | 600 |
No | output | 1 | 300 | 18 | 601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The Little Girl loves problems on games very much. Here's one of them.
Two players have got a string s, consisting of lowercase English letters. They play a game that is described by the following rules:
* The players move in turns; In one move the player can remove an arbitrary letter from string s.
* If the player before his turn can reorder the letters in string s so as to get a palindrome, this player wins. A palindrome is a string that reads the same both ways (from left to right, and vice versa). For example, string "abba" is a palindrome and string "abc" isn't.
Determine which player will win, provided that both sides play optimally well — the one who moves first or the one who moves second.
Input
The input contains a single line, containing string s (1 ≤ |s| ≤ 103). String s consists of lowercase English letters.
Output
In a single line print word "First" if the first player wins (provided that both players play optimally well). Otherwise, print word "Second". Print the words without the quotes.
Examples
Input
aba
Output
First
Input
abca
Output
Second
Submitted Solution:
```
from sys import stdin, stdout
from sys import maxsize
input = stdin.readline
#def print(n):
# stdout.write(str(n)+'\n')
def solve():
pass
test = 1
# test = int(input().strip())
for t in range(0, test):
# n = int(input().strip())
s = input().strip() # String Input, converted to mutable list.
if(s==s[::-1]):
print('First')
else:
if(len(s)%2==0):
print('Second')
else:
print('First')
# n, k = list(map(int, input().strip().split()))
# arr = [int(x) for x in input().strip().split()]
# brr = [list(map(int,input().strip().split())) for i in range(rows)] # 2D array row-wise input
ans = solve()
'''
rows, cols = (5, 5)
arr = [[0]*cols for j in range(rows)] # 2D array initialization
rev_str=s[::-1] # To reverse given string
b=input().strip().split() # list created by spliting about spaces
ch = chr(ord(ch) + 1) # To increment character by one in ASCII
brr = [[int(b[cols*i+j]) for j in range(cols)] for i in range(rows)] # 2D array Linear Input
rows,cols=len(brr),len(brr[0]) # no of rows/cols for 2D array
arr.sort(key = lambda x : x[1]) # sort list of tuples by 2nd element, Default priority - 1st Element then 2nd Element
s=set() # empty set
a=maxsize # initializing infinity
b=-maxsize # initializing -infinity
mapped=list(map(function,input)) # to apply function to list element-wise
try: # Error handling
#code 1
except: # ex. to stop at EOF
#code 2 , if error occurs
'''
``` | instruction | 0 | 301 | 18 | 602 |
No | output | 1 | 301 | 18 | 603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
alice = input()
bob = input()
a = alice.count('1')
b = bob.count('1')
a += (a%2) # if it is odd, then add another one.
print("YES" if a >= b else "NO")
``` | instruction | 0 | 1,149 | 18 | 2,298 |
Yes | output | 1 | 1,149 | 18 | 2,299 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
a,b = input().strip(),input().strip()
x,y = a.count('1'),b.count('1')
if y > x+(x&1):
print('NO')
else:
print('YES')
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,150 | 18 | 2,300 |
Yes | output | 1 | 1,150 | 18 | 2,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a=input()
print('YES'if a.count('1')+(a.count('1')&1)>=input().count('1')else'NO')
``` | instruction | 0 | 1,151 | 18 | 2,302 |
Yes | output | 1 | 1,151 | 18 | 2,303 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a = list(input())
b = list(input())
print(['NO', 'YES'][a.count('1')+a.count('1')%2 >= b.count('1')])
``` | instruction | 0 | 1,152 | 18 | 2,304 |
Yes | output | 1 | 1,152 | 18 | 2,305 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a=input()
b=input()
coa=0
cob=0
for i in a:
if i=='1': coa+=1
for i in b:
if i=='1': cob+=1
if coa%2==1:
print('YES')
else:
if cob>coa: print('NO')
else: print('YES')
``` | instruction | 0 | 1,153 | 18 | 2,306 |
No | output | 1 | 1,153 | 18 | 2,307 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
def play(s, t):
count1, count2 = 0, 0
for elem in s:
if elem == '1':
count1 += 1
for elem in t:
if elem == '1':
count2 += 1
if count2 - count1 > 1 or (count2 - count1 == 0 and count1 % 2 == 0):
return "NO"
return "YES"
a = input()
b = input()
print(play(a, b))
``` | instruction | 0 | 1,154 | 18 | 2,308 |
No | output | 1 | 1,154 | 18 | 2,309 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
a=input()
b=input()
coa=0
cob=0
for i in a:
if i=='1': coa+=1
for i in b:
if i=='1': cob+=1
if coa+(coa&1)>cob:print('YES')
else:print('NO')
``` | instruction | 0 | 1,155 | 18 | 2,310 |
No | output | 1 | 1,155 | 18 | 2,311 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are fishing with polar bears Alice and Bob. While waiting for the fish to bite, the polar bears get bored. They come up with a game. First Alice and Bob each writes a 01-string (strings that only contain character "0" and "1") a and b. Then you try to turn a into b using two types of operations:
* Write parity(a) to the end of a. For example, <image>.
