message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,528 | 0 | 81,056 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import string
def solve():
s = input()
n = int(input())
alphabets = len(string.ascii_lowercase)
count = [[0] * alphabets for _ in range(len(s) + 1)]
for i, c in enumerate(s, 1):
count[i][ord(c) - ord('a')] += 1
for c in range(alphabets):
count[i][c] += count[i - 1][c]
def query(l, r):
if l == r:
return 'Yes'
if s[l - 1] != s[r - 1]:
return 'Yes'
unique = 0
for x, y in zip(count[l - 1], count[r]):
if y - x > 0:
unique += 1
if unique > 2:
return 'Yes'
return 'No'
for _ in range(n):
print(query(*map(int, input().split())))
solve()
``` | output | 1 | 40,528 | 0 | 81,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,529 | 0 | 81,058 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
# -*- coding: utf-8 -*-
import sys
from itertools import accumulate
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
# sys.setrecursionlimit(10 ** 9)
INF = 10 ** 18
MOD = 10 ** 9 + 7
S = [ord(c)-97 for c in input()]
N = len(S)
acc = list2d(26, N+1, 0)
for i, c in enumerate(S):
acc[c][i+1] += 1
for c in range(26):
acc[c] = list(accumulate(acc[c]))
for _ in range(INT()):
l, r = MAP()
l -= 1
ln = r - l
if ln == 1:
Yes()
continue
cnt = 0
for c in range(26):
if acc[c][r] - acc[c][l] >= 1:
cnt += 1
if cnt >= 3:
Yes()
continue
if S[l] != S[r-1]:
Yes()
else:
No()
``` | output | 1 | 40,529 | 0 | 81,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,530 | 0 | 81,060 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
s = input()
n = len(s)
memo = [None]*(n+1)
count = [0]*26
memo[0] = count.copy()
for i,char in enumerate(s):
count[ord(char)-97] += 1
memo[i+1] = count.copy()
def f():
l, r = [int(char) for char in input().split()]
if l == r:
return 'Yes'
# len >= 2
if s[r-1] != s[l-1]:
return 'Yes'
kinds = 0
seg = [memo[r][k] - memo[l-1][k] for k in range(26)]
for c in seg:
if c:
kinds += 1
if kinds >= 3:
return 'Yes'
return 'No'
T = int(input())
for t in range(T):
print(f())
``` | output | 1 | 40,530 | 0 | 81,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,531 | 0 | 81,062 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
import sys
input=sys.stdin.readline
s=input().rstrip('\n')
l=len(s)
ref=[[0]*26]
ref1=[0]*26
for i in range(l):
#print(ord(s[i]))
ref1[ord(s[i])-97]+=1
ref.append(ref1.copy())
#@print(ref)
q=int(input())
for i in range(q):
l,r=map(int,input().split())
if(l==r or s[l-1]!=s[r-1]):
print('Yes')
continue
p=0
f=[ref[r][k]-ref[l-1][k] for k in range(26)]
for i in f:
if(i>0):
p+=1
if(p>=3):
print('Yes')
else:
print('No')
``` | output | 1 | 40,531 | 0 | 81,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,532 | 0 | 81,064 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
s = input()
n = len(s)
dp = [[0] * 26 for _ in range(n + 1)]
for i in range(1, n + 1):
dp[i] = dp[i - 1].copy()
dp[i][ord(s[i - 1]) - ord('a')] += 1
for _ in range(int(input())):
l, r = map(int, input().split())
l -= 1
k = [dp[r][i] - dp[l][i] for i in range(26)]
c = sum((p != 0) for p in k)
k1 = 0
k2 = 0
for j in k:
if j:
k1 = j
k2 += j
if c > 2 or (l + 1 == r) or (c == 2 and s[l] != s[r - 1]):
print("Yes")
else:
print("No")
``` | output | 1 | 40,532 | 0 | 81,065 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,533 | 0 | 81,066 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
ALPHA = 26
def unique(cnt, l, r):
sum = 0
for a,b in zip(cnt[l-1],cnt[r]):
sum += a < b
return sum
s = list(map(lambda c: ord(c)-ord('a'),list(input())))
cnt = [[0 for j in range(ALPHA)]]
for i in range(len(s)):
cnt.append(list(cnt[i]))
cnt[i+1][s[i]] += 1
n = int(input())
for i in range(n):
l, r = map(int,input().split())
if (l == r or s[l-1] != s[r-1] or unique(cnt,l,r) >= 3):
print("Yes")
else:
print("No")
``` | output | 1 | 40,533 | 0 | 81,067 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram. | instruction | 0 | 40,534 | 0 | 81,068 |
Tags: binary search, constructive algorithms, data structures, strings, two pointers
Correct Solution:
```
def first():
s = input()
cases = int(input())
lastChange = [0]*len(s)
lastChange[0] = 0
lc = s[0]
for i in range(1, len(s)):
c = s[i]
if c != lc:
lastChange[i] = i
lc = c
else:
lastChange[i] = lastChange[i-1]
for i in range(cases):
l, r = map(int, input().split())
l -= 1
if r-l == 1:
print('Yes')
continue
if s[l] != s[r-1]:
print('Yes')
continue
answer = 'No'
p = lastChange[r-1]
c1 = s[r-1]
if p > l:
c2 = s[p-1]
while p > l:
print('-',p)
if s[p-1] != c1 and s[p-1] != c2:
answer = 'Yes'
break
p = lastChange[p-1]
print(answer)
if __name__ == '__main__':
s = input()
cases = int(input())
if len(s) > 1:
char3 = []
next3 = [None]*len(s)
p1 = 0
c1 = s[0]
p2 = None
for i in range(1, len(s)):
if s[i] != c1:
p2 = i
c2 = s[i]
break
if p2 != None:
for i in range(p2, len(s)):
if s[i] == c1:
p1 = i
elif s[i] == c2:
p2 = i
else:
if p1 > p2:
p1, c1, p2, c2 = p2, c2, p1, c1
char3.append((p1, i))
p1, c1, p2, c2 = p2, c2, i, s[i]
#print(char3)
if char3 != []:
i = 0
for j in range(len(s)):
while i < len(char3) and j > char3[i][0]:
i += 1
if i < len(char3):
next3[j] = i
#print(next3)
for i in range(cases):
l, r = map(int, input().split())
l -= 1
if r-l == 1:
print('Yes')
continue
if s[l] != s[r-1]:
print('Yes')
continue
if next3[l] != None and char3[next3[l]][1] <= r-1:
print('Yes')
continue
print("No")
``` | output | 1 | 40,534 | 0 | 81,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
s=input()
m=[]
for i in range(len(s)+1):
m.append([])
for j in range(26):
m[-1].append(0)
for i in range(len(s)):
for j in range(26):
if j==ord(s[i])-97:
m[i+1][j]=m[i][j]+1
