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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not. Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him! More formally, after replacing all characters '?', the condition s_i β‰  s_{i+1} should be satisfied for all 1 ≀ i ≀ |s| - 1, where |s| is the length of the string s. Input The first line contains positive integer t (1 ≀ t ≀ 1000) β€” the number of test cases. Next t lines contain the descriptions of test cases. Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'. It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5. Output For each test case given in the input print the answer in the following format: * If it is impossible to create a beautiful string, print "-1" (without quotes); * Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them. Example Input 3 a???cb a??bbc a?b?c Output ababcb -1 acbac Note In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful. In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal. In the third test case, the only answer is "acbac". Submitted Solution: ``` import random t = int(input()) for _ in range(t): txt = input() for i in range(len(txt)-1): chars = ['a', 'b', 'c'] if txt[i] != '?' and txt[i] == txt[i+1]: print(-1) break if txt[i] == '?': if i>0 and txt[i-1] != '?': chars.remove(txt[i-1]) if i<(len(txt)-1) and txt[i+1] != '?' and txt[i+1] in chars: chars.remove(txt[i+1]) k = random.randint(0, len(chars)-1) txt = txt[:i]+chars[k]+txt[i+1:] # txt[i] = chars[k] else: print(txt) ```
instruction
0
103,039
0
206,078
No
output
1
103,039
0
206,079
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not. Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him! More formally, after replacing all characters '?', the condition s_i β‰  s_{i+1} should be satisfied for all 1 ≀ i ≀ |s| - 1, where |s| is the length of the string s. Input The first line contains positive integer t (1 ≀ t ≀ 1000) β€” the number of test cases. Next t lines contain the descriptions of test cases. Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'. It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5. Output For each test case given in the input print the answer in the following format: * If it is impossible to create a beautiful string, print "-1" (without quotes); * Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them. Example Input 3 a???cb a??bbc a?b?c Output ababcb -1 acbac Note In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful. In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal. In the third test case, the only answer is "acbac". Submitted Solution: ``` def getChar(x, y): opts = ['a', 'b', 'c'] if x in opts: opts.remove(x) if y in opts: opts.remove(y) return opts[0] def findBS(s): length = len(s) opts = ['a', 'b', 'c'] if length == 1: if '?' in s: print(-1) else: print(s) return for i in range(len(opts)): if s[0] == '?' and s[1] != opts[i]: s[0] = opts[i] if s[length-1] == '?' and s[length-2] != opts[i]: s[length-1] = opts[i] for i in range(1, length-1): if s[i] == '?': s[i] = getChar(s[i-1], s[i+1]) if s[i] == s[i-1] or s[i] == s[i+1]: print('-1') return if '?' in opts: print(-1) print(''.join(s)) t = int(input()) for _ in range(t): s = list(input().rstrip()) findBS(s) ```
instruction
0
103,040
0
206,080
No
output
1
103,040
0
206,081
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. A string is called beautiful if no two consecutive characters are equal. For example, "ababcb", "a" and "abab" are beautiful strings, while "aaaaaa", "abaa" and "bb" are not. Ahcl wants to construct a beautiful string. He has a string s, consisting of only characters 'a', 'b', 'c' and '?'. Ahcl needs to replace each character '?' with one of the three characters 'a', 'b' or 'c', such that the resulting string is beautiful. Please help him! More formally, after replacing all characters '?', the condition s_i β‰  s_{i+1} should be satisfied for all 1 ≀ i ≀ |s| - 1, where |s| is the length of the string s. Input The first line contains positive integer t (1 ≀ t ≀ 1000) β€” the number of test cases. Next t lines contain the descriptions of test cases. Each line contains a non-empty string s consisting of only characters 'a', 'b', 'c' and '?'. It is guaranteed that in each test case a string s has at least one character '?'. The sum of lengths of strings s in all test cases does not exceed 10^5. Output For each test case given in the input print the answer in the following format: * If it is impossible to create a beautiful string, print "-1" (without quotes); * Otherwise, print the resulting beautiful string after replacing all '?' characters. If there are multiple answers, you can print any of them. Example Input 3 a???cb a??bbc a?b?c Output ababcb -1 acbac Note In the first test case, all possible correct answers are "ababcb", "abcacb", "abcbcb", "acabcb" and "acbacb". The two answers "abcbab" and "abaabc" are incorrect, because you can replace only '?' characters and the resulting string must be beautiful. In the second test case, it is impossible to create a beautiful string, because the 4-th and 5-th characters will be always equal. In the third test case, the only answer is "acbac". Submitted Solution: ``` def solve(p): if not p: return [] if len(p)==1: if p[0] == '?': return ['a'] else: return p s = set() s.add('a') s.add('b') s.add('c') for i in range(1,len(p)-1): if p[i] == '?': rem = list(s.difference(set([p[i-1], p[i+1], '?']))) if not rem: return -1 else: p[i] = rem[0] if p[0] == '?': p[0] = list(s.difference(set([p[1], '?'])))[0] if p[-1] == '?': p[-1] = list(s.difference(set([p[-2], '?'])))[0] for i in range(1, len(p)): if p[i]==p[i-1]: return -1 return p problems = [] for i in range(int(input())): problems += [list(input())] for p in problems: print(p) p = solve(p) if p!= -1: p = ''.join(p) print(p) ```
instruction
0
103,041
0
206,082
No
output
1
103,041
0
206,083
Provide a correct Python 3 solution for this coding contest problem. Problem Jennifer and Marian presented Carla with the string S. However, Carla is not happy to receive the string S. I wanted the string T. The three decided to work together to change the string S to the string T. Jennifer first sorts the letters in any order. Marian then exchanges the lowercase letters of the two alphabets any number of times. This operation exchanges all the same characters in the string, for example: * aab-> Swap a and b-> bba * aab-> Swap a and c-> ccb Finally, Carla replaces one character with another and repeats until T. Jennifer and Marian decided to try to reduce the number of Carla replacements. Find the minimum number of replacements that Carla makes. Constraints * 1 ≀ n ≀ 105 * S and T contain only'a'~'z' * | S | = | T | = n Input n S T The length n of the string is given on the first line. The character string S is given on the second line, and the character string T is given on the third line. Output Print the minimum number of Carla replacements on one line. Examples Input 3 abc xyz Output 0 Input 5 aaabb xyxyz Output 1
instruction
0
103,618
0
207,236