* Remove the first character of a. For example, <image>. You cannot perform this operation if a is empty.
You can use as many operations as you want. The problem is, is it possible to turn a into b?
The parity of a 01-string is 1 if there is an odd number of "1"s in the string, and 0 otherwise.
Input
The first line contains the string a and the second line contains the string b (1 ≤ |a|, |b| ≤ 1000). Both strings contain only the characters "0" and "1". Here |x| denotes the length of the string x.
Output
Print "YES" (without quotes) if it is possible to turn a into b, and "NO" (without quotes) otherwise.
Examples
Input
01011
0110
Output
YES
Input
0011
1110
Output
NO
Note
In the first sample, the steps are as follows: 01011 → 1011 → 011 → 0110
Submitted Solution:
```
print('YES' if (int(input()) + 1) // 2 >= (int(input()) + 1) // 2 else 'NO')
``` | instruction | 0 | 1,156 | 18 | 2,312 |
No | output | 1 | 1,156 | 18 | 2,313 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
INT_MAX = int(1e9 + 7)
def dfs(start_vertex, go_out):
m = 0
last = {}
for key, value in go_out.items():
last[key] = value - 1 # -1for each item from go_out
m += value # And total count of ways
stack = [start_vertex] # Init stack and put first vertex
path = [] # Result path
while stack:
now = stack[-1] # take one vertex
if last[now] == -1: # If all paths have already been gone:
stack.pop() # Delete this vertex from stack...
path.append(now) # And add to the result. Do not forget to reverse the result
continue
# Choosing nex vertex - simply with index las[now]. For this the 2-3 line.
nex = vertexes[now][last[now]]
last[now] -= 1 # This way is gone...
m -= 1 # and here (total count)...
stack.append(nex) # TODO this vertex
return path if m == 0 else [] # If one or more ways have not been gone: path = []
n = int(input())
strings = [input() for _ in range(n)]
parts = set()
# Building vertexes and go_in:
vertexes = {}
for string in strings:
part1 = string[:2]
part2 = string[1:]
parts.add(part1)
parts.add(part2)
try:
vertexes[part1].append(part2)
except Exception:
vertexes[part1] = [part2]
# //////////////////
parts = list(parts)
# Building go_in and go_out:
go_in = {i: 0 for i in parts}
go_out = go_in.copy()
for go_from, go_to in vertexes.items():
go_out[go_from] += len(go_to)
for i in go_to:
go_in[i] += 1
# //////////////////////////
# Serching for vertex with go_out > go_in and for the main answer: YES or NO:
count = 0
start_vertex = 0
for i in range(len(parts)):
if go_out[parts[i]] - go_in[parts[i]] > 1:
count = INT_MAX
break
if go_out[parts[i]] != go_in[parts[i]]:
count += 1
if go_out[parts[i]] > go_in[parts[i]]:
start_vertex = i
# ////////////////////////////////////////
if count > 2:
print('NO')
else:
path = dfs(parts[start_vertex], go_out)
if path:
# Building result
path.reverse()
result = [path[0][0]]
for i in range(len(path)):
result.append(path[i][1])
print('YES')
print(''.join(result))
else:
print('NO')
``` | instruction | 0 | 1,260 | 18 | 2,520 |
Yes | output | 1 | 1,260 | 18 | 2,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
from collections import Counter, defaultdict
def main():
ht = defaultdict(list)
cnt = Counter()
n = int(input())
for _ in range(n):
s = input()
h, t = s[:2], s[1:]
ht[h].append(t)
cnt[h] += 1
cnt[t] -= 1
st = nd = flag = True
for t, i in cnt.items():
if i:
if i == 1:
if st:
st, h = False, t
else:
break
elif i == -1:
if nd:
nd = False
else:
break
else:
break
else:
flag = st != nd
if flag:
print('NO')
return
stack, res = [h], []
while stack:
m = stack[-1]
if ht[m]:
stack.append(ht[m].pop())
else:
del stack[-1]
res.append(m[-1])
if len(res) != n + 1:
print('NO')
return
res.append(h[0])
res.reverse()
print('YES')
print(''.join(res))
if __name__ == '__main__':
main()
``` | instruction | 0 | 1,261 | 18 | 2,522 |
Yes | output | 1 | 1,261 | 18 | 2,523 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
INT_MAX = int(1e9 + 7)
def dfs(start_vertex, go_out):
m = 0
last = {}
for key, value in go_out.items():
last[key] = value - 1 # -1for each item from go_out
m += value # And total count of ways
stack = [start_vertex] # Init stack and put first vertex
path = [] # Result path
while stack:
now = stack[-1] # take one vertex
if last[now] == -1: # If all paths have already been gone:
stack.pop() # Delete this vertex from stack...