else:
m[i+1][j]=m[i][j]
q=int(input())
for _ in range(q):
l,r=map(int,input().split())
if l==r:
print('Yes')
continue
count=0
for i in range(26):
if m[l-1][i]!=m[r][i]:
count+=1
if count==1:
print('No')
else:
if count>=3:
print('Yes')
else:
if s[l-1]==s[r-1]:
print('No')
else:
print('Yes')
``` | instruction | 0 | 40,535 | 0 | 81,070 |
Yes | output | 1 | 40,535 | 0 | 81,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 2/3/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(s, queries):
N = len(s)
charcount = [[0 for _ in range(N+1)] for _ in range(26)]
for i, v in enumerate(s):
for ch in range(26):
charcount[ch][i+1] = charcount[ch][i]
ch = ord(v) - ord('a')
charcount[ch][i+1] += 1
# for i in range(26):
# print(charcount[i])
ans = []
for l, r in queries:
if l == r:
ans.append(True)
elif s[l] != s[r]:
ans.append(True)
else:
c = 0
for ch in range(26):
c += 1 if charcount[ch][r+1] - charcount[ch][l] > 0 else 0
ans.append(True if c > 2 else False)
print('\n'.join(['Yes' if v else 'No' for v in ans]))
s = input()
nq = int(input())
queries = []
for i in range(nq):
l, r = map(int, input().split())
queries.append((l-1, r-1))
solve(s, queries)
``` | instruction | 0 | 40,536 | 0 | 81,072 |
Yes | output | 1 | 40,536 | 0 | 81,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
s = input()
q = int(input())
N = len(s)
prefix_sum = []
for i in range(N + 1):
prefix_sum.append([0] * 26)
for i, ch in enumerate(s, start=1):
tmp = ord(ch)-ord('a')
for j in range(26):
if tmp == j:
prefix_sum[i][j] = prefix_sum[i-1][j]+1
else:
prefix_sum[i][j] = prefix_sum[i-1][j]
for case in range(q):
l, r = list(map(int, input().split()))
if l == r:
print('Yes')
continue
if s[l-1] != s[r-1]:
print('Yes')
else:
sum_ch = 0
for i in range(26):
if prefix_sum[r][i]-prefix_sum[l-1][i] > 0:
sum_ch += 1
print('Yes' if sum_ch > 2 else 'No')
``` | instruction | 0 | 40,537 | 0 | 81,074 |
Yes | output | 1 | 40,537 | 0 | 81,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import io
import os
from collections import Counter, defaultdict, deque
alpha = "abcdefghijklmnopqrstuvwxyz"
assert len(alpha) == 26
def solve(S, Q, queries):
lettersToPref = [[0] for i in range(26)]
for i, x in enumerate(S):
for i, c in enumerate(alpha):
if c == x:
lettersToPref[i].append(lettersToPref[i][-1] + 1)
else:
lettersToPref[i].append(lettersToPref[i][-1] + 0)
ans = []
for l, r in queries:
r += 1
if r - l == 1:
# Single character always work
ans.append("Yes")
continue
numUnique = 0
for i in range(26):
count = lettersToPref[i][r] - lettersToPref[i][l]
# assert count == sum(1 for c in S[l:r] if ord(c) - ord('a') == i)
if count:
numUnique += 1
assert r - l > 1
if numUnique == 1:
ans.append("No")
continue
assert numUnique >= 2
if S[l] != S[r - 1]:
# Can always put last character all in front, won't reach same count for any prefix until end
ans.append("Yes")
continue
assert S[l] == S[r - 1]
if numUnique >= 3:
ans.append("Yes")
continue
ans.append("No")
return "\n".join(ans)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
S = input().decode().rstrip()
(Q,) = [int(x) for x in input().split()]
queries = [
[int(x) - 1 for x in input().split()] for i in range(Q)
] # zero indexed, inclusive both l,r
ans = solve(S, Q, queries)
print(ans)
``` | instruction | 0 | 40,538 | 0 | 81,076 |
Yes | output | 1 | 40,538 | 0 | 81,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
sl = list(input())
al = []
a = 0
before = ""
for s in sl:
if before != s:
a += 1
before = s
al.append(a)
q = int(input())
for _ in range(q):
l,r = map(int,input().split())
if l == r:
print("Yes")
elif al[l-1] == al[r-1]:
print("No")
else:
print("Yes")
``` | instruction | 0 | 40,539 | 0 | 81,078 |
No | output | 1 | 40,539 | 0 | 81,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import string
def solve():
s = input()
n = int(input())
alphabets = len(string.ascii_lowercase)
count = [[0] * alphabets for _ in range(len(s) + 1)]
for i, c in enumerate(s, 1):
count[i][ord(c) - ord('a')] += 1
for c in range(alphabets):
count[i][c] += count[i - 1][c]
print(s)
for c in count:
print(c)
def query(l, r):
if l == r:
return 'Yes'
if s[l - 1] != s[r - 1]:
return 'Yes'
unique = 0
for x, y in zip(count[l - 1], count[r]):
if y - x > 0:
unique += 1
if unique > 2:
return 'Yes'
return 'No'
for _ in range(n):
print(query(*map(int, input().split())))
solve()
``` | instruction | 0 | 40,540 | 0 | 81,080 |
No | output | 1 | 40,540 | 0 | 81,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
sl = list(input())
from collections import defaultdict
dic = defaultdict(list)
for idx, s in enumerate(dic):
dic[s].append(idx)
import bisect
q = int(input())
alpha = set(sl)
for _ in range(q):
l,r = map(int,input().split())
if sl[l-1] != sl[r-1]:
print("Yes")
elif l == r:
print("Yes")
else:
cnt = 0
for a in alpha:
i1 = bisect.bisect_left(dic[a], l-1)
i2 = bisect.bisect_right(dic[a], r-1)
if i1 != i2:
cnt += 1
if cnt <= 2:
print("No")
else:
print("Yes")
``` | instruction | 0 | 40,541 | 0 | 81,082 |
No | output | 1 | 40,541 | 0 | 81,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.
Let's consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k ≥ 2 and 2k non-empty strings s_1, t_1, s_2, t_2, ..., s_k, t_k that satisfy the following conditions:
1. If we write the strings s_1, s_2, ..., s_k in order, the resulting string will be equal to s;
2. If we write the strings t_1, t_2, ..., t_k in order, the resulting string will be equal to t;
3. For all integers i between 1 and k inclusive, s_i and t_i are anagrams of each other.
If such strings don't exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.
