message stringlengths 2 23.8k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 97 109k | cluster float64 0 0 | __index_level_0__ int64 194 217k |
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Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,179 | 0 | 112,358 |
"Correct Solution:
```
# -*- coding: utf-8 -*-
#############
# Libraries #
#############
import sys
input = sys.stdin.readline
import math
#from math import gcd
import bisect
import heapq
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache
#############
# Constants #
#############
MOD = 998244353
INF = float('inf')
AZ = "abcdefghijklmnopqrstuvwxyz"
#############
# Functions #
#############
######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]
######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()
#####Shorten#####
def DD(arg): return defaultdict(arg)
#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)
######Combination######
kaijo_memo = []
def kaijo(n):
if(len(kaijo_memo) > n):
return kaijo_memo[n]
if(len(kaijo_memo) == 0):
kaijo_memo.append(1)
while(len(kaijo_memo) <= n):
kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
return kaijo_memo[n]
gyaku_kaijo_memo = []
def gyaku_kaijo(n):
if(len(gyaku_kaijo_memo) > n):
return gyaku_kaijo_memo[n]
if(len(gyaku_kaijo_memo) == 0):
gyaku_kaijo_memo.append(1)
while(len(gyaku_kaijo_memo) <= n):
gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
return gyaku_kaijo_memo[n]
def nCr(n,r):
if(n == r):
return 1
if(n < r or r < 0):
return 0
ret = 1
ret = ret * kaijo(n) % MOD
ret = ret * gyaku_kaijo(r) % MOD
ret = ret * gyaku_kaijo(n-r) % MOD
return ret
######Factorization######
def factorization(n):
arr = []
temp = n
for i in range(2, int(-(-n**0.5//1))+1):
if temp%i==0:
cnt=0
while temp%i==0:
cnt+=1
temp //= i
arr.append([i, cnt])
if temp!=1:
arr.append([temp, 1])
if arr==[]:
arr.append([n, 1])
return arr
#####MakeDivisors######
def make_divisors(n):
divisors = []
for i in range(1, int(n**0.5)+1):
if n % i == 0:
divisors.append(i)
if i != n // i:
divisors.append(n//i)
return divisors
#####MakePrimes######
def make_primes(N):
max = int(math.sqrt(N))
seachList = [i for i in range(2,N+1)]
primeNum = []
while seachList[0] <= max:
primeNum.append(seachList[0])
tmp = seachList[0]
seachList = [i for i in seachList if i % tmp != 0]
primeNum.extend(seachList)
return primeNum
#####GCD#####
def gcd(a, b):
while b:
a, b = b, a % b
return a
#####LCM#####
def lcm(a, b):
return a * b // gcd (a, b)
#####BitCount#####
def count_bit(n):
count = 0
while n:
n &= n -1
count += 1
return count
#####ChangeBase#####
def base_10_to_n(X, n):
if X//n:
return base_10_to_n(X//n, n)+[X%n]
return [X%n]
def base_n_to_10(X, n):
return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))
#####IntLog#####
def int_log(n, a):
count = 0
while n>=a:
n //= a
count += 1
return count
#############
# Main Code #
#############
S,K = SL()
K = int(K)
data = []
ones = 0
count = 0
for s in S:
if s == "1":
count += 1
ones += 1
else:
data.append(count)
count = 0
data.append(count)
data = data[::-1]
N = len(data)
#print(data)
K = min(K,ones)
dp = [[[0 for j in range(K+1)] for k in range(K+1)] for i in range(N)]
for i in range(N):
dp[i][0][0] = 1
for k in range(min(data[0]+1,K+1)):
dp[0][k][k] = 1
#print(dp[0])
for i in range(1,N):
for k in range(K+1):
for j in range(K+1)[::-1]:
if j == K:
dp[i][k][j] = dp[i-1][k][j]
dp[i][k][j] %= MOD
else:
dp[i][k][j] = dp[i-1][k][j] + dp[i][k][j+1]
dp[i][k][j] %= MOD
a = data[i]
if a:
for k in range(K):
q = deque([dp[i-1][k][0]])
temp = dp[i-1][k][0]
for j in range(1,K+1):
#print(i,k,q)
if k+j > K: break
dp[i][k+j][j] += temp
dp[i][k+j][j] %= MOD
q.append(dp[i-1][k+j][j])
temp += dp[i-1][k+j][j]
if len(q) > a:
temp -= q.popleft()
temp %= MOD
#print(dp[i])
ans = 0
for k in range(K+1):
ans += dp[-1][k][0]
ans %= MOD
print(ans)
``` | output | 1 | 56,179 | 0 | 112,359 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,180 | 0 | 112,360 |
"Correct Solution:
```
s,k = input().split()
K = int(k)
if s[-1] == "1":
s += "0"
n = len(s)
ls = []
cnt = 0
for i in range(n):
if s[i] == "0":
ls.append(cnt)
cnt = 0
else:
cnt += 1
l = len(ls)
sm = sum(ls)
mod = 998244353
dp = [[[0 for i in range(sm+1)] for j in range(sm+1)] for k in range(l+1)]
dp[0][0][0] = 1
smprv = 0
for i in range(1,l+1):
num = ls[i-1]
for j in range(smprv+num,sm+1):
if j == smprv+num:
dp[i][j][0] = 1
for k in range(1,j+1):
x = 0
for m in range(j+1):
if num >= m:
x += dp[i-1][j-m][k]
else:
x += dp[i-1][j-m][k-(m-num)]
dp[i][j][k] = x%mod
smprv += num
ans = 0
for i in range(sm+1):
if i <= K:
ans += dp[-1][-1][i]
ans %= mod
print(ans)
``` | output | 1 | 56,180 | 0 | 112,361 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,181 | 0 | 112,362 |
"Correct Solution:
```
def main():
mod = 998244353
s, k = input().split()
k = int(k)
n = len(s)
one = s.count("1")
cnt = 0
zero_list = []
for i in range(n):
if s[i] == "0":
zero_list.append(cnt)
cnt = 0
else:
cnt += 1
z = 0
mm = min(one, k)
dp = [[0]*(one+1) for _ in [0]*(one+1)]
dp[0][0] = 1
for i in range(len(zero_list)):
dp2 = [[0]*(mm+1) for _ in [0]*(one+1)]
base = zero_list[i]
# j:何個今までに入れたか
for j in range(one+1):