path.append(now) # And add to the result. Do not forget to reverse the result
continue
# Choosing nex vertex - simply with index las[now]. For this the 2-3 line.
nex = vertexes[now][last[now]]
last[now] -= 1 # This way is gone...
m -= 1 # and here (total count)...
stack.append(nex) # TODO this vertex
return path if m == 0 else [] # If one or more ways have not been gone: path = []
n = int(input())
strings = [input() for _ in range(n)]
parts = set()
# Building vertexes and go_in:
vertexes = {}
for string in strings:
part1 = string[:2]
part2 = string[1:]
parts.add(part1)
parts.add(part2)
try:
vertexes[part1].append(part2)
except Exception:
vertexes[part1] = [part2]
# //////////////////
parts = list(parts)
# Building go_in and go_out:
go_in = {i: 0 for i in parts}
go_out = go_in.copy()
for go_from, go_to in vertexes.items():
go_out[go_from] += len(go_to)
for i in go_to:
go_in[i] += 1
# //////////////////////////
# Serching for vertex with go_out > go_in and for the main answer: YES or NO:
count = 0
start_vertex = 0
for i in range(len(parts)):
if go_out[parts[i]] - go_in[parts[i]] > 1:
count = INT_MAX
break
if go_out[parts[i]] != go_in[parts[i]]:
count += 1
if go_out[parts[i]] > go_in[parts[i]]:
start_vertex = i
# ////////////////////////////////////////
if count > 2:
print('NO')
else:
path = dfs(parts[start_vertex], go_out)
if path:
# Building result
path.reverse()
result = [path[0][0]]
for item in path:
result.append(item[1])
print('YES')
print(''.join(result))
else:
print('NO')
``` | instruction | 0 | 1,262 | 18 | 2,524 |
Yes | output | 1 | 1,262 | 18 | 2,525 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
def eulerian():
for i in range(numEdges):
edge = input()
v1 = edge[:2]
v2 = edge[1:]
if i == 0:
firstEdge = edge
if v1 in passwordGraph:
passwordGraph[v1][0].append(v2)
else:
passwordGraph[v1] = [[v2],[]]
if v2 in passwordGraph:
passwordGraph[v2][1].append(v1)
else:
passwordGraph[v2] = [[],[v1]]
oddVertices = 0
vertices = list(passwordGraph)
size = {}
start_node = None
for vertex, adjList in passwordGraph.items():
numInEdges = len(adjList[1])
numOutEdges = len(adjList[0])
size[vertex] = numOutEdges
# if firstEdge == "M1o" and numEdges == 200000:
# print(degree[vertex])
if numOutEdges != numInEdges:
oddVertices += 1
if numInEdges == numOutEdges - 1:
start_node = vertex
if oddVertices > 2:
return "NO"
if oddVertices == 0:
start_node = vertex
elif oddVertices == 1:
return "NO"
if start_node == None:
return "NO"
curr_path = [start_node]
curr_v = curr_path[0]
circuit = []
while len(curr_path):
if (size[curr_v]):
curr_path.append(curr_v)
size[curr_v] -= 1
curr_v = passwordGraph[curr_v][0].pop()
else:
if curr_v in passwordGraph:
del passwordGraph[curr_v]
circuit.append(curr_v)
curr_v = curr_path.pop()
#print(str(len(passwordGraph)) + "\t" + str(len(circuit)))
if len(passwordGraph) > 0:
return "NO"
circuit.reverse()
#print(circuit)
# findpath(start_node, numEdges)
# password.insert(0,start_node[0])
password = circuit[0]
for edge in circuit[1:]:
password += edge[1]
return "YES\n" + str(password)
# def findpath(node, numEdges):
# global password
# while len(passwordGraph[node][0]) != 0:
# vertex = passwordGraph[node][0].pop()
# if(numEdges == 200000):
# print(vertex)
# findpath(vertex, numEdges)
# password.insert(0,node[1])
numEdges = int(input())
passwordGraph = {}
password = []
print(eulerian())
#and start_node == None
``` | instruction | 0 | 1,263 | 18 | 2,526 |
Yes | output | 1 | 1,263 | 18 | 2,527 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
from collections import defaultdict
# arr = ["aaa" for i in range(7)]
# print(arr)
arr = ["aca", "aba", "aba", "cab", "bac"]
n = len(arr)
graph = defaultdict(list)