For example, consider the string s = "gamegame". Then the string t = "megamage" is a reducible anagram of s, we may choose for example s_1 = "game", s_2 = "gam", s_3 = "e" and t_1 = "mega", t_2 = "mag", t_3 = "e":
<image>
On the other hand, we can prove that t = "memegaga" is an irreducible anagram of s.
You will be given a string s and q queries, represented by two integers 1 ≤ l ≤ r ≤ |s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.
Input
The first line contains a string s, consisting of lowercase English characters (1 ≤ |s| ≤ 2 ⋅ 10^5).
The second line contains a single integer q (1 ≤ q ≤ 10^5) — the number of queries.
Each of the following q lines contain two integers l and r (1 ≤ l ≤ r ≤ |s|), representing a query for the substring of s formed by characters from the l-th to the r-th.
Output
For each query, print a single line containing "Yes" (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing "No" (without quotes) otherwise.
Examples
Input
aaaaa
3
1 1
2 4
5 5
Output
Yes
No
Yes
Input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
Output
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is "a", which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain "a". On the other hand, in the second query, the substring is "aaa", which has no irreducible anagrams: its only anagram is itself, and we may choose s_1 = "a", s_2 = "aa", t_1 = "a", t_2 = "aa" to show that it is a reducible anagram.
In the second query of the second sample, the substring is "abb", which has, for example, "bba" as an irreducible anagram.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from decimal import *
getcontext().prec = 25
MOD = pow(10, 9) + 7
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")
# n, k = map(int, input().split(" "))
# l = list(map(int, input().split(" ")))
a = input()
alp = [[0] * 26 for i in range(len(a)+1)]
alp[0][ord(a[0]) - 97] += 1
for i in range(1, len(a)):
alp[i] = alp[i - 1].copy()
alp[i][ord(a[i]) - 97] += 1
alp[-1] = alp[-2].copy()
for i in range(int(input())):
l, r = map(int, input().split())
if l - r == 0:
print("YES")
elif r - l == 1:
print('NO')
elif a[r - 1] != a[l - 1]:
print('YES')
else:
c = 0
for i in range(26):
s = alp[r][i]
s2 = alp[l - 1][i]
if chr(i + 97) == a[l - 1]:
s2 -= 1
if s - s2:
c += 1
if c > 1:
print('YES')
else:
print('NO')
``` | instruction | 0 | 40,542 | 0 | 81,084 |
No | output | 1 | 40,542 | 0 | 81,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, WJMZBMR learned how to answer the query "how many times does a string x occur in a string s" quickly by preprocessing the string s. But now he wants to make it harder.
So he wants to ask "how many consecutive substrings of s are cyclical isomorphic to a given string x". You are given string s and n strings xi, for each string xi find, how many consecutive substrings of s are cyclical isomorphic to xi.
Two strings are called cyclical isomorphic if one can rotate one string to get the other one. 'Rotate' here means 'to take some consecutive chars (maybe none) from the beginning of a string and put them back at the end of the string in the same order'. For example, string "abcde" can be rotated to string "deabc". We can take characters "abc" from the beginning and put them at the end of "de".
Input
The first line contains a non-empty string s. The length of string s is not greater than 106 characters.
The second line contains an integer n (1 ≤ n ≤ 105) — the number of queries. Then n lines follow: the i-th line contains the string xi — the string for the i-th query. The total length of xi is less than or equal to 106 characters.
In this problem, strings only consist of lowercase English letters.
Output
For each query xi print a single integer that shows how many consecutive substrings of s are cyclical isomorphic to xi. Print the answers to the queries in the order they are given in the input.
Examples
Input
baabaabaaa
5
a
ba
baa
aabaa
aaba
Output
7
5
7
3
5
Input
aabbaa
3
aa
aabb
abba
Output
2
3
3
Submitted Solution:
```
from collections import deque,OrderedDict
s = input()
d = {}
for i in range(len(s)):
for j in range(i,len(s)):
d[s[i:j+1]] = d.get(s[i:j+1],0) + 1
n = int(input())
print(d)
for i in range(n):
count = 0
t = {}
s = deque(input())
#count += d.get(''.join(s),0)
t[''.join(s)] = ''.join(s)
for i in range(len(s)):
s.rotate(1)
#count += d.get(''.join(s),0)
t[''.join(s)] = ''.join(s)
for i in t:
count += d.get(i, 0)
print(count)
``` | instruction | 0 | 40,706 | 0 | 81,412 |
No | output | 1 | 40,706 | 0 | 81,413 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Some days ago, WJMZBMR learned how to answer the query "how many times does a string x occur in a string s" quickly by preprocessing the string s. But now he wants to make it harder.
So he wants to ask "how many consecutive substrings of s are cyclical isomorphic to a given string x". You are given string s and n strings xi, for each string xi find, how many consecutive substrings of s are cyclical isomorphic to xi.
Two strings are called cyclical isomorphic if one can rotate one string to get the other one. 'Rotate' here means 'to take some consecutive chars (maybe none) from the beginning of a string and put them back at the end of the string in the same order'. For example, string "abcde" can be rotated to string "deabc". We can take characters "abc" from the beginning and put them at the end of "de".
Input
The first line contains a non-empty string s. The length of string s is not greater than 106 characters.
The second line contains an integer n (1 ≤ n ≤ 105) — the number of queries. Then n lines follow: the i-th line contains the string xi — the string for the i-th query. The total length of xi is less than or equal to 106 characters.
In this problem, strings only consist of lowercase English letters.
Output
For each query xi print a single integer that shows how many consecutive substrings of s are cyclical isomorphic to xi. Print the answers to the queries in the order they are given in the input.