# l:何個入れるか
for l in range(one+1-j):
if l < z+base-j:
continue
ml = max(l-base, 0)
# p:これまでのペナルティ
for p in range(min(one, k)+1):
q = p+ml
if q <= mm:
dp2[j+l][q] = (dp2[j+l][q]+dp[j][p]) % mod
else:
break
z += base
dp = dp2
print(sum([sum(i) for i in dp]) % mod)
main()
``` | output | 1 | 56,181 | 0 | 112,363 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,182 | 0 | 112,364 |
"Correct Solution:
```
S,Kstring = input().split()
K = int(Kstring)
lastZero = -1
segment = []
MOD = 998244353
N = len(S)
for i in range(N):
if S[i] == '0':
segment.append(i - lastZero - 1)
lastZero = i
segment.append(N - 1 - lastZero)
M = len(segment)
No1 = N + 1 - M
partialSumSegment = []
for i in range(M):
if partialSumSegment:
partialSumSegment.append(segment[i] + partialSumSegment[-1])
else:
partialSumSegment.append(segment[i])
DP = []
DPSum = [] # Sum of DP i,j,k ~ i,j,a
for i in range(M):
DP.append([])
DPSum.append([])
for j in range(No1 + 2):
DP[i].append([])
DPSum[i].append([])
for k in range(No1 + 2):
DP[i][j].append(0)
DPSum[i][j].append(0)
# Initialize on the last segment
for j in range(No1 - partialSumSegment[M-2] + 1):
DP[M-1][j][partialSumSegment[M-2] + j] = 1
for j in range(No1 - partialSumSegment[M-2] + 1):
for k in range(partialSumSegment[M-2] + j,-1,-1):
DPSum[M-1][j][k] = DPSum[M-1][j][k+1] + DP[M-1][j][k]
for i in range(M-2,-1,-1):
for j in range(No1 + 1):
Kmin = partialSumSegment[i-1]
if i == 0:
Kmin = 0
for k in range(No1, Kmin - 1, -1):
DPCur = 0
for l in range(1,segment[i] + 1):
if j >= l and k + segment[i] - l <= No1 + 1:
DPCur += DP[i + 1][j - l][k + segment[i] - l]
if k + segment[i] <= No1:
DPCur += DPSum[i + 1][j][k + segment[i]]
DP[i][j][k] = DPCur % MOD
DPSum[i][j][k] = (DPSum[i][j][k+1] + DP[i][j][k]) % MOD
# print(DP)
# print(DPSum)
AnsSum = 0
for j in range(min(K,No1) + 1):
AnsSum += DP[0][j][0]
if M == 1:
print(1)
else:
print(AnsSum % MOD)
``` | output | 1 | 56,182 | 0 | 112,365 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,183 | 0 | 112,366 |
"Correct Solution:
```
s,k = input().split()
mod = 998244353
k = int(k)
k = min(300,k)
raw = [0]
l = len(s)
for i in range(l):
if s[i] == '0':
raw.append(0)
else:
raw[-1] += 1
#print(raw)
now = [[0 for _ in range(k+1)] for _ in range(k+1)]
l = len(raw)
raw.reverse()
#print(raw)
now[0][0] = 1
for x in raw:
last = now[:]
now = []
for i in range(k+1):
use = last[i][:]
use.reverse()
cum = []
for j in range(k+1):
if cum:
cum.append(cum[-1]+use[j])
cum[-1] %= mod
else:
cum = [use[0]]
cum.reverse()
now.append(cum)
#print(cum)
#print('#')
cum2 = []
for i in range(k+1):
cum = [0 for _ in range(i)]
cum.append(last[i][0])
#print('%',cum)
for l in range(i+1,k+1):
cum.append(cum[-1]+last[l][l-i])
#print(cum)
cum2.append(cum)
for i in range(k+1):
for j in range(k+1):
if j > i:
pass
now[i][j] += cum2[i-j][i]
if i - x - 1 >= 0:
now[i][j] -= cum2[i-j][i-x-1]
now[i][j] %= mod
for i in range(k+1):
for j in range(k+1):
now[i][j] -= last[i][j]
now[i][j] %= mod
#print(now)
#print('##')
#print(now)
#print(now)
ans = 0
for i in range(k+1):
ans += now[i][0]
print(ans%mod)
``` | output | 1 | 56,183 | 0 | 112,367 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,184 | 0 | 112,368 |
"Correct Solution:
```
S,K = input().split()
A = list(map(len,S.split("0")))
A[0] = 0
A.append(0)
N = len(A)
K = int(K)
K = min(K,S.count("1"))
MOD = 998244353
dp = [[[0]*(K+1) for _ in [0]*(K+1)] for _ in [0]*N]
dp[-1][0][0] = 1
for i in range(1,N):
i = N-1-i
for k in range(K+1):
for n in range(K+1):
dp[i][k][n] = sum(dp[i+1][k][n:])%MOD
for a in range(1,A[i]+1):
if k-a < 0 or n-a < 0: break
dp[i][k][n] += dp[i+1][k-a][n-a]
dp[i][k][n] %= MOD
ans = sum(dp[0][k][0] for k in range(K+1))
ans %= MOD
print(ans)
``` | output | 1 | 56,184 | 0 | 112,369 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,185 | 0 | 112,370 |
"Correct Solution:
```
def main():
mod = 998244353
s, k = input().split()
k, n, o, cnt, z, zero_list = int(k), len(s), s.count("1")+1, 0, 0, []
for i in range(n):
if s[i] == "0":
zero_list.append(cnt)
cnt = 0
else:
cnt += 1
m = min(o, k+1)
dp = [[0]*m for _ in [0]*o]
dp[0][0] = 1
for i in zero_list:
dp2 = [[0]*m for _ in [0]*o]
dp3 = [[0] for _ in [0]*o]
for x in range(o):
t = 0
for y in range(min(m, o-x)):
t = (t+dp[x+y][y]) % mod
dp3[x].append(t)
dp4 = [[0]*m for _ in [0]*o]
for y in range(m):
t = dp[0][y]
for x in range(1, o):
dp4[x-1][y] = t
t = (t+dp[x][y]) % mod
dp4[o-1][y] = t
for j in range(z+i, o):
for x in range(min(j-i+1, m)):
dp2[j][x] = (dp2[j][x]+dp3[j-x-i][min(o, x+1)]) % mod
for p in range(m):
dp2[j][p] = (dp2[j][p]+dp4[j][p]-dp4[max(j-i, p-1)][p]) % mod
z += i
dp = dp2
print(sum([sum(i) for i in dp]) % mod)
main()
``` | output | 1 | 56,185 | 0 | 112,371 |
Provide a correct Python 3 solution for this coding contest problem.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815 | instruction | 0 | 56,186 | 0 | 112,372 |
"Correct Solution:
```
s,k=input().split();k,o,c,z,L,O=int(k),s.count("1")+1,0,0,[],998244353;M,r=min(o,k+1),range
for i in s:
if i=="0":L+=[c];c=0
else:c+=1
d=[[0]*M for _ in r(o)];d[0][0]=1
for i in L:
D=[[0]*M for _ in r(o)]
for j in r(o):
for l in r(max(z+i-j,0),o-j):
m=max(l-i,0)