"Correct Solution: ``` # AOJ 1568: String Conversion # Python3 2018.7.13 bal4u cha, chzn = ord('a'), ord('z')+1 S, T = [0]*128, [0]*128 input() a = input() for x in a: S[ord(x)] += 1 S = sorted(S[cha:chzn], reverse=True) a = input() for x in a: T[ord(x)] += 1 T = sorted(T[cha:chzn], reverse=True) ans = 0 for i in range(26): ans += abs(S[i]-T[i]) print(ans >> 1) ```
output
1
103,618
0
207,237
Provide a correct Python 3 solution for this coding contest problem. Problem Jennifer and Marian presented Carla with the string S. However, Carla is not happy to receive the string S. I wanted the string T. The three decided to work together to change the string S to the string T. Jennifer first sorts the letters in any order. Marian then exchanges the lowercase letters of the two alphabets any number of times. This operation exchanges all the same characters in the string, for example: * aab-> Swap a and b-> bba * aab-> Swap a and c-> ccb Finally, Carla replaces one character with another and repeats until T. Jennifer and Marian decided to try to reduce the number of Carla replacements. Find the minimum number of replacements that Carla makes. Constraints * 1 ≀ n ≀ 105 * S and T contain only'a'~'z' * | S | = | T | = n Input n S T The length n of the string is given on the first line. The character string S is given on the second line, and the character string T is given on the third line. Output Print the minimum number of Carla replacements on one line. Examples Input 3 abc xyz Output 0 Input 5 aaabb xyxyz Output 1
instruction
0
103,619
0
207,238
"Correct Solution: ``` #!/usr/bin/env python3 # -*- coding: utf-8 -*- from collections import defaultdict N = int(input()) s1 = input() s2 = input() def make_appears(s): chars = list(map(ord, s)) hist = defaultdict(int) for ch in chars: hist[ch] += 1 appears = list(hist.values()) appears.sort() appears.reverse() return appears appears1 = make_appears(s1) appears2 = make_appears(s2) ans = 0 for i, j in zip(appears1, appears2): ans += abs(i - j) shorter = min(len(appears1), len(appears2)) ans += sum(appears1[shorter:]) + sum(appears2[shorter:]) print(ans//2) ```
output
1
103,619
0
207,239
Provide a correct Python 3 solution for this coding contest problem. Problem Jennifer and Marian presented Carla with the string S. However, Carla is not happy to receive the string S. I wanted the string T. The three decided to work together to change the string S to the string T. Jennifer first sorts the letters in any order. Marian then exchanges the lowercase letters of the two alphabets any number of times. This operation exchanges all the same characters in the string, for example: * aab-> Swap a and b-> bba * aab-> Swap a and c-> ccb Finally, Carla replaces one character with another and repeats until T. Jennifer and Marian decided to try to reduce the number of Carla replacements. Find the minimum number of replacements that Carla makes. Constraints * 1 ≀ n ≀ 105 * S and T contain only'a'~'z' * | S | = | T | = n Input n S T The length n of the string is given on the first line. The character string S is given on the second line, and the character string T is given on the third line. Output Print the minimum number of Carla replacements on one line. Examples Input 3 abc xyz Output 0 Input 5 aaabb xyxyz Output 1
instruction
0
103,620
0
207,240
"Correct Solution: ``` from collections import Counter n = int(input()) counter1 = Counter(input()) counter2 = Counter(input()) values1 = [0] * (26 - len(counter1)) + sorted(counter1.values()) values2 = [0] * (26 - len(counter2)) + sorted(counter2.values()) print(sum([abs(i - j) for i, j in zip(values1, values2)]) // 2) ```
output
1
103,620
0
207,241
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Problem Jennifer and Marian presented Carla with the string S. However, Carla is not happy to receive the string S. I wanted the string T. The three decided to work together to change the string S to the string T. Jennifer first sorts the letters in any order. Marian then exchanges the lowercase letters of the two alphabets any number of times. This operation exchanges all the same characters in the string, for example: * aab-> Swap a and b-> bba * aab-> Swap a and c-> ccb Finally, Carla replaces one character with another and repeats until T. Jennifer and Marian decided to try to reduce the number of Carla replacements. Find the minimum number of replacements that Carla makes. Constraints * 1 ≀ n ≀ 105 * S and T contain only'a'~'z' * | S | = | T | = n Input n S T The length n of the string is given on the first line. The character string S is given on the second line, and the character string T is given on the third line. Output Print the minimum number of Carla replacements on one line. Examples Input 3 abc xyz Output 0 Input 5 aaabb xyxyz Output 1 Submitted Solution: ``` #!/usr/bin/env python3 # -*- coding: utf-8 -*- from collections import defaultdict N = int(input()) s1 = input() s2 = input() chars1 = list(map(ord, s1)) chars2 = list(map(ord, s2)) hist1 = defaultdict(int) hist2 = defaultdict(int) for ch in chars1: hist1[ch] += 1 for ch in chars2: hist2[ch] += 1 appears1 = list(hist1.values()) appears2 = list(hist2.values()) ans = 0 for i, j in zip(appears1, appears2): ans += abs(i - j) longer = max(len(appears1), len(appears2)) ans += sum(appears1[longer:]) + sum(appears2[longer:]) print(ans) ```
instruction
0
103,622
0
207,244
No
output
1
103,622
0
207,245
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. You are given two integers l and r. Let's call an integer x modest, if l ≀ x ≀ r. Find a string of length n, consisting of digits, which has the largest possible number of substrings, which make a modest integer. Substring having leading zeros are not counted. If there are many answers, find lexicographically smallest one. If some number occurs multiple times as a substring, then in the counting of the number of modest substrings it is counted multiple times as well. Input The first line contains one integer l (1 ≀ l ≀ 10^{800}). The second line contains one integer r (l ≀ r ≀ 10^{800}). The third line contains one integer n (1 ≀ n ≀ 2 000). Output In the first line, print the maximum possible number of modest substrings. In the second line, print a string of length n having exactly that number of modest substrings. If there are multiple such strings, print the lexicographically smallest of them. Examples Input 1 10 3 Output 3 101 Input 1 11 3 Output 5 111 Input 12345 12346 6 Output 1 012345 Note In the first example, string Β«101Β» has modest substrings Β«1Β», Β«10Β», Β«1Β». In the second example, string Β«111Β» has modest substrings Β«1Β» (3 times) and Β«11Β» (2 times). Submitted Solution: ``` n = int(input()) x = [int(x) for x in input().split()] h = x[0] l = x[0] a = 0 for i in range(0,n): if x[i] > h: h = x[i] a+=1 if x[i] < l: l = x[i] a+=1 print(a) ```
instruction
0
103,731
0
207,462
No
output
1
103,731
0
207,463
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,075
0
208,150
Tags: binary search, dp, strings, two pointers Correct Solution: ``` n,k=map(int, input().split()) s=input() a=0 b=0 ans=0 x=0 for i in range(n): if s[i]=='a': a+=1 else: b+=1 if(min(a,b)<=k): ans=max(ans,a+b) else: if s[x]=='a': a-=1 else: b-=1 x+=1 print(ans) ```
output
1
104,075
0
208,151
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,076
0
208,152