Examples
Input
baabaabaaa
5
a
ba
baa
aabaa
aaba
Output
7
5
7
3
5
Input
aabbaa
3
aa
aabb
abba
Output
2
3
3
Submitted Solution:
```
def perms(lst):
if len(lst)==0:
return []
elif len(lst)==1:
return [lst]
else:
l=[]
for i in range(len(lst)):
x=lst[i]
xs=lst[:i]+lst[i+1:]
for p in perms(xs):
l.append([x]+p)
return l
z=input()
n=int(input())
for i in range(n):
a=input()
y=len(a)
d=0
m=list(a)
k=perms(m)
zl=[]
if y==1:
zl=list(z)
d=zl.count(a)
else:
for j in range(len(z)-y):
zk=list(z[j:j+y])
zl.append(zk)
for stri in zl:
for p in k:
if p==stri:
d+=1
print(d)
``` | instruction | 0 | 40,707 | 0 | 81,414 |
No | output | 1 | 40,707 | 0 | 81,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,779 | 0 | 81,558 |
Tags: greedy
Correct Solution:
```
s=input()
t=input()
n=len(s)
l=[]
dist=0
for i in range(n):
if s[i]!=t[i]:
dist+=1
if dist%2:
print("impossible")
else:
flag=0
for i in range(n):
if s[i]==t[i]:
l.append(s[i])
else:
if flag==0:
l.append(s[i])
flag=1
else:
l.append(t[i])
flag=0
print(''.join(l))
``` | output | 1 | 40,779 | 0 | 81,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,780 | 0 | 81,560 |
Tags: greedy
Correct Solution:
```
n=list(input())
k=list(input())
j=0
for i in range(len(n)):
if n[i]!=k[i]:
j=j+1
if j%2==0:
n[i]=k[i]
if j%2==0:
print(*n,sep='')
else:
print("impossible")
``` | output | 1 | 40,780 | 0 | 81,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,781 | 0 | 81,562 |
Tags: greedy
Correct Solution:
```
s, t = input(), input()
a, w1, w2 = 0, 0, 0
l = len(s)
q = [0 for i in range(l)]
for i in range(0, l):
if s[i] != t[i]:
a += 1
if a%2 == 1:
print("impossible")
else:
for i in range(0, l):
if s[i] != t[i] and w1 < a//2:
q[i] = s[i]
w1 += 1
elif s[i] != t[i] and w2 < a//2:
q[i] = t[i]
w2 += 1
elif s[i] == t[i]:
q[i] = s[i]
print(*q, sep="")
``` | output | 1 | 40,781 | 0 | 81,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,782 | 0 | 81,564 |
Tags: greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
def main():
s=input()
t=input()
n=len(s)
ans=[]
reqs=0
reqt=0
chk=0
for x in range(n):
if s[x]!=t[x]:
if chk==0:
ans.append(str(int(s[x])^1))
reqs+=1
chk=1
else:
ans.append(str(int(t[x])^1))
reqt+=1
chk=0
else:
ans.append(str(int(t[x])^1))
if reqs==reqt:
print("".join(ans))
else:
print("impossible")
#-----------------------------BOSS-------------------------------------!
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | output | 1 | 40,782 | 0 | 81,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,783 | 0 | 81,566 |
Tags: greedy
Correct Solution:
```
a=list(map(int,input()))
b=list(map(int,input()))
n=0
x=[]
ans=a.copy()
for i in range(len(a)):
if a[i]!=b[i]:
n+=1
x.append(i)
if n%2!=0:
print("impossible")
else:
for i in range(n//2):
ans[x[i]]=b[x[i]]
print("".join(map(str,ans)))
``` | output | 1 | 40,783 | 0 | 81,567 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,784 | 0 | 81,568 |
Tags: greedy
Correct Solution:
```
s = list(input())
t = list(input())
n = len(s)
res = []
turn = 0
for i in range (n):
if s[i] == t[i]:
res.append(s[i])
else:
turn += 1
if turn % 2 == 0:
res.append(s[i])
else:
res.append(t[i])
if turn % 2 != 0:
print('impossible')
else:
print(''.join(res))
``` | output | 1 | 40,784 | 0 | 81,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,785 | 0 | 81,570 |
Tags: greedy
Correct Solution:
```
g=input()
h=input()
c=0
for i in range(len(g)):
if int(g[i])^int(h[i])==1:
c+=1
if c%2==0:
for i in range(len(g)):
if int(g[i])^int(h[i])==1:
if c%2==0:
print(g[i],end='')
else:
print(h[i],end='')
c-=1
else:
print(g[i],end='')
print()
else:
print("impossible")
``` | output | 1 | 40,785 | 0 | 81,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101. | instruction | 0 | 40,786 | 0 | 81,572 |
Tags: greedy
Correct Solution:
```
class CodeforcesTask545BSolution:
def __init__(self):
self.result = ''
self.s = ''
self.t = ''
def read_input(self):
self.s = input()
self.t = input()
def process_task(self):
n = len(self.s)
res = [0] * n
cnt = 0
b = False
for x in range(n):
if self.s[x] != self.t[x]:
if b:
res[x] = self.s[x]
else:
res[x] = self.t[x]
b = not b
cnt += 1
else:
res[x] = self.s[x]
if not cnt % 2:
self.result = "".join([str(x) for x in res])
else:
self.result = "impossible"
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask545BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 40,786 | 0 | 81,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
s1 = input()
s2 = input()
s = 0
x = list(s1)
for i in range(len(s1)):
if s1[i] != s2[i]:
s += 1
if s % 2 == 1:
print('impossible')
else:
s = s / 2
i = 0
while s:
if s1[i] != s2[i]:
x[i] = s2[i]
s -= 1
i += 1
print(''.join(x))
``` | instruction | 0 | 40,787 | 0 | 81,574 |
Yes | output | 1 | 40,787 | 0 | 81,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
s = input().strip()
t = input().strip()
dif1 = 0
dif2 = 0
ans = ''
for p in map(lambda x,y: (x,y), s,t):
if p[0] != p[1]:
if dif1 > dif2:
ans += p[0]
dif2+=1
else:
ans += p[1]
dif1+=1
else: ans += p[0]
if dif1 == dif2: print(ans)
else: print("impossible")
``` | instruction | 0 | 40,788 | 0 | 81,576 |
Yes | output | 1 | 40,788 | 0 | 81,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
# Description of the problem can be found at http://codeforces.com/problemset/problem/545/B
s = input()
t = input()
l_d = list()
c = 0
for i in range(len(s)):
if s[i] != t[i] and c % 2 == 0:
l_d.append("0" if "0" == s[i] else "1")
c += 1
elif s[i] != t[i]:
c += 1
l_d.append("1" if "0" == s[i] else "0")
else:
l_d.append("1")
if c % 2 == 1:
print("impossible")
else:
print("".join(l_d))
``` | instruction | 0 | 40,789 | 0 | 81,578 |
Yes | output | 1 | 40,789 | 0 | 81,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
s=input()
t=input()
n=len(s)
c=0
l=[]
for i in range(n):
if(s[i]==t[i]):
l.append(s[i])
else:
c=c+1
if(c%2==1):
l.append(s[i])
else:
l.append(t[i])
if(c%2==1):
print("impossible")
else:
print(''.join(l))
``` | instruction | 0 | 40,790 | 0 | 81,580 |
Yes | output | 1 | 40,790 | 0 | 81,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
s = input()
t = input()
n = len(s)
d1 = d2 = 0
for x, y in zip(s, t):
if x != y:
d2 += 1
s = list(s)
ans = 'impossible'
for i in range(n):
if s[i] == t[i]:
continue
if d1 != d2:
s[i] = '0' if s[i] == '1' else '1'
d1 += 1
if t[i] == s[i]:
d2 -= 1
else:
ans = ''.join(s)
break
print(ans)
``` | instruction | 0 | 40,791 | 0 | 81,582 |
No | output | 1 | 40,791 | 0 | 81,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
#from bisect import bisect_left as bl #c++ lowerbound bl(array,element)
#from bisect import bisect_right as br #c++ upperbound br(array,element)
def main():
s=input()
t=input()
n=len(s)
if n%2!=0:
print("impossible")
else:
ans=[]
reqs=0
reqt=0
chk=0
for x in range(n):
if s[x]!=t[x]:
if chk==0:
ans.append(str(int(s[x])^1))
reqs+=1
chk=1
else:
ans.append(str(int(t[x])^1))
reqt+=1
chk=0
else:
ans.append(str(int(t[x])^1))
if reqs==reqt:
print("".join(ans))
else:
print("impossible")
#-----------------------------BOSS-------------------------------------!