IN = defaultdict(int)
OUT = defaultdict(int)
solution = []
default_start = None
for ele in arr:
first_char, last_char = ele[0:2], ele[1:3]
graph[first_char].append(last_char)
if not last_char in graph:
graph[last_char] = []
OUT[first_char] += 1
IN[last_char] += 1
default_start = first_char
# n = len(graph)
# m = sum(len(graph[node]) for node in graph)
print(graph)
# for node in graph:
# print(node, IN[node], OUT[node])
def hasEulerianPath():
global default_start
start_nodes, end_nodes = 0, 0
for node in graph:
if OUT[node] - IN[node] > 1 and IN[node] - OUT[node] > 1: return False
if OUT[node] - IN[node] == 1:
default_start = node
start_nodes += 1
if IN[node] - OUT[node] == 1: end_nodes += 1
return (start_nodes==0 and end_nodes == 0) or (start_nodes == 1 and end_nodes == 1)
def DFS(at):
while OUT[at] > 0:
OUT[at] -= 1
# print(at, graph[at][OUT[at]])
DFS(graph[at][OUT[at]])
solution.append(at)
def eulerPath():
if not hasEulerianPath():
# print("NO")
return None
# print(default_start)
DFS(default_start)
print(''.join(solution[::-1]))
eulerPath()
``` | instruction | 0 | 1,264 | 18 | 2,528 |
No | output | 1 | 1,264 | 18 | 2,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
def eulerian():
for i in range(numEdges):
edge = input()
v1 = edge[:2]
v2 = edge[1:]
if i == 0:
firstEdge = edge
if v1 in passwordGraph:
passwordGraph[v1][0].append(v2)
else:
passwordGraph[v1] = [[v2],[]]
if v2 in passwordGraph:
passwordGraph[v2][1].append(v1)
else:
passwordGraph[v2] = [[],[v1]]
oddVertices = 0
vertices = list(passwordGraph)
size = {}
degree = {}
start_node = None
for vertex, adjList in passwordGraph.items():
numInEdges = len(adjList[1])
numOutEdges = len(adjList[0])
size[vertex] = numOutEdges
degree[vertex] = numOutEdges+numInEdges
if firstEdge == "M1o" and numEdges == 200000:
print(degree[vertex])
if degree[vertex] % 2 == 1:
oddVertices += 1
if numInEdges == numOutEdges - 1:
start_node = vertex
if oddVertices > 2:
return "NO"
if oddVertices == 0:
start_node = vertex
elif oddVertices == 1:
return "NO"
if start_node == None:
return "NO"
curr_path = [start_node]
curr_v = curr_path[0]
circuit = []
while len(curr_path):
if (size[curr_v]):
curr_path.append(curr_v)
size[curr_v] -= 1
curr_v = passwordGraph[curr_v][0].pop()
else:
circuit.append(curr_v)
curr_v = curr_path.pop()
#print(str(len(passwordGraph)) + "\t" + str(len(circuit)))
if (len(passwordGraph) - len(circuit)) > 0:
return "NO"
circuit.reverse()
#print(circuit)
# findpath(start_node, numEdges)
# password.insert(0,start_node[0])
password = circuit[0]
for edge in circuit[1:]:
password += edge[1]
return "YES\n" + str(password)
# def findpath(node, numEdges):
# global password
# while len(passwordGraph[node][0]) != 0:
# vertex = passwordGraph[node][0].pop()
# if(numEdges == 200000):
# print(vertex)
# findpath(vertex, numEdges)
# password.insert(0,node[1])
numEdges = int(input())
passwordGraph = {}
password = []
print(eulerian())
#and start_node == None
``` | instruction | 0 | 1,265 | 18 | 2,530 |
No | output | 1 | 1,265 | 18 | 2,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
numEdges = int(input())
passwordGraph = {}
def eulerian():
for i in range(numEdges):
edge = input()
v1 = edge[:2]
v2 = edge[1:]
if v1 in passwordGraph:
passwordGraph[v1][0].append(v2)
else:
passwordGraph[v1] = [[v2],[]]
if v2 in passwordGraph:
passwordGraph[v2][1].append(v1)
else:
passwordGraph[v2] = [[],[v1]]
oddVertices = 0
for vertex in passwordGraph:
numInEdges = len(passwordGraph[vertex][1])
numOutEdges = len(passwordGraph[vertex][0])
if numInEdges < numOutEdges:
start_node = vertex
oddVertices += 1
elif numInEdges > numOutEdges:
oddVertices += 1
if oddVertices > 2:
return "NO"
if oddVertices == 0:
start_node = vertex
elif oddVertices == 1:
return "NO"
password = start_node
vertex = start_node
while len(passwordGraph[vertex][0]) != 0:
#print (passwordGraph)
vertex = passwordGraph[vertex][0].pop(0)
password += vertex[1:]
#print(str(start_node) + "\t" + str(passwordGraph[start_node]))
#print(password)
return "YES\n" + password
print(eulerian())
#and start_node == None
``` | instruction | 0 | 1,266 | 18 | 2,532 |
No | output | 1 | 1,266 | 18 | 2,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
While dad was at work, a little girl Tanya decided to play with dad's password to his secret database. Dad's password is a string consisting of n + 2 characters. She has written all the possible n three-letter continuous substrings of the password on pieces of paper, one for each piece of paper, and threw the password out. Each three-letter substring was written the number of times it occurred in the password. Thus, Tanya ended up with n pieces of paper.
Then Tanya realized that dad will be upset to learn about her game and decided to restore the password or at least any string corresponding to the final set of three-letter strings. You have to help her in this difficult task. We know that dad's password consisted of lowercase and uppercase letters of the Latin alphabet and digits. Uppercase and lowercase letters of the Latin alphabet are considered distinct.
Input
The first line contains integer n (1 ≤ n ≤ 2·105), the number of three-letter substrings Tanya got.
Next n lines contain three letters each, forming the substring of dad's password. Each character in the input is a lowercase or uppercase Latin letter or a digit.
Output
If Tanya made a mistake somewhere during the game and the strings that correspond to the given set of substrings don't exist, print "NO".
If it is possible to restore the string that corresponds to given set of substrings, print "YES", and then print any suitable password option.
Examples
Input
5
aca
aba
aba
cab
bac
Output
YES
abacaba
Input
4
abc
bCb
cb1
b13
Output
NO
Input
7
aaa
aaa
aaa
aaa
aaa
aaa
aaa
Output
YES
aaaaaaaaa
Submitted Solution:
```
# from debug import debug
from collections import defaultdict, Counter, deque
import sys; input = sys.stdin.readline
n = int(input())
graph = defaultdict(list)
d = Counter()
for i in range(n):
s = input().strip()
a, b = s[:2], s[1:3]
graph[a].append(b)
d[a] += 1
d[b] -= 1
start, stop = -1, -1
for k,v in d.items():
if v>1: print("NO"); sys.exit()
if v == 1 and start == -1: start = k
elif v == 1 and start != -1: print("NO"); sys.exit()
if v == -1 and stop == -1: stop = k
elif v == -1 and stop != -1: print("NO"); sys.exit()
e = k
if start == -1: start = e
# finding eular path
stack = [start]
ans = deque()
while stack:
node = stack[-1]
if graph[node]: stack.append(graph[node].pop())
else: ans.appendleft(stack.pop()[1])
ans.appendleft(start[0])
print("YES")
print(*ans, sep="")
``` | instruction | 0 | 1,267 | 18 | 2,534 |
No | output | 1 | 1,267 | 18 | 2,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string S of length N consisting of `A`, `C`, `G` and `T`. Answer the following Q queries:
* Query i (1 \leq i \leq Q): You will be given integers l_i and r_i (1 \leq l_i < r_i \leq N). Consider the substring of S starting at index l_i and ending at index r_i (both inclusive). In this string, how many times does `AC` occurs as a substring?
Constraints
* 2 \leq N \leq 10^5
* 1 \leq Q \leq 10^5
* S is a string of length N.
* Each character in S is `A`, `C`, `G` or `T`.
* 1 \leq l_i < r_i \leq N
Input
Input is given from Standard Input in the following format:
N Q
S
l_1 r_1
:
l_Q r_Q
Output
Print Q lines. The i-th line should contain the answer to the i-th query.