Tags: binary search, dp, strings, two pointers Correct Solution: ``` def executa(n, k, letra, s): r = 0 tk = k p = 0 for i in range(n): while r < n and tk >= 0: if a[r] == letra: if tk == 0: break tk -= 1 r += 1 p += 1 s = max(s, p) p -= 1 if a[i] == letra: tk += 1 if r == n: break return s if __name__ == '__main__': entrada = input() entrada_str = list(entrada.split(" ")) entrada_int = list(map(int, entrada_str)) n = entrada_int[0] k = entrada_int[1] a = input() soma = executa(n, k, "b", 0) soma = executa(n, k, "a", soma) print(soma) ```
output
1
104,076
0
208,153
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,077
0
208,154
Tags: binary search, dp, strings, two pointers Correct Solution: ``` n, k = list(map(int, input().split())) s = input() res = 0 count_a = 0 count_b = 0 left = 0 right = 0 while True: if s[right] == 'a': count_a += 1 else: count_b += 1 if count_a <= k or count_b <= k: if right - left + 1 > res: res = right - left + 1 else: if s[left] == 'a': count_a -= 1 else: count_b -= 1 left += 1 right += 1 if n - left < res or right == n: break print(res) ```
output
1
104,077
0
208,155
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,078
0
208,156
Tags: binary search, dp, strings, two pointers Correct Solution: ``` def answer(n,k,A): if n==1: return 1 dp=[0]*(n+1) for i in range(1,n+1): if A[i-1]=="a": dp[i]=dp[i-1]+1 else: dp[i]=dp[i-1] l=0;r=0 maxi=0 while r>=l and r<n: x=dp[r+1]-dp[l] if min(x, r+1-l-x)<=k: r+=1 else: maxi=max(maxi,r-l) l+=1 maxi=max(maxi,r-l) return maxi n,k=map(int,input().split()) A=input() print(answer(n,k,A)) ```
output
1
104,078
0
208,157
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,079
0
208,158
Tags: binary search, dp, strings, two pointers Correct Solution: ``` le, ch = map(int, input().split()) st = input() ca, cb, si, mx = [0] * 4 for x in st: if x == 'a': ca += 1 else: cb += 1 if min(ca, cb) > ch: if st[si] == 'a': ca -= 1 else: cb -= 1 si += 1 else: mx += 1 print(mx) ```
output
1
104,079
0
208,159
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,080
0
208,160
Tags: binary search, dp, strings, two pointers Correct Solution: ``` from sys import stdin,stdout def fn(a,ch): # print(a) ans=0 for i in range(n): l,r=i,n-1 while l<=r: mid=(l+r)>>1 req=a[i]+k-int(s[i]!=ch) if a[mid]<=req:l=mid+1 else:r=mid-1 # print(i,r) ans=max(ans,r-i+1) return ans for _ in range(1):#int(stdin.readline())): # n=int(stdin.readline()) n,k=list(map(int,stdin.readline().split())) s=input() make_a=[];make_b=[] make_a+=[int(s[0]=='a')] make_b+=[int(s[0]=='b')] for i in range(1,n): make_a+=[make_a[-1]+int(s[i]=='a')] make_b+=[make_b[-1]+int(s[i]=='b')] # print(make_a) # print(make_b) ans=max(fn(make_a,'b'),fn(make_b,'a')) print(ans) ''' 10 1 bbabbabbba ''' ```
output
1
104,080
0
208,161
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,081
0
208,162
Tags: binary search, dp, strings, two pointers Correct Solution: ``` num = input() n, k = num.split() n = int(n) k = int(k) string = input() count_a = 0 count_b = 0 def check(ch): i = 0 j = 0 count = 0 answer = 0 for i in range(n): if(string[i] == ch): count += 1 if(count > k): while(count > k): if(string[j] == ch): count -= 1 j += 1 answer = max(answer, i-j +1) return answer max_a = check('a') max_b = check('b') print(max(max_a, max_b)) ```
output
1
104,081
0
208,163
Provide tags and a correct Python 3 solution for this coding contest problem. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa".
instruction
0
104,082
0
208,164
Tags: binary search, dp, strings, two pointers Correct Solution: ``` read = lambda: map(int, input().split()) n, k = read() s = input() i = j = 0 cur = k while cur and j < n: if s[j] == 'a': cur -= 1 j += 1 while j < n and s[j] == 'b': j += 1 ans = j while j < n: while i < n and s[i] == 'b': i += 1 i += 1 j += 1 while j < n and s[j] == 'b': j += 1 ans = max(ans, j - i) i = j = 0 cur = k while cur and j < n: if s[j] == 'b': cur -= 1 j += 1 while j < n and s[j] == 'a': j += 1 ans = max(ans, j) while j < n: while i < n and s[i] == 'a': i += 1 i += 1 j += 1 while j < n and s[j] == 'a': j += 1 ans = max(ans, j - i) print(ans) ```
output
1
104,082
0
208,165
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` z,zz=input,lambda:list(map(int,z().split())) zzz=lambda:[int(i) for i in stdin.readline().split()] szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz()) from string import * from re import * from collections import * from queue import * from sys import * from collections import * from math import * from heapq import * from itertools import * from bisect import * from collections import Counter as cc from math import factorial as f from bisect import bisect as bs from bisect import bisect_left as bsl from itertools import accumulate as ac def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2)) def prime(x): p=ceil(x**.5)+1 for i in range(2,p): if (x%i==0 and x!=2) or x==0:return 0 return 1 def dfs(u,visit,graph): visit[u]=True for i in graph[u]: if not visit[i]: dfs(i,visit,graph) ###########################---Test-Case---################################# """ """ ###########################---START-CODING---############################## n,k=zz() s=z() m=k a=s[:k].count('a') b=k-a l=0 for i in range(k,n): if s[i]=='a':a+=1 else:b+=1 while min(a,b)>k: if s[l]=='a':a-=1 else:b-=1 l+=1 m=max(m,i-l+1) print(m) ```
instruction
0
104,083
0
208,166
Yes
output
1
104,083
0
208,167
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` n, m = input().split(' ') n, m = int(n), int(m) ans = 0 s = input() for i in range(2): left, used = 0, 0 for right in range(n): if ord(s[right]) != ord('a') + i: used += 1 while used > m: if ord(s[left]) != ord('a') + i: used -= 1 left += 1 ans = max(ans, right - left + 1) print(ans) ```
instruction
0
104,084
0
208,168
Yes
output
1
104,084
0
208,169
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` def main(): n, k = liee() s = input() l, ans, h = 1, 0, n while l <= h: m = (l + h) // 2 if fun(m, n, s, k): ans, l = m, m + 1 else: h = m - 1 print(ans) def fun(m, n, s, k): a, b = s[0:m].count('a'), s[0:m].count('b') if max(a, b) + k >= m: return 1 for i in range(m, n): if s[i - m] == 'a': a -= 1 else: b -= 1 if s[i] == 'a': a += 1 else: b += 1 if max(a, b) + k >= m: return 1 return 0 def fun2(s): cnt, num = 0, s.count('0') for i in s[1:]: if i == '1': cnt += 1 elif num == 1: cnt += 1 break else: break return cnt from sys import * import inspect import re from math import * import threading from collections import * from pprint import pprint as pp mod = 998244353 MAX = 10**5 def lie(): return int(input()) def liee(): return map(int, input().split()) def array(): return list(map(int, input().split())) def deb(p): for line in inspect.getframeinfo(inspect.currentframe().f_back)[3]: m = re.search(r'\bdeb\s*\(\s*([A-Za-z_][A-Za-z0-9_]*)\s*\)', line) print('%s %d' % (m.group(1), p)) def vector(size, val=0): vec = [val for i in range(size)] return vec def matrix(rowNum, colNum, val=0): mat = [] for i in range(rowNum): collumn = [val for j in range(colNum)] mat.append(collumn) return mat def dmain(): setrecursionlimit(100000000) threading.stack_size(40960000) thread = threading.Thread(target=main) thread.start() if __name__ == '__main__': # main() dmain() ```