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
main()
``` | instruction | 0 | 40,792 | 0 | 81,584 |
No | output | 1 | 40,792 | 0 | 81,585 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
a,b=input(),input()
d=0
s=''
j=0
if a==b:
print(a)
else:
for i in range(len(a)):
if a[i]!=b[i]:
if j%2==0:
s+='1'
j+=1
else:
s+='0'
j+=1
else:
s+='0'
if j%2==0:
print(s)
else:
print('impossible')
``` | instruction | 0 | 40,793 | 0 | 81,586 |
No | output | 1 | 40,793 | 0 | 81,587 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie loves strings. Today she calculates distances between them. As Susie is a small girl after all, her strings contain only digits zero and one. She uses the definition of Hamming distance:
We will define the distance between two strings s and t of the same length consisting of digits zero and one as the number of positions i, such that si isn't equal to ti.
As besides everything else Susie loves symmetry, she wants to find for two strings s and t of length n such string p of length n, that the distance from p to s was equal to the distance from p to t.
It's time for Susie to go to bed, help her find such string p or state that it is impossible.
Input
The first line contains string s of length n.
The second line contains string t of length n.
The length of string n is within range from 1 to 105. It is guaranteed that both strings contain only digits zero and one.
Output
Print a string of length n, consisting of digits zero and one, that meets the problem statement. If no such string exist, print on a single line "impossible" (without the quotes).
If there are multiple possible answers, print any of them.
Examples
Input
0001
1011
Output
0011
Input
000
111
Output
impossible
Note
In the first sample different answers are possible, namely — 0010, 0011, 0110, 0111, 1000, 1001, 1100, 1101.
Submitted Solution:
```
s = input()
t = input()
D = {}
for i in range(len(s)):
if s[i] != t[i]:
D[i] = 1
'''print(D)
for i in range(len(s)):
if i in D:
print(i)'''
if len(D) %2 == 1:
print('impossible')
else :
for i in range(len(s)):
if i not in D:
print(s[i],end = '')
else :
if i + 1<= len(D)//2:
print(s[i],end = '')
else :
print(t[i],end = '')
``` | instruction | 0 | 40,794 | 0 | 81,588 |
No | output | 1 | 40,794 | 0 | 81,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1. | instruction | 0 | 40,896 | 0 | 81,792 |
Tags: games, greedy
Correct Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a <= b)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a <= b) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a <= b)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a <= b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | output | 1 | 40,896 | 0 | 81,793 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1. | instruction | 0 | 40,897 | 0 | 81,794 |
Tags: games, greedy
Correct Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a <= b)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a <= b) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a <= b)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a <= b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
# Made By Mostafa_Khaled
``` | output | 1 | 40,897 | 0 | 81,795 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1. | instruction | 0 | 40,898 | 0 | 81,796 |
Tags: games, greedy
Correct Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
#In abab segment
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a <= b)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a <= b) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a <= b)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a <= b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | output | 1 | 40,898 | 0 | 81,797 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
Submitted Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a == b)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a == b) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a == b)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a + max(a - b, 0)
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a == b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | instruction | 0 | 40,899 | 0 | 81,798 |
No | output | 1 | 40,899 | 0 | 81,799 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
Submitted Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a <= b)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a <= b) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a <= b)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a + max(a - b, 0)
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a <= b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | instruction | 0 | 40,900 | 0 | 81,800 |
No | output | 1 | 40,900 | 0 | 81,801 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
Submitted Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
#abababab
newL = qL * (2 * a + 2 * b) + 1
newR = (qR + 1) * (2 * a + 2 * b)
if 1 <= rL <= a + b and a + b + 1 <= rR:
return a + max(a - b, 0) + int(a == b or a == 1)
if a + b + 1 <= rL <= 2 * (a + b) and (2 * a + 2 * b) + 1 <= rR <= a + b:
return main(a, b, l - (a + b), r - (a + b))
if 1 <= rL <= a and 1 <= rR <= a:
return a + max(a - b, 0) + int(a == b or a == 1) + rR - max(rR - rL + 1, 0)
if 1 <= rL <= a and a + 1 <= rR <= a + b:
return a + max(a - b, 0) + int(a == b or a == 1)
if a + 1 <= rL <= a + b and 1 <= rR <= a:
return 1 + a + max(a - b, 0)
if a + 1 <= rL <= a + b and a + 1 <= rR <= a + b:
return 1 + a + max(a - b, 0)
return main(a, b, l - (a + b), r - (a + b))
else:
return a + max(a - b, 0) + int(a == b or a == 1) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | instruction | 0 | 40,901 | 0 | 81,802 |
No | output | 1 | 40,901 | 0 | 81,803 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sometimes Mister B has free evenings when he doesn't know what to do. Fortunately, Mister B found a new game, where the player can play against aliens.
All characters in this game are lowercase English letters. There are two players: Mister B and his competitor.
Initially the players have a string s consisting of the first a English letters in alphabetical order (for example, if a = 5, then s equals to "abcde").
The players take turns appending letters to string s. Mister B moves first.
Mister B must append exactly b letters on each his move. He can arbitrary choose these letters. His opponent adds exactly a letters on each move.
Mister B quickly understood that his opponent was just a computer that used a simple algorithm. The computer on each turn considers the suffix of string s of length a and generates a string t of length a such that all letters in the string t are distinct and don't appear in the considered suffix. From multiple variants of t lexicographically minimal is chosen (if a = 4 and the suffix is "bfdd", the computer chooses string t equal to "aceg"). After that the chosen string t is appended to the end of s.