for p in r(min(j+1,M-m)):D[j+l][p+m]=(D[j+l][p+m]+d[j][p])%O
z+=i;d=D
print(sum([sum(i)for i in d])%O)
``` | output | 1 | 56,186 | 0 | 112,373 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**7) #再帰関数の上限,10**5以上の場合python
import math
from copy import copy, deepcopy
from copy import deepcopy as dcp
from operator import itemgetter
from bisect import bisect_left, bisect, bisect_right#2分探索
#bisect_left(l,x), bisect(l,x)#aはソート済みである必要あり。aの中からx未満の要素数を返す。rightだと以下
from collections import deque
#deque(l), pop(), append(x), popleft(), appendleft(x)
##listでqueの代用をするとO(N)の計算量がかかってしまうので注意
from collections import Counter#文字列を個数カウント辞書に、
#S=Counter(l),S.most_common(x),S.keys(),S.values(),S.items()
from itertools import accumulate,combinations,permutations#累積和
#list(accumulate(l))
from heapq import heapify,heappop,heappush
#heapify(q),heappush(q,a),heappop(q) #q=heapify(q)としないこと、返り値はNone
#import fractions#古いatcoderコンテストの場合GCDなどはここからimportする
from functools import lru_cache#pypyでもうごく
#@lru_cache(maxsize = None)#maxsizeは保存するデータ数の最大値、2**nが最も高効率
from decimal import Decimal
def input():
x=sys.stdin.readline()
return x[:-1] if x[-1]=="\n" else x
def printl(li): _=print(*li, sep="\n") if li else None
def argsort(s, return_sorted=False):
inds=sorted(range(len(s)), key=lambda k: s[k])
if return_sorted: return inds, [s[i] for i in inds]
return inds
def alp2num(c,cap=False): return ord(c)-97 if not cap else ord(c)-65
def num2alp(i,cap=False): return chr(i+97) if not cap else chr(i+65)
def matmat(A,B):
K,N,M=len(B),len(A),len(B[0])
return [[sum([(A[i][k]*B[k][j]) for k in range(K)]) for j in range(M)] for i in range(N)]
def matvec(M,v):
N,size=len(v),len(M)
return [sum([M[i][j]*v[j] for j in range(N)]) for i in range(size)]
def T(M):
n,m=len(M),len(M[0])
return [[M[j][i] for j in range(n)] for i in range(m)]
def main():
mod = 998244353
#w.sort(key=itemgetter(1),reversed=True) #二個目の要素で降順並び替え
#N = int(input())
#N, K = map(int, input().split())
#A = tuple(map(int, input().split())) #1行ベクトル
#L = tuple(int(input()) for i in range(N)) #改行ベクトル
#S = tuple(tuple(map(int, input().split())) for i in range(N)) #改行行列
s,K=input().split()
K=int(K)
l=len(s)
if K>l:
K=l
sep=[]
count=0
for i in range(l):
if s[i]=="1": count+=1
else:
sep.append(count)
count=0
sep.append(count)
dp=[[[0]*(l+1) for _ in range(K+1)] for _ in range(len(sep))]
#dp[i][j][k] 末尾からi番目の領域、j個の0をストック,k回の操作
dp[0][0][0]=1
for i in range(1,min(sep[-1]+1,K+1)):
dp[0][i][i]=1
dp20=[[0]*(l+1) for _ in range(K+1)]
for i in range(1,len(sep)):
ori=sep[-1-i]
dp2=deepcopy(dp20)
for k in range(K+1):
dp[i][k][-1]=dp[i-1][k][-1]
for j in range(1,l+1):
dp[i][k][-1-j]=(dp[i][k][-j]+dp[i-1][k][-1-j])%mod
if k==0 or j==0: continue
dp2[k][j]=(dp2[k-1][j-1]+dp[i-1][k-1][j-1])%mod
if j>ori and k>ori:
dp2[k][j]-=dp[i-1][k-ori-1][j-ori-1]
dp2[k][j]%=mod
for k in range(K+1):
for j in range(1,l+1):
dp[i][k][j]=(dp[i][k][j]+dp2[k][j])%mod
ans=0
for k in range(K+1):
ans+=dp[-1][k][0]
ans%=mod
print(ans)
#print(dp)
if __name__ == "__main__":
main()
``` | instruction | 0 | 56,187 | 0 | 112,374 |
Yes | output | 1 | 56,187 | 0 | 112,375 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Given is a string S consisting of `0` and `1`. Find the number of strings, modulo 998244353, that can result from applying the following operation on S between 0 and K times (inclusive):
* Choose a pair of integers i, j (1\leq i < j\leq |S|) such that the i-th and j-th characters of S are `0` and `1`, respectively. Remove the j-th character from S and insert it to the immediate left of the i-th character.
Constraints
* 1 \leq |S| \leq 300
* 0 \leq K \leq 10^9
* S consists of `0` and `1`.
Input
Input is given from Standard Input in the following format:
S K
Output
Find the number of strings, modulo 998244353, that can result from applying the operation on S between 0 and K times (inclusive).
Examples
Input
0101 1
Output
4
Input
01100110 2
Output
14
Input
1101010010101101110111100011011111011000111101110101010010101010101 20
Output
113434815
Submitted Solution:
```
# coding: utf-8
# Your code here!
import sys
read = sys.stdin.read
readline = sys.stdin.readline
#a,b,c,d = map(int,readline().split())
s,k = readline().split()
a = [len(i) for i in s.split("0")]
while a and a[-1] == 0: a.pop()
if not a:
print(1)
exit()
MOD = 998244353
M = sum(a)+1
k = min(int(k),M)
dp = [[0]*M for _ in range(k+1)] # j 使って（上限 k）、l 余ってる
dp[0][0] = 1
#print(a)
for ai in a[::-1]:
ndp = [[0]*M for _ in range(k+1)] # j 使って（上限 k）、l 余ってる
for j in range(k+1):
for l in range(M):
for ll in range(l):
ndp[j][ll] += dp[j][l]
ndp[j][ll] %= MOD
V = min(M-l,k-j+1,ai+1)
for i in range(V):
#if j+i > k: break
ndp[j+i][l+i] += dp[j][l]
ndp[j+i][l+i] %= MOD
dp = ndp
#print(dp)
ans = 0
for jj in range(k+1):
ans += dp[jj][0]
print(ans%MOD)
``` | instruction | 0 | 56,188 | 0 | 112,376 |
Yes | output | 1 | 56,188 | 0 | 112,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The length of the longest common prefix of two strings s=s_1 s_2 … s_n and t = t_1 t_2 … t_m is defined as the maximum k ≤ min(n, m) such that s_1 s_2 … s_k equals t_1 t_2 … t_k. Let's denote the longest common prefix of two strings s and t as lcp(s,t).