instruction
0
104,085
0
208,170
Yes
output
1
104,085
0
208,171
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` import os import sys from io import BytesIO, IOBase BUFSIZE = 8192 class FastIO(IOBase): newlines = 0 def __init__(self, file): self._fd = file.fileno() self.buffer = BytesIO() self.writable = "x" in file.mode or "r" not in file.mode self.write = self.buffer.write if self.writable else None def read(self): while True: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) if not b: break ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines = 0 return self.buffer.read() def readline(self): while self.newlines == 0: b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE)) self.newlines = b.count(b"\n") + (not b) ptr = self.buffer.tell() self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr) self.newlines -= 1 return self.buffer.readline() def flush(self): if self.writable: os.write(self._fd, self.buffer.getvalue()) self.buffer.truncate(0), self.buffer.seek(0) class IOWrapper(IOBase): def __init__(self, file): self.buffer = FastIO(file) self.flush = self.buffer.flush self.writable = self.buffer.writable self.write = lambda s: self.buffer.write(s.encode("ascii")) self.read = lambda: self.buffer.read().decode("ascii") self.readline = lambda: self.buffer.readline().decode("ascii") sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout) input = lambda: sys.stdin.readline().rstrip("\r\n") #################################################################################### n,k=map(int,input().split()) from bisect import bisect s=input() l1=[] l2=[] l3=[] l4=[] c,d=0,0 for i in range(n): if s[i]=='a': l1.append(i) d+=1 else: c+=1 l3.append(i) l2.append(c) l4.append(d) if len(l1)!=0 and c!=0: ans=0 ans1=0 for i in l1: a=bisect(l2,l2[i]+k) ans=max(ans,a-i) for i in l3: a=bisect(l4,l4[i]+k) ans1=max(ans1,a-i) if l1[0]==0: a=bisect(l4,k) ans1=max(ans1,a) if l3[0]==0: a=bisect(l2,k) ans=max(ans,a) else: ans=len(s) ans1=len(s) print(max(ans,ans1)) ```
instruction
0
104,086
0
208,172
Yes
output
1
104,086
0
208,173
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` aplha=[chr(i) for i in range(ord('a'),ord('z')+1)] #for _ in range(int(input())): n,k=map(int,input().split()) s=list(input()) out=0 for i in aplha: l=0 r=0 t=k while l<n and r<n: if s[r]==i: r+=1 else: if t>0: t-=1 r+=1 continue else: if s[l]==i: l+=1 else: t+=1 l+=1 out=max(out,r-l) print(out) ```
instruction
0
104,087
0
208,174
No
output
1
104,087
0
208,175
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` # !/bin/env python3 # encoding: UTF-8 # βœͺ H4WK3yEδΉ‘ # Mohd. Farhan Tahir # Indian Institute Of Information Technology and Management,Gwalior # Question Link # https://codeforces.com/problemset/problem/676/C # # ///==========Libraries, Constants and Functions=============/// import sys inf = float("inf") mod = 1000000007 def get_array(): return list(map(int, sys.stdin.readline().split())) def get_ints(): return map(int, sys.stdin.readline().split()) def input(): return sys.stdin.readline() # ///==========MAIN=============/// def solve(c, s, k): ans = 0 r = 0 balance = 0 for i in range(len(s)): while r < len(s) and (s[r] == c or balance < k): if s[r] != c: balance += 1 r += 1 ans = max(ans, r-i) if s[i] != c: balance -= 1 return ans def main(): n, k = get_ints() s = input() print(max(solve('a', s, k), solve('b', s, k))) if __name__ == "__main__": main() ```
instruction
0
104,088
0
208,176
No
output
1
104,088
0
208,177
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` from collections import Counter def good(s, k, l): c = Counter() for i in range(len(s)): c[s[i]] += 1 if i - l >= 0: c[s[i - l]] -= 1 if i + 1 >= l: if min(c.values()) <= k: return True return False def solve(s, k): low = 1 high = 1 while good(s, k, high): high *= 2 while low + 1 < high: mid = (low + high) // 2 if good(s, k, mid): low = mid else: high = mid return low n, k = map(int, input().split()) s = input() print(solve(s, k)) ```
instruction
0
104,089
0
208,178
No
output
1
104,089
0
208,179
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. High school student Vasya got a string of length n as a birthday present. This string consists of letters 'a' and 'b' only. Vasya denotes beauty of the string as the maximum length of a substring (consecutive subsequence) consisting of equal letters. Vasya can change no more than k characters of the original string. What is the maximum beauty of the string he can achieve? Input The first line of the input contains two integers n and k (1 ≀ n ≀ 100 000, 0 ≀ k ≀ n) β€” the length of the string and the maximum number of characters to change. The second line contains the string, consisting of letters 'a' and 'b' only. Output Print the only integer β€” the maximum beauty of the string Vasya can achieve by changing no more than k characters. Examples Input 4 2 abba Output 4 Input 8 1 aabaabaa Output 5 Note In the first sample, Vasya can obtain both strings "aaaa" and "bbbb". In the second sample, the optimal answer is obtained with the string "aaaaabaa" or with the string "aabaaaaa". Submitted Solution: ``` n, k = [int(i) for i in input().split()] s = input() def beauty (x): ret = 0 cnt = 0 r = 0 for l in range(n): while r < n and (cnt < k or s[r] != x): if s[r] == x: cnt += 1 r += 1 if r == x: cnt -= 1 ret = max(ret, r-l) return ret print(max(beauty('a'), beauty('b'))) exit(0) ```
instruction
0
104,090
0
208,180
No
output
1
104,090
0
208,181
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,136
0
208,272
Tags: brute force, dp, hashing, strings Correct Solution: ``` s = input() n = len(s) max_palin = [[0 for i in range(n + 1)] for j in range(n + 1)] count = [0 for i in range(n + 1)] for sub_len in range(1, n + 1): for left in range(0, n - sub_len + 1): right = left + sub_len - 1 if sub_len == 1: max_palin[left][right] = 1 elif sub_len == 2: if s[left] == s[right]: max_palin[left][right] = 2 else: max_palin[left][right] = 0 else: if s[left] == s[right] and max_palin[left + 1][right - 1] > 0: mid = (left + right) // 2 if sub_len % 2 == 0: max_palin[left][right] = max_palin[left][mid] + 1 else: max_palin[left][right] = max_palin[left][mid - 1] + 1 count[max_palin[left][right]] += 1 for i in range(n - 1, 0, -1): count[i] += count[i + 1] for i in range(1, n + 1): print(count[i], end=' ') print() ```
output
1
104,136
0
208,273
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,137
0
208,274
Tags: brute force, dp, hashing, strings Correct Solution: ``` class HString: def __init__(self, string, base=257, modulo=1000000007): self.__base, self.__modulo = base, modulo self.__prefix_hash, self.__base_pow, self.__size = [], [1], 0 self += string def __add__(self, string): for ch in string: self.__base_pow.append((self.__base_pow[-1] * self.__base) % self.__modulo) if self.__size == 0: self.__prefix_hash.append(ord(ch)) else: self.__prefix_hash.append((self.__prefix_hash[-1] * self.__base + ord(ch)) % self.__modulo) self.__size += 1 return self def size(self): return self.__size def getModulo(self): return self.__modulo def getHashValue(self, st, en): value = self.__prefix_hash[en] if st > 0: value -= ((self.__prefix_hash[st-1] * self.__base_pow[en-st+1]) % self.__modulo) if value < 0: value += self.__modulo return value def palindromic_characteristics(s): n, org, rev = len(s), HString(s), HString(s[::-1]) palindrome_level = [[0 for _ in range(n)] for _ in range(n)] palindrome_level_count = [0 for _ in range(n + 1)] for i in range(n): for j in range(i, n): if org.getHashValue(i, j) == rev.getHashValue(n-1-j, n-1-i): mid = (i + j) // 2 + (i + j) % 2 if i > mid-1: palindrome_level[i][j] = 1 else: palindrome_level[i][j] = palindrome_level[i][mid-1] + 1 palindrome_level_count[palindrome_level[i][j]] += 1 for i in range(n-1, 0, -1): palindrome_level_count[i] += palindrome_level_count[i+1] return palindrome_level_count[1:] s = input() print(' '.join(map(str, palindromic_characteristics(s)))) ```