Mister B soon found the game boring and came up with the following question: what can be the minimum possible number of different letters in string s on the segment between positions l and r, inclusive. Letters of string s are numerated starting from 1.
Input
First and only line contains four space-separated integers: a, b, l and r (1 ≤ a, b ≤ 12, 1 ≤ l ≤ r ≤ 109) — the numbers of letters each player appends and the bounds of the segment.
Output
Print one integer — the minimum possible number of different letters in the segment from position l to position r, inclusive, in string s.
Examples
Input
1 1 1 8
Output
2
Input
4 2 2 6
Output
3
Input
3 7 4 6
Output
1
Note
In the first sample test one of optimal strategies generate string s = "abababab...", that's why answer is 2.
In the second sample test string s = "abcdbcaefg..." can be obtained, chosen segment will look like "bcdbc", that's why answer is 3.
In the third sample test string s = "abczzzacad..." can be obtained, chosen, segment will look like "zzz", that's why answer is 1.
Submitted Solution:
```
def main(a, b, l, r):
qL = (l - 1) // (2 * a + 2 * b)
rL = (l - 1) % (2 * a + 2 * b) + 1
qR = (r - 1) // (2 * a + 2 * b)
rR = (r - 1) % (2 * a + 2 * b) + 1
#print(qL, qR, rL, rR)
if qL == qR:
#In b segment
if a < rL <= a + b and a < rR <= a + b:
return 1
if 2 * a + b < rL and 2 * a + b < rR:
return 1
#In a segment
if 1 <= rL <= a and 1 <= rR <= a:
return rR - rL + 1
if a + b < rL <= 2 * a + b and a + b < rR <= 2 * a + b:
return rR - rL + 1
#In a + b segment
if 1 <= rL <= a + b and 1 <= rR <= a + b:
return a - rL + 1
if a + b < rL and a + b < rR:
return (2 * a + b) - rL + 1
if a < rL <= a + b and a + b < rR <= 2 * a + b:
return 1 + rR - (a + b)
if a < rL <= a + b and 2 * a + b < rR:
return 1 + a
if 1 <= rL <= a and a + b < rR <= 2 * a + b:
ans = a - rL + 1 + max(rR - (a + b + b), 0) + min(b, rR) - max(min(rR, b) - rL + 1, 0)
return ans
if 1 <= rL <= a and 2 * a + b < rR:
return a - rL + 1 + a - max(b - rL + 1, 0)
elif qL == qR - 1:
return main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
else:
return a + max(a - b, 0) + int(a == b) # + main(a, b, l, (qL + 1) * (2 * a + 2 * b)) + main(a, b, qR * (2 * a + 2 * b) + 1, r)
a, b, l, r = [int(item) for item in input().split()]
print(main(a, b, l, r))
``` | instruction | 0 | 40,902 | 0 | 81,804 |
No | output | 1 | 40,902 | 0 | 81,805 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G： 検閲により置換 (Censored String)
Problem Statement
You are given the string S, and the set of strings \mathcal P whose size is N. Now we consider applying the following operation to S:
* Choose exactly one character in S, and replace it with '*'.
Let s_i be a i-th character of S, S_{ij} be a consecutive substring in S such that S = s_i s_{i+1} $\cdots$ s_j, and p_k as the k-th string of P. Your task is to calculate the minimum number of operations for satisfying the following condition.
* For all pairs (S_{ij}, p_k), it is satisfied that S_{ij} \neq p_k.
Input
An input is given in the following format.
S
N
p_1
p_2
$\vdots$
p_N
* In line 1, you are given the string S.
* In line 2, given an integer N. N means the size of P.
* From line 3 to the end, given p_i, the i-th string of P.
Constraints
* 1 \leq |S| \leq 10^5
* 1 \leq N \leq 10^5
* 1 \leq |p_1| + |p_2| + $\cdots$ + |p_N| \leq 10^5
* If i \neq j, it is guaranteed that p_i \neq p_j
* S, and p_i include lowercase letters only
Output
Print the minimum number of operations in one line.
Sample Input 1
brainfuck
1
fuck
Sample Output 1
1
For example, if you change "brainfuck" into "brainf*ck", you can satisfy the condition. You have to perform operation at least once, and it is the minimum value.
Sample Input 2
aaa
1
a
Sample Output 2
3
You have to operate for all characters.
Sample Input 3
hellshakeyano
3
hell
shake
key
Sample Output 3
2
In this case, for example, if you change "hellshakeyano" into "h*llsha*eyano", you can satisfy the condition. Notice if you replace one character, it might happen that several strings in the set P, which appear as the consecutive substring of S, simultaneously disappear.
Example
Input
brainfuck
1
fuck
Output
1
Submitted Solution:
```
S = input()
n = int(input())
P = [input() for i in range(n)]
ans = 0
for p in P:
i = 0
while i < len(S):
for q in p:
if S[i] != q:
break
i += 1
if i > len(S):
break
else:
S = S[:i-1] + '*' + S[i:]
ans += 1
continue
i += 1
print(ans)
``` | instruction | 0 | 41,166 | 0 | 82,332 |
No | output | 1 | 41,166 | 0 | 82,333 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G： 検閲により置換 (Censored String)
Problem Statement
You are given the string S, and the set of strings \mathcal P whose size is N. Now we consider applying the following operation to S:
* Choose exactly one character in S, and replace it with '*'.
Let s_i be a i-th character of S, S_{ij} be a consecutive substring in S such that S = s_i s_{i+1} $\cdots$ s_j, and p_k as the k-th string of P. Your task is to calculate the minimum number of operations for satisfying the following condition.
* For all pairs (S_{ij}, p_k), it is satisfied that S_{ij} \neq p_k.
Input
An input is given in the following format.
S
N
p_1
p_2
$\vdots$
p_N
* In line 1, you are given the string S.
* In line 2, given an integer N. N means the size of P.
* From line 3 to the end, given p_i, the i-th string of P.
Constraints
* 1 \leq |S| \leq 10^5
* 1 \leq N \leq 10^5
* 1 \leq |p_1| + |p_2| + $\cdots$ + |p_N| \leq 10^5
* If i \neq j, it is guaranteed that p_i \neq p_j
* S, and p_i include lowercase letters only
Output
Print the minimum number of operations in one line.
Sample Input 1
brainfuck
1
fuck
Sample Output 1
1
For example, if you change "brainfuck" into "brainf*ck", you can satisfy the condition. You have to perform operation at least once, and it is the minimum value.
Sample Input 2
aaa
1
a
Sample Output 2
3
You have to operate for all characters.