Z-function of a string s_1 s_2 ... s_n is a sequence of integers z_1, z_2, …, z_n, where z_i = lcp(s_1 s_2 … s_n,\ \ s_i s_{i+1} ... s_n). Ж-function of a string s is defined as z_1 + z_2 + … + z_n.
You're given a string s=s_1 s_2 … s_n and q queries. Each query is described by two integers l_i and r_i, where 1 ≤ l_i ≤ r_i ≤ n. The answer for the query is defined as Ж-function of the string s_{l_i} s_{l_i +1} … s_{r_i}.
Input
The first line contains the string s, consisting of lowercase English letters (1 ≤ |s| ≤ 200 000). The second line contains one integer q — the number of queries (1 ≤ q ≤ 200 000). Each of the following q lines contains two integers l_i and r_i, describing the query (1 ≤ l_i ≤ r_i ≤ |s|).
Output
For every query output one integer: the value of Ж-function of the corresponding substring.
Examples
Input
abbd
4
2 3
1 3
3 3
1 4
Output
3
3
1
4
Input
bbaaa
5
2 4
1 5
1 5
3 3
1 2
Output
3
6
6
1
3
Note
In the first sample case there are four queries:
* the first query corresponds to the substring bb, and its Ж-function equals 2 + 1 = 3;
* the second query corresponds to the substring abb, and its Ж-function equals 3 + 0 + 0 = 3;
* the third query corresponds to the substring b, and its Ж-function equals 1.
* the fourth query corresponds to the substring abdd, and its Ж-function equals 4 + 0 + 0 + 0= 4.
Submitted Solution:
```
from sys import stdin, stdout
s = stdin.readline().rstrip()
q = int(stdin.readline().rstrip())
arr = []
for i in range(q):
arr.append([int(x) for x in stdin.readline().rstrip().split()])
def lcp(a,b):
if a[:len(b)]==b:
return len(b)
else:
return 0
def g(s):
res = 0
for i in range(len(s)):
res+=lcp(s,s[i:])
return res
def substr(s,arr):
res = []
for diap in arr:
res.append(s[diap[0]-1:diap[1]])
return res
for i in substr(s,arr):
print(g(i))
``` | instruction | 0 | 56,424 | 0 | 112,848 |
No | output | 1 | 56,424 | 0 | 112,849 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The length of the longest common prefix of two strings s=s_1 s_2 … s_n and t = t_1 t_2 … t_m is defined as the maximum k ≤ min(n, m) such that s_1 s_2 … s_k equals t_1 t_2 … t_k. Let's denote the longest common prefix of two strings s and t as lcp(s,t).
Z-function of a string s_1 s_2 ... s_n is a sequence of integers z_1, z_2, …, z_n, where z_i = lcp(s_1 s_2 … s_n,\ \ s_i s_{i+1} ... s_n). Ж-function of a string s is defined as z_1 + z_2 + … + z_n.
You're given a string s=s_1 s_2 … s_n and q queries. Each query is described by two integers l_i and r_i, where 1 ≤ l_i ≤ r_i ≤ n. The answer for the query is defined as Ж-function of the string s_{l_i} s_{l_i +1} … s_{r_i}.
Input
The first line contains the string s, consisting of lowercase English letters (1 ≤ |s| ≤ 200 000). The second line contains one integer q — the number of queries (1 ≤ q ≤ 200 000). Each of the following q lines contains two integers l_i and r_i, describing the query (1 ≤ l_i ≤ r_i ≤ |s|).
Output
For every query output one integer: the value of Ж-function of the corresponding substring.
Examples
Input
abbd
4
2 3
1 3
3 3
1 4
Output
3
3
1
4
Input
bbaaa
5
2 4
1 5
1 5
3 3
1 2
Output
3
6
6
1
3
Note
In the first sample case there are four queries:
* the first query corresponds to the substring bb, and its Ж-function equals 2 + 1 = 3;
* the second query corresponds to the substring abb, and its Ж-function equals 3 + 0 + 0 = 3;
* the third query corresponds to the substring b, and its Ж-function equals 1.
* the fourth query corresponds to the substring abdd, and its Ж-function equals 4 + 0 + 0 + 0= 4.
Submitted Solution:
```
I=input
I()
_=int(I())
for x in range(_):
x=list(map(int,I().split()))
if x[0] == x[1]:print(1)
elif _%2==0:print(max(x))
else:
if x[0]%2==0 and x[1]%2==0:print(x[0]+1)
else:print(sum(x))
``` | instruction | 0 | 56,425 | 0 | 112,850 |
No | output | 1 | 56,425 | 0 | 112,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The length of the longest common prefix of two strings s=s_1 s_2 … s_n and t = t_1 t_2 … t_m is defined as the maximum k ≤ min(n, m) such that s_1 s_2 … s_k equals t_1 t_2 … t_k. Let's denote the longest common prefix of two strings s and t as lcp(s,t).
Z-function of a string s_1 s_2 ... s_n is a sequence of integers z_1, z_2, …, z_n, where z_i = lcp(s_1 s_2 … s_n,\ \ s_i s_{i+1} ... s_n). Ж-function of a string s is defined as z_1 + z_2 + … + z_n.
You're given a string s=s_1 s_2 … s_n and q queries. Each query is described by two integers l_i and r_i, where 1 ≤ l_i ≤ r_i ≤ n. The answer for the query is defined as Ж-function of the string s_{l_i} s_{l_i +1} … s_{r_i}.
Input
The first line contains the string s, consisting of lowercase English letters (1 ≤ |s| ≤ 200 000). The second line contains one integer q — the number of queries (1 ≤ q ≤ 200 000). Each of the following q lines contains two integers l_i and r_i, describing the query (1 ≤ l_i ≤ r_i ≤ |s|).
Output
For every query output one integer: the value of Ж-function of the corresponding substring.
Examples
Input
abbd
4
2 3
1 3
3 3
1 4
Output
3
3
1
4
Input
bbaaa
5
2 4
1 5
1 5
3 3
1 2
Output
3
6
6
1
3
Note
In the first sample case there are four queries:
* the first query corresponds to the substring bb, and its Ж-function equals 2 + 1 = 3;
* the second query corresponds to the substring abb, and its Ж-function equals 3 + 0 + 0 = 3;
* the third query corresponds to the substring b, and its Ж-function equals 1.
* the fourth query corresponds to the substring abdd, and its Ж-function equals 4 + 0 + 0 + 0= 4.