Example
Input
8 3
ACACTACG
3 7
2 3
1 8
Output
2
0
3
Submitted Solution:
```
n, q = map(int, input().split())
s = input()
# 先に回数を数える。累積和的に
hoge = [s[0:(i + 1)].count('AC') for i in range(n)]
for _ in range(q + 1):
l, r = map(int, input().split())
print(hoge[r - 1] - hoge[l - 1])
``` | instruction | 0 | 1,505 | 18 | 3,010 |
No | output | 1 | 1,505 | 18 | 3,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
a,b=input().split();print(['=<'[a<b],'>'][a>b])
``` | instruction | 0 | 1,546 | 18 | 3,092 |
Yes | output | 1 | 1,546 | 18 | 3,093 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In programming, hexadecimal notation is often used.
In hexadecimal notation, besides the ten digits 0, 1, ..., 9, the six letters `A`, `B`, `C`, `D`, `E` and `F` are used to represent the values 10, 11, 12, 13, 14 and 15, respectively.
In this problem, you are given two letters X and Y. Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
When X and Y are seen as hexadecimal numbers, which is larger?
Constraints
* Each X and Y is `A`, `B`, `C`, `D`, `E` or `F`.
Input
Input is given from Standard Input in the following format:
X Y
Output
If X is smaller, print `<`; if Y is smaller, print `>`; if they are equal, print `=`.
Examples
Input
A B
Output
<
Input
E C
Output
>
Input
F F
Output
=
Submitted Solution:
```
s1=input()
s2=input()
s1=s1.lower()
s2=s2.lower()
flag=0
if(len(s1)>len(s2)):
a=len(s2)
else:
a=len(s1)
for i in range(0,a):
if(s1[i]==s2[i]):
flag=flag+1
elif(s1[i]>s2[i]):
print('>')
break;
elif(s1[i]<s2[i]):
print('<')
break;
if(flag==a):
if(len(s1)>len(s2)):
print(">")
elif(len(s1)<len(s2)):
print("<")
else:
print("=")
``` | instruction | 0 | 1,550 | 18 | 3,100 |
No | output | 1 | 1,550 | 18 | 3,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.
Let's help Vitaly solve this easy problem!
Input
The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.
The second line contains string t (|t| = |s|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.
Output
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Examples
Input
a
c
Output
b
Input
aaa
zzz
Output
kkk
Input
abcdefg
abcdefh
Output
No such string
Note
String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.
Submitted Solution:
```
s = str(input())
t = str(input())
r = ""
l = len(s)
for i in range(l):
if s[i] == t[i]:
r += s[i]
continue
else:
if ord(s[i]) + 1 < ord(t[i]):
r += chr(ord(s[i])+1)
r += "a"*(l - len(r))
print(r)
exit()
else:
x = r
y = r
x += s[i]
x += "z"*(l - len(x))
y += t[i]
y += "a"*(l - len(y))
if x == s:
if y == t:
print("No such string")
exit()
else:
print(y)
exit()
else:
print(x)
exit()
``` | instruction | 0 | 2,014 | 18 | 4,028 |
Yes | output | 1 | 2,014 | 18 | 4,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.
Let's help Vitaly solve this easy problem!
Input
The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.
The second line contains string t (|t| = |s|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.
Output
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Examples
Input
a
c
Output
b
Input
aaa
zzz
Output
kkk
Input
abcdefg
abcdefh
Output
No such string
Note
String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.
Submitted Solution:
```
def f(a,b):
if a==b:return False
if ord(a[0])<ord(b[0]):return True
if ord(a[0])>ord(b[0]):return False
return f(a[1:],b[1:])
def g(a):
if a[-1]=='z':return g(a[:-1])+'a'
return a[:-1]+chr(ord(a[-1])+1)
a=g(input())
b=input()
print(a if f(a,b)else'No such string')
``` | instruction | 0 | 2,015 | 18 | 4,030 |
Yes | output | 1 | 2,015 | 18 | 4,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.
Let's help Vitaly solve this easy problem!
Input
The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.
The second line contains string t (|t| = |s|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.
Output
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Examples
Input
a
c
Output
b
Input
aaa
zzz
Output
kkk
Input
abcdefg
abcdefh
Output
No such string
Note
String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.
Submitted Solution:
```
s = input()
t = input()
fi = s
ok = False
for i in range(len(s) - 1, -1, -1):
if s[i] == 'z': continue
tm = len(s)
s = s[:i] + chr(ord(s[i]) + 1)
for j in range(i + 1, tm):
s += 'a'
if fi < s and s < t:
print(s)
exit()
print("No such string")
``` | instruction | 0 | 2,016 | 18 | 4,032 |
Yes | output | 1 | 2,016 | 18 | 4,033 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.