Sample Input 3
hellshakeyano
3
hell
shake
key
Sample Output 3
2
In this case, for example, if you change "hellshakeyano" into "h*llsha*eyano", you can satisfy the condition. Notice if you replace one character, it might happen that several strings in the set P, which appear as the consecutive substring of S, simultaneously disappear.
Example
Input
brainfuck
1
fuck
Output
1
Submitted Solution:
```
S = input()
n = int(input())
P = [input() for i in range(n)]
ans = 0
for p in P:
i = 0
while i < len(S):
for q in p:
if S[i] != q:
break
i += 1
else:
S = S[:i-1] + '*' + S[i:]
ans += 1
continue
i += 1
print(ans)
``` | instruction | 0 | 41,167 | 0 | 82,334 |
No | output | 1 | 41,167 | 0 | 82,335 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G： 検閲により置換 (Censored String)
Problem Statement
You are given the string S, and the set of strings \mathcal P whose size is N. Now we consider applying the following operation to S:
* Choose exactly one character in S, and replace it with '*'.
Let s_i be a i-th character of S, S_{ij} be a consecutive substring in S such that S = s_i s_{i+1} $\cdots$ s_j, and p_k as the k-th string of P. Your task is to calculate the minimum number of operations for satisfying the following condition.
* For all pairs (S_{ij}, p_k), it is satisfied that S_{ij} \neq p_k.
Input
An input is given in the following format.
S
N
p_1
p_2
$\vdots$
p_N
* In line 1, you are given the string S.
* In line 2, given an integer N. N means the size of P.
* From line 3 to the end, given p_i, the i-th string of P.
Constraints
* 1 \leq |S| \leq 10^5
* 1 \leq N \leq 10^5
* 1 \leq |p_1| + |p_2| + $\cdots$ + |p_N| \leq 10^5
* If i \neq j, it is guaranteed that p_i \neq p_j
* S, and p_i include lowercase letters only
Output
Print the minimum number of operations in one line.
Sample Input 1
brainfuck
1
fuck
Sample Output 1
1
For example, if you change "brainfuck" into "brainf*ck", you can satisfy the condition. You have to perform operation at least once, and it is the minimum value.
Sample Input 2
aaa
1
a
Sample Output 2
3
You have to operate for all characters.
Sample Input 3
hellshakeyano
3
hell
shake
key
Sample Output 3
2
In this case, for example, if you change "hellshakeyano" into "h*llsha*eyano", you can satisfy the condition. Notice if you replace one character, it might happen that several strings in the set P, which appear as the consecutive substring of S, simultaneously disappear.
Example
Input
brainfuck
1
fuck
Output
1
Submitted Solution:
```
S = input()
n = int(input())
P = [input() for i in range(n)]
ans = 0
for p in P:
i = 0
while i < len(S):
for q in p:
if i >= len(S) or S[i] != q:
break
i += 1
else:
S = S[:i-1] + '*' + S[i:]
ans += 1
continue
i += 1
print(ans)
``` | instruction | 0 | 41,168 | 0 | 82,336 |
No | output | 1 | 41,168 | 0 | 82,337 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
G： 検閲により置換 (Censored String)
Problem Statement
You are given the string S, and the set of strings \mathcal P whose size is N. Now we consider applying the following operation to S:
* Choose exactly one character in S, and replace it with '*'.
Let s_i be a i-th character of S, S_{ij} be a consecutive substring in S such that S = s_i s_{i+1} $\cdots$ s_j, and p_k as the k-th string of P. Your task is to calculate the minimum number of operations for satisfying the following condition.
* For all pairs (S_{ij}, p_k), it is satisfied that S_{ij} \neq p_k.
Input
An input is given in the following format.
S
N
p_1
p_2
$\vdots$
p_N
* In line 1, you are given the string S.
* In line 2, given an integer N. N means the size of P.
* From line 3 to the end, given p_i, the i-th string of P.
Constraints
* 1 \leq |S| \leq 10^5
* 1 \leq N \leq 10^5
* 1 \leq |p_1| + |p_2| + $\cdots$ + |p_N| \leq 10^5
* If i \neq j, it is guaranteed that p_i \neq p_j
* S, and p_i include lowercase letters only
Output
Print the minimum number of operations in one line.
Sample Input 1
brainfuck
1
fuck
Sample Output 1
1
For example, if you change "brainfuck" into "brainf*ck", you can satisfy the condition. You have to perform operation at least once, and it is the minimum value.
Sample Input 2
aaa
1
a
Sample Output 2
3
You have to operate for all characters.
Sample Input 3
hellshakeyano
3
hell
shake
key
Sample Output 3
2
In this case, for example, if you change "hellshakeyano" into "h*llsha*eyano", you can satisfy the condition. Notice if you replace one character, it might happen that several strings in the set P, which appear as the consecutive substring of S, simultaneously disappear.
Example
Input
brainfuck
1
fuck
Output
1
Submitted Solution:
```
S = input()
n = int(input())
P = [input() for i in range(n)]
ans = 0
for p in P:
i = 0
while i < len(S):
for q in p:
if S[i] != q:
break
i += 1
if i >= len(S):
break
else:
S = S[:i-1] + '*' + S[i:]
ans += 1
continue
i += 1
print(ans)
``` | instruction | 0 | 41,169 | 0 | 82,338 |
No | output | 1 | 41,169 | 0 | 82,339 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,449 | 0 | 82,898 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
import sys
s = input().strip()
N = len(s)
if len(s) == 1:
print(1, s[0])
sys.exit()
X = [s[-1], s[-2]+s[-1] if s[-2]!=s[-1] else ""]
Y = [1, 2 if s[-2]!=s[-1] else 0]
for i in range(N-3, -1, -1):
c = s[i]
k1 = c+X[-1]
ng = Y[-1]+1
if ng > 10:
k1 = k1[:5] + "..." + k1[-2:]
if c == s[i+1] and k1 > X[-2]:
k1 = X[-2]
ng = Y[-2]
X.append(k1)
Y.append(ng)
for i in range(N-1, -1, -1):
print(Y[i], X[i])
``` | output | 1 | 41,449 | 0 | 82,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,450 | 0 | 82,900 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
import io
import os
from collections import deque
def solve(S):
N = len(S)
ans = []
suff = deque()
suffUnique = deque()
count = 0
for i in reversed(range(N)):
x = S[i]
if i == N - 1 or S[i] != S[i + 1]:
count = 1
else:
count += 1
if (suffUnique and x < suffUnique[0]) or (
len(suffUnique) >= 2 and x == suffUnique[0] and x < suffUnique[1]
):
# Want to keep all repeats of this character
suff.appendleft(x)
if not suffUnique or x != suffUnique[0]:
suffUnique.appendleft(x)
else:
# Want to keep as few repeats of this character as possible
if count % 2 == 1:
suff.appendleft(x)
if not suffUnique or x != suffUnique[0]:
suffUnique.appendleft(x)
else:
suff.popleft()
if suffUnique and (not suff or suffUnique[0] != suff[0]):
suffUnique.popleft()
if len(suff) <= 10:
ans.append(str(len(suff)) + " " + "".join(suff))
else:
line = [str(len(suff)), " "]
for i in range(5):
line.append(suff[i])
line.append("...")