Submitted Solution:
```
lstr = list(input())
for _ in range(int(input())):
l,r = map(int,input().split())
counter = 0
lst1 = lstr[l-1:r]
for i in range(l-1,r):
lst2 = lstr[i:r]
for k in range(len(lst2)):
print(lst2[k],lst1[k])
if lst2[k] != lst1[k]:
break
else:
counter+=1
print(counter)
``` | instruction | 0 | 56,426 | 0 | 112,852 |
No | output | 1 | 56,426 | 0 | 112,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,471 | 0 | 112,942 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
s = input()[::-1]
ans=''
c=0
for i in s:
if i=='0':
ans+='0'
c+=1
elif c>0:
ans+='1'
c-=1
else:
ans+='0'
print(ans[::-1])
``` | output | 1 | 56,471 | 0 | 112,943 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,472 | 0 | 112,944 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
''' Hey stalker :) '''
INF = 10**10
def main():
print = out.append
''' Cook your dish here! '''
st = list(input())
stk = []
for index, i in enumerate(st):
if i=='0' and len(stk)>0 and stk[-1][0]=='1':
stk.pop()
else: stk.append([i, index])
for li in stk:
st[li[1]] = '0'
print("".join(st))
''' Pythonista fLite 1.1 '''
import sys
#from collections import defaultdict, Counter
#from bisect import bisect_left, bisect_right
#from functools import reduce
#import math
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
out = []
get_int = lambda: int(input())
get_list = lambda: list(map(int, input().split()))
main()
#[main() for _ in range(int(input()))]
print(*out, sep='\n')
``` | output | 1 | 56,472 | 0 | 112,945 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,473 | 0 | 112,946 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
import sys
input = sys.stdin.readline
def solve():
i = 0
s = list(input().strip())
st = []
for c in s:
if c == '1':
st.append(i)
elif len(st) > 0:
st.pop()
i += 1
for i in st:
s[i] = '0'
print(''.join(s))
solve()
``` | output | 1 | 56,473 | 0 | 112,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,474 | 0 | 112,948 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
s=[int(x) for x in list(input())]
n=len(s)
b=[0]*n
counter=0
for i in range(n-1,-1,-1):
if s[i]==0:
counter+=1
elif counter>0:
counter-=1
else:
s[i]=0
arr=''
for item in s:
arr+=str(item)
print(arr)
``` | output | 1 | 56,474 | 0 | 112,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,475 | 0 | 112,950 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
s=input()
a,b,c=[],[],[]
count=0
for i in range(len(s)-1,-1,-1):
a.append(s[i])
for i in range(len(a)):
if(a[i]=='0'):
count+=1
b.append('0')
elif(a[i]=='1' and count>0):
count-=1
b.append('1')
elif(a[i]=='1' and count==0):
b.append('0')
for i in range(len(b)-1,-1,-1):
c.append(b[i])
ans=''.join(c)
print(ans)
``` | output | 1 | 56,475 | 0 | 112,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,476 | 0 | 112,952 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
s = input()
t = list(s)
stack =[]
for i in range(len(s)):
if t[i]=='1':
stack.append(i)
elif len(stack):
stack.pop()
for i in range(len(stack)):
t[stack[i]] = '0'
print(''.join(t))
``` | output | 1 | 56,476 | 0 | 112,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,477 | 0 | 112,954 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
s=input()
ans=['0' for i in range(len(s))]
st=[]
for i in range(len(s)):
if s[i]=='0':
if len(st):
ans[st.pop()]='1'
else:
st.append(i)
print(''.join(ans))
``` | output | 1 | 56,477 | 0 | 112,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one. | instruction | 0 | 56,478 | 0 | 112,956 |
Tags: data structures, greedy, math, strings
Correct Solution:
```
S = input()
N = len(S)
p = q = 0
prv = S[0]
c = 0
C = []
if S[0] == '1':
C.append(0)
for ch in S:
if ch != prv:
C.append(c)
c = 1
prv = ch
else:
c += 1
C.append(c)
ans = []
r = 0
L = len(C)
for i in range(L-1, -1, -1):
c = C[i]
if i % 2:
r += c
if i != L-1:
m = max(min(c-1, r), 0)
else:
m = max(min(c, r), 0)
if m:
r -= m
ans.append("0"*m + "1"*(c-m))
else:
ans.append("1"*c)
else:
r -= c
ans.append("0"*c)
ans.reverse()
print(*ans, sep='')
``` | output | 1 | 56,478 | 0 | 112,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
S = list(map(int,input().strip()))
N = len(S)
stack = []
for i in range(N):
s = S[i]
if s == 0 and stack and stack[-1][0] == 1:
stack.pop()
else:
stack.append((s, i))
T = S[:]
if stack:
for i in tuple(map(list, zip(*stack)))[1]:
T[i] = 0
print(''.join(map(str, T)))
``` | instruction | 0 | 56,479 | 0 | 112,958 |
Yes | output | 1 | 56,479 | 0 | 112,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
s = str(input().strip())
t = list(s[::-1])
cnt = 0
for i,v in enumerate(t):
if v == '0':
cnt += 1
else:
if cnt:
cnt -= 1
else:
t[i] = '0'
print("".join(t[::-1]))
``` | instruction | 0 | 56,480 | 0 | 112,960 |
Yes | output | 1 | 56,480 | 0 | 112,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
s=input()
n=len(s)
tot=0
S=list(s)
for i in range(n-1,-1,-1):
if s[i]=='0':
tot+=1
elif tot:
tot-=1
else:
S[i]='0'
print(''.join(S))
``` | instruction | 0 | 56,481 | 0 | 112,962 |
Yes | output | 1 | 56,481 | 0 | 112,963 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
pp = input()
if len(pp) == 1:
print('0')
exit(0)
z = 1 if pp[0]=='0' else 0
zc = [z]
l = 1
lndl = [l]
for p in pp[1:]:
l = max(z + 1, l + (1 if p == '1' else 0))
z += 1 if p == '0' else 0
lndl.append(l)
zc.append(z)
lnda = lndl[-1]
o = 1 if pp[-1]=='1' else 0
oc = [o]
l = 1
lndr = [l]
for p in reversed(pp[:-1]):
l = max(o + 1, l + (1 if p == '0' else 0))
o += 1 if p == '1' else 0
lndr.append(l)
oc.append(o)
oc.reverse()
lndr.reverse()
qq = []
ez = 0
if pp[0] == '1':
if max(oc[1], lndr[1] + 1) != lnda :
qq.append('1')
else:
qq.append('0')
ez += 1
else:
qq.append('0')
for p, l, o, z, r in zip(pp[1:-1],lndl, oc[2:], zc, lndr[2:]):
if p == '1':
if max(l + o, z + ez + 1 + r) != lnda:
qq.append('1')
else:
qq.append('0')
ez += 1
else:
qq.append('0')
qq.append('0')
print(''.join(qq))
``` | instruction | 0 | 56,482 | 0 | 112,964 |
Yes | output | 1 | 56,482 | 0 | 112,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
s=input()
a,b=[],[]
for i in s:
a.append(int(i))
for i in range(len(a)-1):
if(a[i]<=a[i+1]):
b.append('0')
else:
b.append('1')
b.append('0')
c=''.join(b)
print(c)
``` | instruction | 0 | 56,483 | 0 | 112,966 |
No | output | 1 | 56,483 | 0 | 112,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
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")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#-------------------------------------------------------------------------
prime = [True for i in range(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=0
while (left <= right):
mid = (right + left)//2
if (arr[mid][0] > key):
right = mid-1
else:
res=mid
left = mid + 1
return res
#---------------------------------running code------------------------------------------
s=input()
n=len(s)
a=[int(i) for i in s]
one=[]
zero=[]
for i in range (n):
if a[i]==0:
zero.append(i)
else:
one.append(i)
j=len(zero)-1
for i in one:
if j>=0 and zero[j]>i:
j-=1
else:
a[i]=0
print(*a,sep='')
``` | instruction | 0 | 56,484 | 0 | 112,968 |
No | output | 1 | 56,484 | 0 | 112,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
s = input()
res = ""
min_dif = 0
for i in range(len(s)):
if s[len(s)-i-1] == "0":
res = "0"+res
min_dif = min([-1, min_dif-1])
else:
if min_dif >= 0: res = "0"+res
else: res = "1"+res
min_dif = min([1, min_dif+1])
print(res)
``` | instruction | 0 | 56,485 | 0 | 112,970 |
No | output | 1 | 56,485 | 0 | 112,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the length of the string. You can hack this problem if you solve it. But you can hack the previous problem only if you solve both problems.