output
1
104,137
0
208,275
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,138
0
208,276
Tags: brute force, dp, hashing, strings Correct Solution: ``` def PanlidromicCharacteristics(string): n = len(string) res = [[0 for i in range (n)] for j in range (n)] count = [0 for i in range (n + 1)] # for i in range (n): # res[i][i] = 1 # count[1] += 1 for length in range (1, n + 1): for i in range (n-length + 1): j = i + length - 1 if length == 1: res[i][j] = 1 elif length == 2 and string[i] == string[j]: res[i][j] = 2 elif string[i] == string[j] and res[i + 1][j - 1] > 0: res[i][j] = res[i][i + length//2 - 1] + 1 count[res[i][j]] += 1 # k-palindrome is also a (k - 1)-palindrome for i in range (len(count) - 1, 0, -1): count[i - 1] += count[i] for i in range (1, len(count)): print(count[i], end = " ") return string = input() PanlidromicCharacteristics(string) ```
output
1
104,138
0
208,277
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,139
0
208,278
Tags: brute force, dp, hashing, strings Correct Solution: ``` s = input() n = len(s) isPalindrome = [[False for i in range(n)] for i in range(n)] for i in range(n): isPalindrome[i][i] = True for i in range(n - 1, -1, -1): for j in range(i + 1, n): if (s[i] == s[j] and (i + 1 == j or isPalindrome[i + 1][j - 1] == True)): isPalindrome[i][j] = True degreePalindrome = [[0 for i in range(n)] for i in range(n)] #degreePalindrome[i][j] = degreePalindrome[i][mid] + 1 res = [0] * (n + 1) for i in range(n): for j in range(i, n): if (i == j): degreePalindrome[i][j] = 1 elif isPalindrome[i][j]: mid = (i + j - 1) // 2 degreePalindrome[i][j] = degreePalindrome[i][mid] + 1 res[degreePalindrome[i][j]] += 1 for i in range( n - 2, 0, -1): res[i] += res[i + 1] print(*res[1::]) ```
output
1
104,139
0
208,279
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,140
0
208,280
Tags: brute force, dp, hashing, strings Correct Solution: ``` s = input() n = len(s) dp = [[0 for i in range(n - le + 1)] for le in range(n + 1)] ans = [0 for i in range(n + 1)] for le in range(1, n + 1): for l in range(0, n - le + 1): r = l + le if s[l] != s[r - 1]: continue if le == 1: dp[1][l] = 1 ans[1] += 1 elif le == 2: ans[2] += 1 dp[2][l] = 2 elif dp[le - 2][l + 1]: v = 1 m = (l + r) // 2 st = m + 1 if le & 1 else m le2 = m - l q = dp[le2][l] if q: v = q + 1 ans[v] += 1 dp[le][l] = v for i in range(n - 1, 0, -1): ans[i] += ans[i + 1] print(*ans[1:]) ```
output
1
104,140
0
208,281
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,141
0
208,282
Tags: brute force, dp, hashing, strings Correct Solution: ``` s = input() size = len(s) dp = [[0 for l in range(size)] for li in range(size)] ans = [0]*(size+1) for i in range(1, size+1): if i == 1: for j in range(0, size): dp[j][j] = 1 ans[1] += 1 elif i == 2: for j in range(0, size-1): if s[j+1] == s[j]: dp[j][j+1] = 2 ans[1] += 1 ans[2] += 1 else: dp[j][j+1] = 0 else: for j in range(0, size-i+1): if s[j] != s[j+i-1] or dp[j+1][j+i-2] == 0: dp[j][j+i-1] = 0 else: dp[j][j+i-1] = dp[j][int((j+j+i-2)/2)] + 1 for p in range(1, dp[j][j+i-1]+1): ans[p] += 1 for i in range(1, size): print(ans[i], end="") print(" ", end="") print(ans[size]) ```
output
1
104,141
0
208,283
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,142
0
208,284
Tags: brute force, dp, hashing, strings Correct Solution: ``` def main(): s = input() n = len(s) isPalindrome = [[False for i in range(n + 1)] for i in range(n + 1)] for i in range(n): isPalindrome[i][i] = True for i in range(n - 1, -1, -1): for j in range(i + 1, n): if (s[i] == s[j] and (i + 1 == j or isPalindrome[i + 1][j - 1] == True)): isPalindrome[i][j] = True degreePalindrome = [[0 for i in range(n)] for i in range(n)] #degreePalindrome[i][j] = degreePalindrome[i][mid] + 1 res = [0] * (n + 1) for i in range(n): for j in range(i, n): if (i == j): degreePalindrome[i][j] = 1 elif isPalindrome[i][j]: mid = (i + j - 1) // 2 degreePalindrome[i][j] = degreePalindrome[i][mid] + 1 res[degreePalindrome[i][j]] += 1 for i in range( n - 2, 0, -1): res[i] += res[i + 1] print(*res[1::]) if __name__ == "__main__": main() ```
output
1
104,142
0
208,285
Provide tags and a correct Python 3 solution for this coding contest problem. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here.
instruction
0
104,143
0
208,286
Tags: brute force, dp, hashing, strings Correct Solution: ``` s = input() dp = [[0]*5005 for _ in range(5005)] n = len(s) ans = [0 for _ in range(5005)] for length in range(1,n+1): for l in range(n-length+1): r = l+length if(length == 1): dp[l][r] = 1 continue elif(length == 2): dp[l][r] = 2 if(s[l] == s[r-1]) else 0 continue if(s[l] != s[r-1] or dp[l+1][r-1] == 0): continue dp[l][r] = 1 m = (l+r) // 2 if(length&1): if(dp[l][m] and dp[m+1][r]): dp[l][r] = dp[l][m]+1 else: if(dp[l][m] and dp[m][r]): dp[l][r] = dp[l][m]+1 for length in range(1,n+1): for l in range(n-length+1): ans[dp[l][l+length]] += 1 for i in range(n-1,0,-1): ans[i] += ans[i+1] for i in range(1,n+1): print(ans[i],end=' ') print() ```
output
1
104,143
0
208,287
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` class HString: def __init__(self, string, base=257, modulo=1000000007): self.__base, self.__modulo = base, modulo self.__prefix_hash, self.__base_pow, self.__size = [], [1], 0 self += string def __add__(self, string): for ch in string: self.__base_pow.append((self.__base_pow[-1] * self.__base) % self.__modulo) if self.__size == 0: self.__prefix_hash.append(ord(ch)) else: self.__prefix_hash.append((self.__prefix_hash[-1] * self.__base + ord(ch)) % self.__modulo) self.__size += 1 return self def size(self): return self.__size def getModulo(self): return self.__modulo def getHashValue(self, st, en): value = self.__prefix_hash[en] if st > 0: value -= ((self.__prefix_hash[st-1] * self.__base_pow[en-st+1]) % self.__modulo) if value < 0: value += self.__modulo return value def palindromic_characteristics(s): n, org, rev = len(s), HString(s), HString(s[::-1]) palindrome_level = [[0 for _ in range(n)] for _ in range(n)] palindrome_level_count = [0 for _ in range(n + 1)] i, j = 0, 0 while i < n: j = i while j < n: if org.getHashValue(i, j) == rev.getHashValue(n-1-j, n-1-i): mid = (i + j) // 2 + (i + j) % 2 if i > mid-1: palindrome_level[i][j] = 1 else: palindrome_level[i][j] = palindrome_level[i][mid-1] + 1 palindrome_level_count[palindrome_level[i][j]] += 1 j += 1 i += 1 for i in range(n-1, 0, -1): palindrome_level_count[i] += palindrome_level_count[i+1] return palindrome_level_count[1:] s = input() print(' '.join(map(str, palindromic_characteristics(s)))) ```