line.append(suff[-2])
line.append(suff[-1])
ans.append("".join(line))
return "\n".join(ans[::-1])
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
S = input().decode().rstrip()
ans = solve(S)
print(ans)
``` | output | 1 | 41,450 | 0 | 82,901 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,451 | 0 | 82,902 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
def main(s):
if len(s) == 0:
return []
if len(s) == 1:
return [[s], [1]]
s = s[::-1]
sres = [s[0]]
nres = [1]
if s[0] == s[1]:
sres.append('')
nres.append(0)
else:
sres.append(s[1] + s[0])
nres.append(2)
for i in range(2, len(s)):
st = s[i] + sres[-1]
nt = nres[-1] + 1
if nt > 10:
st = st[:5] + '...' + st[-2:]
if s[i] == s[i-1] and st > sres[-1]:
st = sres[-2]
nt = nres[-2]
sres.append(st)
nres.append(nt)
return [sres, nres]
s = input()
sres, nres = main(s)
for i in range(len(s)-1, -1, -1):
print(nres[i], sres[i])
``` | output | 1 | 41,451 | 0 | 82,903 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,452 | 0 | 82,904 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
s = input()
n = len(s)
res = ['',s[-1]]
l = [1]
flag = True
last = ['']
for i in range(n-2,-1,-1):
tmp = res[-1]
if flag and s[i]==tmp[0] and s[i]>last[-1]:
res.append(res[-2])
l.append(l[-1]-1)
flag = False
if l[-1]>0 and res[-1][0]!=s[i]:
last.pop()
else:
l.append(l[-1]+1)
tmp = s[i]+tmp
flag = True
if len(tmp)>10:
res.append(tmp[:5]+'...'+tmp[-2:])
else:
res.append(tmp)
if len(res[-1])>1 and res[-1][0]!=res[-1][1]:
last.append(res[-1][1])
res.pop(0)
for i in range(n-1,-1,-1):
print(l[i],res[i])
``` | output | 1 | 41,452 | 0 | 82,905 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,453 | 0 | 82,906 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
s = input()
cs = ['']
ans = []
last = ['']
poped = False
for c in s[::-1]:
if not poped and c == cs[-1] and c > last[-2]:
cs.pop()
if (not cs) or c != cs[-1]:
last.pop()
poped = True
else:
poped = False
if c != cs[-1]:
last.append(c)
cs.append(c)
if len(cs) <= 11:
ans.append((len(cs) - 1, ''.join(cs[::-1])))
else:
ans.append((len(cs) - 1, (''.join(cs[:3]) + '...' + ''.join(cs[-5:]))[::-1]))
for a, b in ans[::-1]:
print(a, b)
``` | output | 1 | 41,453 | 0 | 82,907 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Some time ago Lesha found an entertaining string s consisting of lowercase English letters. Lesha immediately developed an unique algorithm for this string and shared it with you. The algorithm is as follows.
Lesha chooses an arbitrary (possibly zero) number of pairs on positions (i, i + 1) in such a way that the following conditions are satisfied:
* for each pair (i, i + 1) the inequality 0 ≤ i < |s| - 1 holds;
* for each pair (i, i + 1) the equality s_i = s_{i + 1} holds;
* there is no index that is contained in more than one pair.
After that Lesha removes all characters on indexes contained in these pairs and the algorithm is over.
Lesha is interested in the lexicographically smallest strings he can obtain by applying the algorithm to the suffixes of the given string.
Input
The only line contains the string s (1 ≤ |s| ≤ 10^5) — the initial string consisting of lowercase English letters only.
Output
In |s| lines print the lengths of the answers and the answers themselves, starting with the answer for the longest suffix. The output can be large, so, when some answer is longer than 10 characters, instead print the first 5 characters, then "...", then the last 2 characters of the answer.
Examples
Input
abcdd
Output
3 abc
2 bc
1 c
0
1 d
Input
abbcdddeaaffdfouurtytwoo
Output
18 abbcd...tw
17 bbcdd...tw
16 bcddd...tw
15 cddde...tw
14 dddea...tw
13 ddeaa...tw
12 deaad...tw
11 eaadf...tw
10 aadfortytw
9 adfortytw
8 dfortytw
9 fdfortytw
8 dfortytw
7 fortytw
6 ortytw
5 rtytw
6 urtytw
5 rtytw
4 tytw
3 ytw
2 tw
1 w
0
1 o
Note
Consider the first example.
* The longest suffix is the whole string "abcdd". Choosing one pair (4, 5), Lesha obtains "abc".
* The next longest suffix is "bcdd". Choosing one pair (3, 4), we obtain "bc".
* The next longest suffix is "cdd". Choosing one pair (2, 3), we obtain "c".
* The next longest suffix is "dd". Choosing one pair (1, 2), we obtain "" (an empty string).
* The last suffix is the string "d". No pair can be chosen, so the answer is "d".
In the second example, for the longest suffix "abbcdddeaaffdfouurtytwoo" choose three pairs (11, 12), (16, 17), (23, 24) and we obtain "abbcdddeaadfortytw" | instruction | 0 | 41,454 | 0 | 82,908 |
Tags: dp, greedy, implementation, strings
Correct Solution:
```
import sys
z=sys.stdin.readline
s=z().strip()
l=len(s)
b=[' ']
o=[]
m=[' ']
for i in range(l):
c=s[-1-i]
if c==b[-1]:
if c>m[-1].lower() or (c==m[-1].lower() and c>m[-2].lower()):
b.pop()
if b[-1].lower()!=m[-1].lower():m.pop()
b[-1]=b[-1].upper()
m[-1]=m[-1].upper()
else:b+=[c]
else:
b+=[c]
if b[-1]!=m[-1].lower():m+=[b[-1]]
if len(b)>11:o+=[(str(len(b)-1)+' '+''.join(b[:-6:-1])+'...'+b[2]+b[1]).lower()]
elif len(b)>1:o+=[(str(len(b)-1)+' '+''.join(b[-1::-1])).lower()]
else:o.append('0')
print('\n'.join(o[::-1]))
``` | output | 1 | 41,454 | 0 | 82,909 |
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