Kirk has a binary string s (a string which consists of zeroes and ones) of length n and he is asking you to find a binary string t of the same length which satisfies the following conditions:
* For any l and r (1 ≤ l ≤ r ≤ n) the length of the longest non-decreasing subsequence of the substring s_{l}s_{l+1} … s_{r} is equal to the length of the longest non-decreasing subsequence of the substring t_{l}t_{l+1} … t_{r};
* The number of zeroes in t is the maximum possible.
A non-decreasing subsequence of a string p is a sequence of indices i_1, i_2, …, i_k such that i_1 < i_2 < … < i_k and p_{i_1} ≤ p_{i_2} ≤ … ≤ p_{i_k}. The length of the subsequence is k.
If there are multiple substrings which satisfy the conditions, output any.
Input
The first line contains a binary string of length not more than 10^5.
Output
Output a binary string which satisfied the above conditions. If there are many such strings, output any of them.
Examples
Input
110
Output
010
Input
010
Output
010
Input
0001111
Output
0000000
Input
0111001100111011101000
Output
0011001100001011101000
Note
In the first example:
* For the substrings of the length 1 the length of the longest non-decreasing subsequnce is 1;
* For l = 1, r = 2 the longest non-decreasing subsequnce of the substring s_{1}s_{2} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{2} is 01;
* For l = 1, r = 3 the longest non-decreasing subsequnce of the substring s_{1}s_{3} is 11 and the longest non-decreasing subsequnce of the substring t_{1}t_{3} is 00;
* For l = 2, r = 3 the longest non-decreasing subsequnce of the substring s_{2}s_{3} is 1 and the longest non-decreasing subsequnce of the substring t_{2}t_{3} is 1;
The second example is similar to the first one.
Submitted Solution:
```
'''
Author : thekushalghosh
'''
a = input()
q = []
s = ""
for i in range(len(a) - 1):
if a[i] == "1" and a[i + 1] == "1":
s = s + "0"
else:
s = s + a[i]
s = s + a[i + 1]
print(s)
``` | instruction | 0 | 56,486 | 0 | 112,972 |
No | output | 1 | 56,486 | 0 | 112,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a string s of lowercase English letters. A cursor is positioned on one of the characters. The cursor can be moved with the following operation: choose a letter c and a direction (left or right). The cursor is then moved to the closest occurence of c in the chosen direction. If there is no letter c in that direction, the cursor stays in place. For example, if s = abaab with the cursor on the second character (a[b]aab), then:
* moving to the closest letter a to the left places the cursor on the first character ([a]baab);
* moving to the closest letter a to the right places the cursor the third character (ab[a]ab);
* moving to the closest letter b to the right places the cursor on the fifth character (abaa[b]);
* any other operation leaves the cursor in place.
Let dist(i, j) be the smallest number of operations needed to move the cursor from the i-th character to the j-th character. Compute \displaystyle ∑_{i = 1}^n ∑_{j = 1}^n dist(i, j).
Input
The only line contains a non-empty string s of at most 10^5 lowercase English letters.
Output
Print a single integer \displaystyle ∑_{i = 1}^n ∑_{j = 1}^n dist(i, j).
Examples
Input
abcde
Output
20
Input
abacaba
Output
58
Note
In the first sample case, dist(i, j) = 0 for any pair i = j, and 1 for all other pairs.
Submitted Solution:
```
def min_dis(string, sub, start):
l_part = start - string.find(sub, 0, start - 1) * (string.find(sub, 0, start - 1) > 0)
r_part = string.find(sub, start) *(string.find(sub, start) > 0) - start
if l_part > 0 and r_part > 0:
n = min(l_part, r_part)
elif l_part == 0 or r_part == 0:
n = 0
else:
n = max(l_part, r_part)
if n > 0:
return n
else:
return 0
s=input ("Please enter the string:\n")
# s = "abacaba"
al = set()
for i in range(len(s)):
al.add(s[i])
al = list(al)
# print(al)
count = 0
if len(s) == len(al):
print(len(s) * (len(s) - 1))
else:
for i in range(len(s)):
for j in range(len(al)):
count += min_dis(s, al[j], i)
print(count)
``` | instruction | 0 | 56,503 | 0 | 113,006 |
No | output | 1 | 56,503 | 0 | 113,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,592 | 0 | 113,184 |
Tags: constructive algorithms, greedy
Correct Solution:
```
for _ in range(int(input())):
#n = int(input())
n,k=map(int, input().split())
#arr=list(map(int, input().split()))
s="abc"*(n//3)
if n%3==1:
s+="a"
elif n%3==2:
s+="ab"
print(s)
``` | output | 1 | 56,592 | 0 | 113,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,593 | 0 | 113,186 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
for i in range(n):
if i%3 == 1:
print("a",end="")
if i%3 == 2:
print("b",end="")
if i%3 == 0:
print("c",end="")
print()
``` | output | 1 | 56,593 | 0 | 113,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,594 | 0 | 113,188 |
Tags: constructive algorithms, greedy
Correct Solution:
```
"""
// Author : snape_here - Susanta Mukherjee
"""
from __future__ import division, print_function
import os,sys
from io import BytesIO, IOBase
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
def ii(): return int(input())
def fi(): return float(input())
def si(): return input()
def msi(): return map(str,input().split())
def mi(): return map(int,input().split())
def li(): return list(mi())
def read():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def gcd(x, y):
while y:
x, y = y, x % y
return x
def lcm(x, y):
return (x*y)//(gcd(x,y))
mod=1000000007
def modInverse(b,m):
g = gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
# def ceil(x,y):
# if x%y==0:
# return x//y
# else:
# return x//y+1
def modu(a,b,m):
a = a % m
inv = modInverse(b,m)
if(inv == -1):
return -999999999
else:
return (inv*a)%m
from math import log,sqrt,factorial,cos,tan,sin,radians,floor,ceil,log2
import bisect
from decimal import *
getcontext().prec = 8
abc="abcdefghijklmnopqrstuvwxyz"
pi=3.141592653589793238
def main():