instruction
0
104,144
0
208,288
Yes
output
1
104,144
0
208,289
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` s = input() n = len(s) max_palin = [[0 for i in range(n + 1)] for j in range(n + 1)] count = [0 for i in range(n + 1)] for sub_len in range(1, n + 1): for left in range(0, n - sub_len + 1): right = left + sub_len - 1 if sub_len == 1: max_palin[left][right] = 1 elif sub_len == 2: if s[left] == s[right]: max_palin[left][right] = 2 else: max_palin[left][right] = 0 else: if s[left] == s[right] and max_palin[left + 1][right - 1] > 0: max_palin[left][right] = max_palin[left][left + sub_len // 2 - 1] + 1 count[max_palin[left][right]] += 1 for i in range(n - 1, 0, -1): count[i] += count[i + 1] for i in range(1, n + 1): print(count[i], end=' ') print() ```
instruction
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104,145
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208,290
Yes
output
1
104,145
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208,291
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` def main(): s = input() n = len(s) dp = [[0 for i in range(n + 1)] for j in range(n + 1)] count = [0 for i in range(n + 1)] for sub_len in range(1, n + 1): for left in range(0, n - sub_len + 1): right = left + sub_len - 1 if sub_len == 1: dp[left][right] = 1 elif sub_len == 2: if s[left] == s[right]: dp[left][right] = 2 else: if s[left] == s[right] and dp[left + 1][right - 1] > 0: dp[left][right] = dp[left][left + sub_len // 2 - 1] + 1 count[dp[left][right]] += 1 for i in range(n - 1, 0, -1): count[i] += count[i + 1] for i in range(1, n + 1): print(count[i], end=' ') print() if __name__ == "__main__": main() ```
instruction
0
104,146
0
208,292
Yes
output
1
104,146
0
208,293
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` P = 311 # we use two mods to reduce the chance of collision MOD1 = int(1e9) + 7 MOD2 = int(1e9) + 9 def main(): s = input() n = len(s) # Pre-compute power_1 = [0 for i in range(n + 1)] power_2 = [0 for i in range(n + 1)] mod_inv_1 = [0 for i in range(n + 1)] mod_inv_2 = [0 for i in range(n + 1)] power_1[0] = 1 power_2[0] = 1 mod_inv_1[0] = 1 mod_inv_2[0] = 1 for i in range(1, n + 1): power_1[i] = power_1[i - 1] * P % MOD1 power_2[i] = power_2[i - 1] * P % MOD1 mod_inv_1[i] = bin_exp(power_1[i], MOD1 - 2, MOD1) mod_inv_2[i] = bin_exp(power_2[i], MOD2 - 2, MOD2) # Compute hash values hash_1 = 0 hash_2 = 0 forward_hash_1 = [0 for i in range(n + 1)] forward_hash_2 = [0 for i in range(n + 1)] for i in range(1, n + 1): hash_1 += ord(s[i - 1]) * power_1[i] hash_2 += ord(s[i - 1]) * power_2[i] hash_1 %= MOD1 hash_2 %= MOD2 forward_hash_1[i] = hash_1 forward_hash_2[i] = hash_2 hash_1 = 0 hash_2 = 0 backward_hash_1 = [0 for i in range(n + 1)] backward_hash_2 = [0 for i in range(n + 1)] for i in range(1, n + 1): hash_1 += ord(s[n - i]) * power_1[i] hash_2 += ord(s[n - i]) * power_2[i] hash_1 %= MOD1 hash_2 %= MOD2 backward_hash_1[i] = hash_1 backward_hash_2[i] = hash_2 dp = [[0 for i in range(n + 1)] for j in range(n + 1)] count = [0 for i in range(n + 1)] for sub_len in range(1, n + 1): for left in range(0, n - sub_len + 1): right = left + sub_len - 1 if sub_len == 1: dp[left][right] = 1 elif sub_len == 2: if s[left] == s[right]: dp[left][right] = 2 else: if s[left] == s[right] and dp[left + 1][right - 1] > 0: dp[left][right] = dp[left][left + sub_len // 2 - 1] + 1 count[dp[left][right]] += 1 for i in range(n - 1, 0, -1): count[i] += count[i + 1] for i in range(1, n + 1): print(count[i], end=' ') print() def bin_exp(a, x, mod): res = 1 while x > 0: if x & 1: res *= a res %= mod a *= a a %= mod x >>= 1 return res def get_forward_hash(forward_hash, mod_inv, left, right, mod): return (forward_hash[right + 1] - forward_hash[left] + mod) * mod_inv[left] % mod def get_backward_hash(backward_hash, mod_inv, n, left, right): r_left = n - left - 1 r_right = n - right - 1 return (backward_hash[r_left + 1] - backward_hash[r_right] + mod) * mod_inv[r_right] % mod if __name__ == "__main__": main() ```
instruction
0
104,147
0
208,294
Yes
output
1
104,147
0
208,295
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` def palh(s): if s[:(len(s)//2)]==s[-(len(s)//2):]: return palh(s[:(len(s)//2)])+1 else: return 0 S=input() N=len(S) R=[0 for i in range(N+1)] for l in range(0, N): for r in range(l+1, N): for i in range(0, palh(S[l:r])+1): R[i]+=1 print(*R[:-1]) ```
instruction
0
104,148
0
208,296
No
output
1
104,148
0
208,297
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes. A string is 1-palindrome if and only if it reads the same backward as forward. A string is k-palindrome (k > 1) if and only if: 1. Its left half equals to its right half. 2. Its left and right halfs are non-empty (k - 1)-palindromes. The left half of string t is its prefix of length ⌊|t| / 2βŒ‹, and right half β€” the suffix of the same length. ⌊|t| / 2βŒ‹ denotes the length of string t divided by 2, rounded down. Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times. Input The first line contains the string s (1 ≀ |s| ≀ 5000) consisting of lowercase English letters. Output Print |s| integers β€” palindromic characteristics of string s. Examples Input abba Output 6 1 0 0 Input abacaba Output 12 4 1 0 0 0 0 Note In the first example 1-palindromes are substring Β«aΒ», Β«bΒ», Β«bΒ», Β«aΒ», Β«bbΒ», Β«abbaΒ», the substring Β«bbΒ» is 2-palindrome. There are no 3- and 4-palindromes here. Submitted Solution: ``` s = input() size = len(s) dp = [[0 for l in range(size)] for l in range(size)] ans = [0]*(size+1) for i in range(1, size+1): if i == 1: for j in range(0, size): dp[j][j] = 1 ans[1] += 1 elif i == 2: for j in range(0, size-1): if s[j+1] == s[j]: dp[j][j+1] = 2 ans[1] += 1 ans[2] += 1 else: dp[j][j+1] = 0 else: for j in range(0, size-i+1): if s[j] != s[j+i-1] or dp[j+1][j+i-2] == 0: dp[j][j+i-1] = 0 else: dp[j][j+i-1] = dp[j][int((j+j+i-2)/2)] + 1 ans[dp[j][j+i-1]] += 1 for i in range(1, size): print(ans[i], end="") print(" ", end="") print(ans[size]) ```
instruction
0
104,149
0
208,298
No
output
1
104,149
0
208,299
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Suppose you have two strings s and t, and their length is equal. You may perform the following operation any number of times: choose two different characters c1 and c2, and replace every occurence of c1 in both strings with c2. Let's denote the distance between strings s and t as the minimum number of operations required to make these strings equal. For example, if s is abcd and t is ddcb, the distance between them is 2 β€” we may replace every occurence of a with b, so s becomes bbcd, and then we may replace every occurence of b with d, so both strings become ddcd. You are given two strings S and T. For every substring of S consisting of |T| characters you have to determine the distance between this substring and T. Input The first line contains the string S, and the second β€” the string T (1 ≀ |T| ≀ |S| ≀ 125000). Both strings consist of lowercase Latin letters from a to f. Output Print |S| - |T| + 1 integers. The i-th of these integers must be equal to the distance between the substring of S beginning at i-th index with length |T| and the string T. Example Input abcdefa