Let's help Vitaly solve this easy problem!
Input
The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.
The second line contains string t (|t| = |s|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.
Output
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Examples
Input
a
c
Output
b
Input
aaa
zzz
Output
kkk
Input
abcdefg
abcdefh
Output
No such string
Note
String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.
Submitted Solution:
```
from string import ascii_lowercase
m = {v: i for i, v in enumerate(ascii_lowercase)}
a = input()
b = input()
s, ans = "", False
for i in range(len(a)):
if (m[b[i]] - m[a[i]] > 1):
s += ascii_lowercase[m[a[i]] + 1] + a[i+1:]
ans = True
break
elif (m[b[i]] - m[a[i]] == 1):
s += a[i]
all_z = True
for c in a[i+1:]:
if c != 'z':
s += ascii_lowercase[m[c]+1]
all_z = False
else:
s += c
if not all_z:
ans = True
break
else:
s += a[i]
if ans:
print(s)
else:
print("No such string")
``` | instruction | 0 | 2,019 | 18 | 4,038 |
No | output | 1 | 2,019 | 18 | 4,039 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.
During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.
Let's help Vitaly solve this easy problem!
Input
The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.
The second line contains string t (|t| = |s|), consisting of lowercase English letters.
It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.
Output
If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).
If such string exists, print it. If there are multiple valid strings, you may print any of them.
Examples
Input
a
c
Output
b
Input
aaa
zzz
Output
kkk
Input
abcdefg
abcdefh
Output
No such string
Note
String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.
Submitted Solution:
```
a=input()
b=input()
l=len(a)
p=0
s=""
for i in range(l):
if ord(b[i])-ord(a[i])>1:
s=s+chr(ord(a[i])+1)
else:
p=1
break
if p!=1:
print(s)
else:
print("No such string")
``` | instruction | 0 | 2,020 | 18 | 4,040 |
No | output | 1 | 2,020 | 18 | 4,041 |
Provide a correct Python 3 solution for this coding contest problem.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko | instruction | 0 | 2,423 | 18 | 4,846 |
"Correct Solution:
```
moji = input()
flag = 0
answer = []
for i in range(0,len(moji)):
if flag == 1:
flag = 0
continue
if moji[i] == "2":
answer.append("k")
elif moji[i] == "3":
answer.append("s")
elif moji[i] == "4":
answer.append("t")
elif moji[i] == "5":
answer.append("n")
elif moji[i] == "6":
answer.append("h")
elif moji[i] == "7":
answer.append("m")
elif moji[i] == "8":
answer.append("y")
elif moji[i] == "9":
answer.append("r")
elif moji[i] == "0":
i+=1
flag = 1
if moji[i] == "T":
answer.append("wa")
elif moji[i] == "D":
answer.append("wo")
elif moji[i] == "U":
answer.append("nn")
elif moji[i] == "T":
answer.append("a")
elif moji[i] == "L":
answer.append("i")
elif moji[i] == "U":
answer.append("u")
elif moji[i] == "R":
answer.append("e")
elif moji[i] == "D":
answer.append("o")
for i in answer:
print(i, end="")
print()
``` | output | 1 | 2,423 | 18 | 4,847 |
Provide a correct Python 3 solution for this coding contest problem.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko | instruction | 0 | 2,424 | 18 | 4,848 |
"Correct Solution:
```
flicks = {
"T": "a",
"L": "i",
"U": "u",
"R": "e",
"D": "o"
}
buttons = {
"1": "",
"2": "k",
"3": "s",
"4": "t",
"5": "n",
"6": "h",
"7": "m",
"8": "y",
"9": "r",
"0": "w"
}
def get_word(button, flick):
if button == "0" and flick == "U":
return "nn"
return buttons[button] + flicks[flick]
cmds = input()
result = ""
for i in range(0, len(cmds), 2):
result += get_word(cmds[i], cmds[i+1])
print(result)
``` | output | 1 | 2,424 | 18 | 4,849 |
Provide a correct Python 3 solution for this coding contest problem.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko | instruction | 0 | 2,425 | 18 | 4,850 |
"Correct Solution:
```
c = {"1":"", "2":"k", "3":"s", "4":"t", "5":"n",
"6":"h","7":"m", "8":"y", "9":"r", "0":"w"}
m = {"T":"a", "L":"i", "U":"u", "R":"e", "D":"o"}
s = input()
ans = ""
for i in range(0, len(s), 2):
com = s[i:i+2]
if com == "0U":ans += "nn"
else:ans += c[com[0]] + m[com[1]]
print(ans)
``` | output | 1 | 2,425 | 18 | 4,851 |
Provide a correct Python 3 solution for this coding contest problem.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko | instruction | 0 | 2,426 | 18 | 4,852 |
"Correct Solution:
```
c="w kstnhmyr"
v="aiueo"
d=["T","L","U","R","D"]
f=[]
a=input()
for i in range(0,len(a),2):
b=int(a[i])
e=d.index(a[i+1])
if b==0 and e==2: f+='nn'
elif b==1: f+=v[e]
else:f+=c[b]+v[e]
print("".join(f))
``` | output | 1 | 2,426 | 18 | 4,853 |
Provide a correct Python 3 solution for this coding contest problem.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko | instruction | 0 | 2,427 | 18 | 4,854 |
"Correct Solution:
```
c,v,d,f,a="w kstnhmyr","aiueo",["T","L","U","R","D"],"",input()
for i in range(0,len(a),2):
b=int(a[i])
e=d.index(a[i+1])
if b==0 and e==2: f+='nn'
elif b==1: f+=v[e]
else:f+=c[b]+v[e]
print(f)
``` | output | 1 | 2,427 | 18 | 4,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A: Flick input
A college student who loves 2D, commonly known as 2D, replaced his long-used Galapagos mobile phone with a smartphone. 2D, who operated the smartphone for the first time, noticed that the character input method was a little different from that of old mobile phones. On smartphones, a method called flick input was adopted by taking advantage of the fact that the screen can be touch-flicked (flick means "the operation of sliding and flipping a finger on the screen"). Flick input is a character input method as shown below.
In flick input, as shown in Fig. A (a), input is performed using 10 buttons 0-9 (# and * are omitted this time). Each button is assigned one row-row row as shown in the figure.
<image>
---
Figure A: Smartphone button details
One button can be operated by the method shown in Fig. A (b). In other words
* If you just "touch" a button, the character "Adan" corresponding to that button will be output.
* Press the button "flick to the left" to output "Idan"
* When you press the button "flick upward", "Udan" is output.
* Press the button "flick to the right" to output "Edan"
* Press the button "flick down" to output "step"
That's what it means. Figure A (c) shows how to input flicks for all buttons. Note that you flick button 0 upwards to output "n".
Your job is to find the string that will be output as a result of a flick input operation, given a string that indicates that operation.
Input
A character string representing a flick input operation is given on one line (2 to 1000 characters). This character string consists of the numbers '0'-'9', which represent the buttons to be operated, and the five types of characters,'T',' L',' U',' R', and'D', which represent the flick direction. It consists of a combination.
The characters that represent the flick direction have the following meanings.
*'T': Just touch
*'L': Flick left
*'U': Flick up
*'R': Flick right
*'D': Flick down
For example, "2D" represents the operation of "flicking the button of 2 downward", so "ko" can be output.
In the given character string, one number and one character indicating the flick direction appear alternately. Also, it always starts with one number and ends with one letter indicating the flick direction. Furthermore, in Fig. A (c), flicking is not performed toward a place without characters (there is no input such as "8L").
Output
Output the character string representing the result of the flick input operation in Roman characters.
For romaji consonants,
* Or line:'k'
* Sayuki:'s'
* Ta line:'t'
* Line:'n'
* Is the line:'h'
* Ma line:'m'
* Ya line:'y'
* Line:'r'
* Wa line:'w'
to use. Romaji, which represents one hiragana character, should be output as a set of the consonants and vowels ('a','i','u','e','o') represented above. "shi" and "fu" are wrong because they violate this condition. However, the character "A line" should output only one vowel character, and "n" should output "n" twice in a row as "nn".
Output a line break at the end of the character string.
Sample Input 1
5R2D
Sample Output 1
neko
Sample Input 2
8U9U6U0T
Sample Output 2
yuruhuwa
Sample Input 3
9L4U7R1L2D0U4R3U4D
Sample Output 3
ritumeikonntesuto
Example
Input
5R2D
Output
neko
Submitted Solution:
```
c = {"1":"", "2":"k", "3":"s", "4":"t", "5":"n",
"6":"h","7":"m", "8":"y", "9":"r", "0":"w"}
m = {"T":"a", "L":"i", "U":"u", "R":"e", "D":"o"}
s = input()
ans = ""
for i in range(0, len(s), 2):
if com == "0U":ans += "nn"
else:ans += c[com[0]] + m[com[1]]
print(ans)
``` | instruction | 0 | 2,428 | 18 | 4,856 |
No | output | 1 | 2,428 | 18 | 4,857 |
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