for _ in range(ii()):
n,k=mi()
s="abc"
r=n%3
q=n//3
ans=s*q+s[0:r]
print(ans)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
#read()
main()
``` | output | 1 | 56,594 | 0 | 113,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,595 | 0 | 113,190 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
n,k=map(int,input().split())
for i in range(n):
print(chr(ord('a')+i%3),end='')
print()
``` | output | 1 | 56,595 | 0 | 113,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,596 | 0 | 113,192 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t = int(input())
def solve():
n,k = map(int, input().split())
ans = []
i = 0
done = False
while(True):
if(done):
break
for c in "abc":
ans.append(c)
i+=1
if(i == n):
done = True
break
print("".join(ans))
for i in range(t):
solve()
``` | output | 1 | 56,596 | 0 | 113,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,597 | 0 | 113,194 |
Tags: constructive algorithms, greedy
Correct Solution:
```
t=int(input())
import math
for i in range(0,t):
nm = input().split()
n = int(nm[0])
k = int(nm[1])
k_1 = 3*k
x= math.ceil(n/k_1)
string= ""
output_string = ""
out_1 = ""
for i in range(0,k):
string = string+"a"
for i in range(k,2*k):
string = string+"b"
for i in range(2*k , 3*k):
string = string+"c"
for i in range(0,x):
output_string += string
out_1 = output_string[:n]
print(out_1)
``` | output | 1 | 56,597 | 0 | 113,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,598 | 0 | 113,196 |
Tags: constructive algorithms, greedy
Correct Solution:
```
def read_int():
return int(input())
def read_ints():
return map(int, input().split(' '))
t = read_int()
for case_num in range(t):
n, k = read_ints()
print(('abc' * n)[:n])
``` | output | 1 | 56,598 | 0 | 113,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one. | instruction | 0 | 56,599 | 0 | 113,198 |
Tags: constructive algorithms, greedy
Correct Solution:
```
import sys
import math
import bisect
from sys import stdin, stdout
from math import gcd, floor, sqrt, log
from collections import defaultdict as dd
from bisect import bisect_left as bl, bisect_right as br
from collections import Counter
from collections import defaultdict as dd
# sys.setrecursionlimit(100000000)
flush = lambda: stdout.flush()
stdstr = lambda: stdin.readline()
stdint = lambda: int(stdin.readline())
stdpr = lambda x: stdout.write(str(x))
stdmap = lambda: map(int, stdstr().split())
stdarr = lambda: list(map(int, stdstr().split()))
mod = 1000000007
for _ in range(stdint()):
n,k = stdmap()
l = ["a", "b", "c"]
p = 0
res = []
curr = 0
for i in range(n):
if(curr == k):
p = (p+1)%3
curr = 0
res.append(l[p])
curr += 1
print("".join(res))
``` | output | 1 | 56,599 | 0 | 113,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
t=int(input())
for _ in range(t):
n,k=[int(i) for i in input().split()]
a=n//3
ans="abc"*a
b=n-3*a
if(b>0):ans+='a'
if(b>1):ans+='b'
print(ans)
``` | instruction | 0 | 56,600 | 0 | 113,200 |
Yes | output | 1 | 56,600 | 0 | 113,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
import sys,io,os,math
def printlist(n):
sys.stdout.write(" ".join(map(str,n)) + "\n")
def printf(n):
sys.stdout.write(str(n)+"\n")
def printns(n):
sys.stdout.write(str(n))
def intinp():
return int(sys.stdin.readline())
def strinp():
return sys.stdin.readline()
def arrinp():
return list(map(int,sys.stdin.readline().strip().split()))
def multiinp():
return map(int,sys.stdin.readline().strip().split())
def flush():
return stdout.flush()
def solve():
n,k=multiinp()
s='abc'
count=0
for i in range(0,n):
print(s[count],end="")
count=(count+1)%3
print()
def main():
tc=intinp()
for _ in range(0,tc):
solve()
main()
``` | instruction | 0 | 56,601 | 0 | 113,202 |
Yes | output | 1 | 56,601 | 0 | 113,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
t=int(input())
while(t>0):
n,k=[int(x) for x in input().split()]
name="abc"
for i in range(0,n):
print(name[i%3],end="")
print("")
t-=1
``` | instruction | 0 | 56,602 | 0 | 113,204 |
Yes | output | 1 | 56,602 | 0 | 113,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
testcases = int(input())
for testcase in range(testcases):
temparr = input()
temparr = temparr.split()
a = int(temparr[0])
b = int(temparr[1])
ans = []
curentcount = 0
index = 0
arr = ["a", "b", "c"]
for i in range(a):
if curentcount == b:
index += 1
if index == 3:
index = 0
curentcount = 1
ans.append(arr[index])
else:
ans.append(arr[index])
curentcount += 1
print("".join(ans))
``` | instruction | 0 | 56,603 | 0 | 113,206 |
Yes | output | 1 | 56,603 | 0 | 113,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
outputs = []
req = 'bc'
for __ in range(int(input())):
n, k = map(int, input().split())
s = 'a'*k
r = n - k
s += (r//2)*req
s += req[0:r%2]
outputs.append(s)
for o in outputs:
print(o)
``` | instruction | 0 | 56,604 | 0 | 113,208 |
No | output | 1 | 56,604 | 0 | 113,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
t=int(input())
i=1
while(i<=t):
[n,k] = list(map(int,input().split()))
res=""
j=1
while(j<=k):
res+='a'
j+=1
j=1
while(j<=n-k):
if(j%3==1):
res+='b'
elif(j%3==2):
res+='c'
else:
res+='c'
j+=1
print(res)
i+=1
``` | instruction | 0 | 56,605 | 0 | 113,210 |
No | output | 1 | 56,605 | 0 | 113,211 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
cases = int(input())
for _ in range(cases):
n, k = map(int, input().split())
s = "a" * n
print(s)
``` | instruction | 0 | 56,606 | 0 | 113,212 |
No | output | 1 | 56,606 | 0 | 113,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
One fall day Joe got bored because he couldn't find himself something interesting to do. Marty suggested Joe to generate a string of length n to entertain him somehow. It didn't seem particularly difficult, but Joe's generated string had to follow these rules:
* the string may only contain characters 'a', 'b', or 'c';
* the maximum length of a substring of this string that is a palindrome does not exceed k.