ddcb Output 2 3 3 3 Submitted Solution: ``` class UnionFind: def __init__(self, n): self.parent = [-1] * n self.cnt = n def root(self, x): if self.parent[x] < 0: return x else: self.parent[x] = self.root(self.parent[x]) return self.parent[x] def merge(self, x, y): x = self.root(x) y = self.root(y) if x == y: return if self.parent[x] > self.parent[y]: x, y = y, x self.parent[x] += self.parent[y] self.parent[y] = x self.cnt -= 1 def is_same(self, x, y): return self.root(x) == self.root(y) def get_cnt(self): return self.cnt s = [ord(char) - 97 for char in input()] t = [ord(char) - 97 for char in input()][::-1] s_ = [0] * 6 t_ = [0] * 6 st = [[0] * 6 for _ in range(6)] for i, val in enumerate(s): s_[val] |= 1 << i for i, val in enumerate(t): t_[val] |= 1 << i for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue st[ind_s][ind_t] = bin(s_[ind_s] * t_[ind_t])[2:][::-1] + "0" * (len(s) + len(t) + 100) length = len(t) - 1 ans = [] for i in range(len(s) - len(t) + 1): shift = length + i uf = UnionFind(6) for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue if st[ind_s][ind_t][shift] == "1": uf.merge(ind_s, ind_t) ans.append(6 - uf.get_cnt()) print(*ans) ```
instruction
0
104,168
0
208,336
No
output
1
104,168
0
208,337
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Suppose you have two strings s and t, and their length is equal. You may perform the following operation any number of times: choose two different characters c1 and c2, and replace every occurence of c1 in both strings with c2. Let's denote the distance between strings s and t as the minimum number of operations required to make these strings equal. For example, if s is abcd and t is ddcb, the distance between them is 2 β€” we may replace every occurence of a with b, so s becomes bbcd, and then we may replace every occurence of b with d, so both strings become ddcd. You are given two strings S and T. For every substring of S consisting of |T| characters you have to determine the distance between this substring and T. Input The first line contains the string S, and the second β€” the string T (1 ≀ |T| ≀ |S| ≀ 125000). Both strings consist of lowercase Latin letters from a to f. Output Print |S| - |T| + 1 integers. The i-th of these integers must be equal to the distance between the substring of S beginning at i-th index with length |T| and the string T. Example Input abcdefa ddcb Output 2 3 3 3 Submitted Solution: ``` import sys input = sys.stdin.readline class UnionFind: def __init__(self, n): self.parent = [-1] * n self.cnt = n def root(self, x): if self.parent[x] < 0: return x else: self.parent[x] = self.root(self.parent[x]) return self.parent[x] def merge(self, x, y): x = self.root(x) y = self.root(y) if x == y: return if self.parent[x] > self.parent[y]: x, y = y, x self.parent[x] += self.parent[y] self.parent[y] = x self.cnt -= 1 def is_same(self, x, y): return self.root(x) == self.root(y) def get_cnt(self): return self.cnt s = [ord(char) - 97 for char in input()[:-1]] t = [ord(char) - 97 for char in input()[:-1]][::-1] s_ = [0] * 6 t_ = [0] * 6 st = [[0] * 6 for i in range(6)] for i, val in enumerate(s): s_[val] |= 1 << i for i, val in enumerate(t): t_[val] |= 1 << i for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue st[ind_s][ind_t] = bin(s_[ind_s] * t_[ind_t])[2:][::-1] + "0" * (len(s) + len(t)) length = len(t) - 1 ans = [] for i in range(len(s) - len(t) + 1): shift = length + i uf = UnionFind(6) for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue if st[ind_s][ind_t][shift] == "1": uf.merge(ind_s, ind_t) ans.append(6 - uf.get_cnt()) print(*ans) ```
instruction
0
104,169
0
208,338
No
output
1
104,169
0
208,339
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. Suppose you have two strings s and t, and their length is equal. You may perform the following operation any number of times: choose two different characters c1 and c2, and replace every occurence of c1 in both strings with c2. Let's denote the distance between strings s and t as the minimum number of operations required to make these strings equal. For example, if s is abcd and t is ddcb, the distance between them is 2 β€” we may replace every occurence of a with b, so s becomes bbcd, and then we may replace every occurence of b with d, so both strings become ddcd. You are given two strings S and T. For every substring of S consisting of |T| characters you have to determine the distance between this substring and T. Input The first line contains the string S, and the second β€” the string T (1 ≀ |T| ≀ |S| ≀ 125000). Both strings consist of lowercase Latin letters from a to f. Output Print |S| - |T| + 1 integers. The i-th of these integers must be equal to the distance between the substring of S beginning at i-th index with length |T| and the string T. Example Input abcdefa ddcb Output 2 3 3 3 Submitted Solution: ``` class UnionFind: def __init__(self, n): self.parent = [-1] * n self.cnt = n def root(self, x): if self.parent[x] < 0: return x else: self.parent[x] = self.root(self.parent[x]) return self.parent[x] def merge(self, x, y): x = self.root(x) y = self.root(y) if x == y: return if self.parent[x] > self.parent[y]: x, y = y, x self.parent[x] += self.parent[y] self.parent[y] = x self.cnt -= 1 def is_same(self, x, y): return self.root(x) == self.root(y) def get_cnt(self): return self.cnt s = [ord(char) - 97 for char in input()] t = [ord(char) - 97 for char in input()][::-1] s_ = [0] * 6 t_ = [0] * 6 st = [[0] * 6 for _ in range(6)] for i, val in enumerate(s): s_[val] |= 1 << i for i, val in enumerate(t): t_[val] |= 1 << i for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue st[ind_s][ind_t] = bin(s_[ind_s] * t_[ind_t])[2:][::-1] + "0" * (len(s) + len(t) + 100) length = len(t) - 1 ans = [] for i in range(len(s) - len(t) + 1): shift = length + i uf = UnionFind(6) for ind_s in range(6): for ind_t in range(6): if ind_s == ind_t: continue if st[ind_s][ind_t][shift] == "1": uf.merge(ind_s, ind_t) ans.append(6 - uf.get_cnt()) if len(s) > 100: print(len(s)) print(len(t)) print(*ans) ```
instruction
0
104,170
0
208,340
No
output
1
104,170
0
208,341
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,539
0
209,078
Tags: implementation, math Correct Solution: ``` """ NTC here """ from sys import stdin # import threading # setrecursionlimit(10**6) # threading.stack_size(2**26) def iin(): return int(stdin.readline()) def lin(): return list(map(int, stdin.readline().split())) # range = xrange # input = raw_input md=10**9+7 def main(): t=iin() while t: t-=1 x=iin() s=list(input()) l=len(s) ans=l i=0 while i<x: i+=1 ch=int(s[i-1]) ans=(i+ch*(ans-i))%md if l>=x:continue incr=l ss=[] for _ in range(ch): for j in range(i,l): if incr<ans and incr<x: ss.append(s[j]) incr+=1 else: break l=incr for k in ss:s.append(k) #print(ans, l, s) print(ans%md) main() #threading.Thread(target=main).start() ```
output
1
104,539
0
209,079
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,540
0
209,080
Tags: implementation, math Correct Solution: ``` M = int(1e9+7) n = int(input()) while n>0: n -= 1 x = int(input()) s = input() idx = 0 length = len(s) while length < x: t = ord(s[idx])-ord('1') s += s[idx+1:]*t length += (length-idx-1)*t idx += 1 while idx != x: t = ord(s[idx])-ord('1') length += ((length-idx-1)*t)%M idx += 1 print(length%M) ```
output
1
104,540
0
209,081
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,541
0
209,082
Tags: implementation, math Correct Solution: ``` for _ in range(int(input())): x = int(input()) s = list(input()) l = 0 ans = len(s) while(l<x): if(ans>=x): break temp = len(s) ans+=(int(s[l])-1)*(temp-l-1) for i in range(int(s[l])-1): if(len(s)>x): break for j in range(l+1,temp): s.append(s[j]) if(len(s)>x): break l+=1 while(l<x): temp = ans ans+=(int(s[l])-1)*(temp-l-1) ans = ans%(10**9+7) l+=1 print(ans) ```
output
1
104,541
0
209,083
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,542
0
209,084
Tags: implementation, math Correct Solution: ``` M = 1000000007 t = int(input()) while t: t += -1 x = int(input()) s = input() i = 0 while len(s) < x: s += s[i + 1: ] * (int(s[i]) - 1) i += 1 n = len(s) for j in range(i, x): tmp = n - (j + 1) n = j + 1 + tmp * (int(s[j])) n = n % M print(n % M) ```
output
1
104,542
0
209,085
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,543
0
209,086
Tags: implementation, math Correct Solution: ``` def main(): TT = int(input()) for _ in range(TT): x = int(input()) s = [int(c) for c in input().strip()] ans = len(s) for l in range(0, x): ans += (ans - l - 1) * (s[l] - 1) ans %= 1000000007 end = len(s) for _ in range(1, s[l]): if len(s) > x + 1: break for c in range(l + 1, end): if len(s) > x + 1: break s.append(s[c]) print(ans) if __name__ == '__main__': main() ```
output
1
104,543
0
209,087
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,544
0
209,088
Tags: implementation, math Correct Solution: ``` ttt = int(input()) mod = 10**9 + 7 s = "#"*(10**6 + 1500) s=list(s) for bo in range(ttt): x=int(input()) st=input() s[1:len(st)+1]=list(st) siz=len(st) p = 0 while siz < x: p += 1 whe = siz for de4d in range(int(s[p])-1): for k in range(p+1,whe+1): siz += 1 s[siz] = s[k] if (siz>x): break if (siz>x): break if (siz>x): break ans = len(st) for i in range(1,x+1): ans = i + (ans-i)*int(s[i]) ans = ans%mod print(ans) ```
output
1
104,544
0
209,089
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,545
0
209,090
Tags: implementation, math Correct Solution: ``` t = int(input()) ans = [] arr = [None] * (10 ** 6) cur = 0 MOD = 10 ** 9 + 7 for _ in range(t): x, s = int(input()), list(map(int, str(input()))) max_p = cur + x for i in range(cur, min(max_p, cur + len(s))): arr[i] = s[i - cur] def copy(start, length, times): b = start + length for i in range(times): for j in range(length): if b >= max_p: return arr[b] = arr[start + j] b += 1 val = len(s) cnt = 1 while x: if arr[max_p - 1] is None and arr[cur] > 1: copy(cur + 1, val - cnt, arr[cur] - 1) val = (cnt + (val - cnt) * arr[cur]) % MOD cnt += 1 cur += 1 x -= 1 ans.append(val) for a in ans: print(a) ```
output
1
104,545
0
209,091
Provide tags and a correct Python 3 solution for this coding contest problem. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25.
instruction
0
104,546
0
209,092
Tags: implementation, math Correct Solution: ``` def solve(s,x): n = len(s) mod = 10**9 + 7 for i in range(x): n += (n-i-1)*(int(s[i])-1) n %= mod if len(s) < x: s += s[i+1:]*(int(s[i])-1) return n for _ in range(int(input())): x = int(input()) s = input() print(solve(s,x)) ```
output
1
104,546
0
209,093
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response. We start with a string s consisting only of the digits 1, 2, or 3. The length of s is denoted by |s|. For each i from 1 to |s|, the i-th character of s is denoted by s_i. There is one cursor. The cursor's location β„“ is denoted by an integer in \{0, …, |s|\}, with the following meaning: * If β„“ = 0, then the cursor is located before the first character of s. * If β„“ = |s|, then the cursor is located right after the last character of s. * If 0 < β„“ < |s|, then the cursor is located between s_β„“ and s_{β„“+1}. We denote by s_left the string to the left of the cursor and s_right the string to the right of the cursor. We also have a string c, which we call our clipboard, which starts out as empty. There are three types of actions: * The Move action. Move the cursor one step to the right. This increments β„“ once. * The Cut action. Set c ← s_right, then set s ← s_left. * The Paste action. Append the value of c to the end of the string s. Note that this doesn't modify c. The cursor initially starts at β„“ = 0. Then, we perform the following procedure: 1. Perform the Move action once. 2. Perform the Cut action once. 3. Perform the Paste action s_β„“ times. 4. If β„“ = x, stop. Otherwise, return to step 1. You're given the initial string s and the integer x. What is the length of s when the procedure stops? Since this value may be very large, only find it modulo 10^9 + 7. It is guaranteed that β„“ ≀ |s| at any time. Input The first line of input contains a single integer t (1 ≀ t ≀ 1000) denoting the number of test cases. The next lines contain descriptions of the test cases. The first line of each test case contains a single integer x (1 ≀ x ≀ 10^6). The second line of each test case consists of the initial string s (1 ≀ |s| ≀ 500). It is guaranteed, that s consists of the characters "1", "2", "3". It is guaranteed that the sum of x in a single file is at most 10^6. It is guaranteed that in each test case before the procedure will stop it will be true that β„“ ≀ |s| at any time. Output For each test case, output a single line containing a single integer denoting the answer for that test case modulo 10^9 + 7. Example Input 4 5 231 7 2323 6 333 24 133321333 Output 25 1438 1101 686531475 Note Let's illustrate what happens with the first test case. Initially, we have s = 231. Initially, β„“ = 0 and c = \varepsilon (the empty string). The following things happen if we follow the procedure above: * Step 1, Move once: we get β„“ = 1. * Step 2, Cut once: we get s = 2 and c = 31. * Step 3, Paste s_β„“ = 2 times: we get s = 23131. * Step 4: β„“ = 1 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 2. * Step 2, Cut once: we get s = 23 and c = 131. * Step 3, Paste s_β„“ = 3 times: we get s = 23131131131. * Step 4: β„“ = 2 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 3. * Step 2, Cut once: we get s = 231 and c = 31131131. * Step 3, Paste s_β„“ = 1 time: we get s = 23131131131. * Step 4: β„“ = 3 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 4. * Step 2, Cut once: we get s = 2313 and c = 1131131. * Step 3, Paste s_β„“ = 3 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 4 not= x = 5, so we return to step 1. * Step 1, Move once: we get β„“ = 5. * Step 2, Cut once: we get s = 23131 and c = 13113111311311131131. * Step 3, Paste s_β„“ = 1 times: we get s = 2313113113111311311131131. * Step 4: β„“ = 5 = x, so we stop. At the end of the procedure, s has length 25. Submitted Solution: ``` for _ in range(int(input())): x=int(input()) a=input() l=len(a) a=[i for i in a] ans=l p=0 m=10**9+7 while ans<x: for j in range(int(a[p])-1): for i in range(p+1,ans): a.append(a[i]) ans+=(int(a[p])-1)*((ans-p-1+m)%m) p+=1 for j in range(p,x): ans+=(int(a[j])-1)*((ans-j-1+m)%m) ans%=m print(ans) ```
instruction
0
104,547
0
209,094
Yes
output
1
104,547
0
209,095