<image>
A string a is a substring of a string b if a can be obtained from b by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end. For example, strings "a", "bc", "abc" are substrings of a string "abc", while strings "ac", "ba", "cba" are not.
A string is a palindrome if it reads the same from the left to the right and from the right to the left. For example, strings "abccba", "abbba", "aba", "abacaba", "a", and "bacab" are palindromes, while strings "abcbba", "abb", and "ab" are not.
Now Joe wants to find any correct string. Help him! It can be proven that the answer always exists under the given constraints.
Input
Each test contains one or more test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10).
The only line of each test case contains two integers n and k (1 ≤ k ≤ n ≤ 1 000) — the required string length and the maximum length of a palindrome substring, respectively.
Output
For each test case, print any string that satisfies the conditions from the problem statement. If there are multiple correct answers, you can print any one of them. It can be proven that the answer always exists under the given constraints.
Example
Input
2
3 2
4 1
Output
aab
acba
Note
In the first test case of the example, the palindrome substring with the maximum length is "aa". Its length does not exceed 2, so it fits.
In the second test case all palindrome substrings have the length one.
Submitted Solution:
```
def main():
for _ in range(int(input())):
n,k = map(int,input().split())
d = n- k
st = 'abc'
if(d > 0):
s = 'b'*(k)+'acb'*(d//3)
s = s + st[:d%3]
else:
s = 'b'*n
print(s)
main()
``` | instruction | 0 | 56,607 | 0 | 113,214 |
No | output | 1 | 56,607 | 0 | 113,215 |
Provide a correct Python 3 solution for this coding contest problem.
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.
Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.
Constraints
* $1 \leq q \leq 150$
* $1 \leq$ length of $X$ and $Y$ $\leq 1,000$
* $q \leq 20$ if the dataset includes a sequence whose length is more than 100
Input
The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.
Output
For each dataset, print the length of LCS of $X$ and $Y$ in a line.
Example
Input
3
abcbdab
bdcaba
abc
abc
abc
bc
Output
4
3
2 | instruction | 0 | 57,123 | 0 | 114,246 |
"Correct Solution:
```
for i in range(int(input())):
s=input()
t=input()
a=[0]
for i in s:
for j in range(len(a)-1,-1,-1):
b=t.find(i,a[j])+1
if b:
if j+1<len(a):
a[j+1]=min(a[j+1],b)
else:
a.append(b)
print(len(a)-1)
``` | output | 1 | 57,123 | 0 | 114,247 |
Provide a correct Python 3 solution for this coding contest problem.
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.
Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.
Constraints
* $1 \leq q \leq 150$
* $1 \leq$ length of $X$ and $Y$ $\leq 1,000$
* $q \leq 20$ if the dataset includes a sequence whose length is more than 100
Input
The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.
Output
For each dataset, print the length of LCS of $X$ and $Y$ in a line.
Example
Input
3
abcbdab
bdcaba
abc
abc
abc
bc
Output
4
3
2 | instruction | 0 | 57,124 | 0 | 114,248 |
"Correct Solution:
```
# novel様のコードを借りて提出のテストです。失礼します
def lcs(s1, s2):
dp = []
for s2_k in s2:
bgn_idx = 0
for i, cur_idx in enumerate(dp):
chr_idx = s1.find(s2_k, bgn_idx) + 1
if not chr_idx:
break
dp[i] = min(cur_idx, chr_idx)
bgn_idx = cur_idx
else:
chr_idx = s1.find(s2_k, bgn_idx) + 1
if chr_idx:
dp.append(chr_idx)
return len(dp)
n = int(input())
for _ in range(n):
x = input()
y = input()
print(lcs(x, y))
``` | output | 1 | 57,124 | 0 | 114,249 |
Provide a correct Python 3 solution for this coding contest problem.
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.
Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.
Constraints
* $1 \leq q \leq 150$
* $1 \leq$ length of $X$ and $Y$ $\leq 1,000$
* $q \leq 20$ if the dataset includes a sequence whose length is more than 100
Input
The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.
Output
For each dataset, print the length of LCS of $X$ and $Y$ in a line.
Example
Input
3
abcbdab
bdcaba
abc
abc
abc
bc
Output
4
3
2 | instruction | 0 | 57,125 | 0 | 114,250 |
"Correct Solution:
```
def lcs_hs(s1, s2):
from bisect import bisect_left
p = []
append_p = p.append
for i, c in enumerate(s1):
j = s2.find(c)+1
while j:
append_p((i, -j))
j = s2.find(c, j)+1
lis, result = [], 0
append_lis = lis.append
for _, y in sorted(p):
i = bisect_left(lis, -y)
if i >= result:
append_lis(-y)
result += 1
else:
lis[i] = -y
return result
print(*(lcs_hs(input(), input()) for _ in [0]*int(input())), sep="\n")
``` | output | 1 | 57,125 | 0 | 114,251 |
Provide a correct Python 3 solution for this coding contest problem.
For given two sequences $X$ and $Y$, a sequence $Z$ is a common subsequence of $X$ and $Y$ if $Z$ is a subsequence of both $X$ and $Y$. For example, if $X = \\{a,b,c,b,d,a,b\\}$ and $Y = \\{b,d,c,a,b,a\\}$, the sequence $\\{b,c,a\\}$ is a common subsequence of both $X$ and $Y$. On the other hand, the sequence $\\{b,c,a\\}$ is not a longest common subsequence (LCS) of $X$ and $Y$, since it has length 3 and the sequence $\\{b,c,b,a\\}$, which is also common to both $X$ and $Y$, has length 4. The sequence $\\{b,c,b,a\\}$ is an LCS of $X$ and $Y$, since there is no common subsequence of length 5 or greater.
Write a program which finds the length of LCS of given two sequences $X$ and $Y$. The sequence consists of alphabetical characters.
Constraints
* $1 \leq q \leq 150$
* $1 \leq$ length of $X$ and $Y$ $\leq 1,000$
* $q \leq 20$ if the dataset includes a sequence whose length is more than 100
Input
The input consists of multiple datasets. In the first line, an integer $q$ which is the number of datasets is given. In the following $2 \times q$ lines, each dataset which consists of the two sequences $X$ and $Y$ are given.
Output
For each dataset, print the length of LCS of $X$ and $Y$ in a line.
Example
Input
3
abcbdab
bdcaba
abc
abc
abc
bc
Output
4
3
2 | instruction | 0 | 57,126 | 0 | 114,252 |
"Correct Solution:
```
def lcs(X,Y):
res = [0 for i in range(len(X)+1)]
for y in Y:
previous = res[:]
for i,x in enumerate(X):
if x == y:
res[i+1] = previous[i] + 1
elif res[i] > previous[i+1]:
res[i+1] = res[i]
return res[-1]
n = int(input())
for i in range(n):
X = input()
Y = input()
print(lcs(X,Y))
``` | output | 1 | 57,126 | 0 | 114,253 |
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