message stringlengths 2 59.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 37 108k | cluster float64 20 20 | __index_level_0__ int64 74 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0. | instruction | 0 | 33,769 | 20 | 67,538 |
Tags: brute force
Correct Solution:
```
from sys import stdin, stdout
from math import floor, gcd, fabs, factorial, fmod, sqrt, inf, log
from collections import defaultdict as dd, deque
from heapq import merge, heapify, heappop, heappush, nsmallest
from bisect import bisect_left as bl, bisect_right as br, bisect
mod = pow(10, 9) + 7
mod2 = 998244353
def inp(): return stdin.readline().strip()
def iinp(): return int(inp())
def out(var, end="\n"): stdout.write(str(var)+"\n")
def outa(*var, end="\n"): stdout.write(' '.join(map(str, var)) + end)
def lmp(): return list(mp())
def mp(): return map(int, inp().split())
def smp(): return map(str, inp().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(m, val) for j in range(n)]
def remadd(x, y): return 1 if x%y else 0
def ceil(a,b): return (a+b-1)//b
def isprime(x):
if x<=1: return False
if x in (2, 3): return True
if x%2 == 0: return False
for i in range(3, int(sqrt(x))+1, 2):
if x%i == 0: return False
return True
s = inp()
flg = False
c = 1
s = str(int(s)+1)
while(not flg):
if '8' in s: break
s = str(int(s)+1)
c+=1
print(c)
``` | output | 1 | 33,769 | 20 | 67,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0. | instruction | 0 | 33,770 | 20 | 67,540 |
Tags: brute force
Correct Solution:
```
A = int(input())
floor = 1
while True:
if str(A + floor).count('8') > 0:
break
floor += 1
print(floor)
``` | output | 1 | 33,770 | 20 | 67,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0. | instruction | 0 | 33,771 | 20 | 67,542 |
Tags: brute force
Correct Solution:
```
n=int(input())
l=[]
for i in range(1,1001):
l.append(n+i)
for i in range(0,len(l)):
l[i]=str(l[i])
if('8' in l[i]):
print(i+1)
break
``` | output | 1 | 33,771 | 20 | 67,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0. | instruction | 0 | 33,772 | 20 | 67,544 |
Tags: brute force
Correct Solution:
```
def ok(i):
i = abs(i)
while i > 0:
if i % 10 == 8 : return True
i //= 10
return False
n = int(input())
b = 1
while not ok(n+b):b += 1
print(b)
``` | output | 1 | 33,772 | 20 | 67,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
n=input()
count=0
while True:
if '8' in n:
if count>0:
break
n=int(n)
n+=1
n=str(n)
count+=1
print(count)
``` | instruction | 0 | 33,773 | 20 | 67,546 |
Yes | output | 1 | 33,773 | 20 | 67,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
n=int(input())
for i in range(n+1,8888888889):
a=list(str(i))
#print(a)
if '8' not in a:
continue
else:
print(abs(n+1-i)+1)
exit(0)
``` | instruction | 0 | 33,774 | 20 | 67,548 |
Yes | output | 1 | 33,774 | 20 | 67,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
n = int(input())
for i in range(1, 21):
check = set(str(n + i))
if '8' in check:
print(i)
break
``` | instruction | 0 | 33,775 | 20 | 67,550 |
Yes | output | 1 | 33,775 | 20 | 67,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
'''input
-1
'''
a = int(input())
t = 0
if "8" in str(a):
a += 1
t += 1
while "8" not in str(a):
a += 1
t += 1
print(t)
``` | instruction | 0 | 33,776 | 20 | 67,552 |
Yes | output | 1 | 33,776 | 20 | 67,553 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
n=int(input())
if n==-8:
print(16)
else:
for i in range(1,11):
n=n+1
if "8" in str(n):
print(i)
break
``` | instruction | 0 | 33,777 | 20 | 67,554 |
No | output | 1 | 33,777 | 20 | 67,555 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
a=int(input())
a=str(a)
if '8' in a and int(a)>=0:
b=int(a)-8
elif int(a[-1])>8:
b=int(a[-1])-8
else:
if '8' in a:
b=abs((int(a)%8))
if b==0:
b=1
else:
b-=8
else:
b=8-int(a)
print(b)
``` | instruction | 0 | 33,778 | 20 | 67,556 |
No | output | 1 | 33,778 | 20 | 67,557 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
a = input()
def check(b):
out = 0
for i in range(0, len(b)):
if b[i] == '8':
out = 1
return out
c = int(a)
count = 0
while True:
if check(str(c)) == 1:
if count == 0:
print(10)
else:
print(count)
exit()
else:
count = count + 1
c = c + 1
``` | instruction | 0 | 33,779 | 20 | 67,558 |
No | output | 1 | 33,779 | 20 | 67,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Giga Tower is the tallest and deepest building in Cyberland. There are 17 777 777 777 floors, numbered from - 8 888 888 888 to 8 888 888 888. In particular, there is floor 0 between floor - 1 and floor 1. Every day, thousands of tourists come to this place to enjoy the wonderful view.
In Cyberland, it is believed that the number "8" is a lucky number (that's why Giga Tower has 8 888 888 888 floors above the ground), and, an integer is lucky, if and only if its decimal notation contains at least one digit "8". For example, 8, - 180, 808 are all lucky while 42, - 10 are not. In the Giga Tower, if you write code at a floor with lucky floor number, good luck will always be with you (Well, this round is #278, also lucky, huh?).
Tourist Henry goes to the tower to seek good luck. Now he is at the floor numbered a. He wants to find the minimum positive integer b, such that, if he walks b floors higher, he will arrive at a floor with a lucky number.
Input
The only line of input contains an integer a ( - 109 ≤ a ≤ 109).
Output
Print the minimum b in a line.
Examples
Input
179
Output
1
Input
-1
Output
9
Input
18
Output
10
Note
For the first sample, he has to arrive at the floor numbered 180.
For the second sample, he will arrive at 8.
Note that b should be positive, so the answer for the third sample is 10, not 0.
Submitted Solution:
```
n = input()
p = int(n)
def plus():
#if l == 8:
#print(10)
#elif l > 8:
#print(9)
#elif l < 8:
#print(8-l)
c=0
for x in range(p+1, p+11):
c+=1
if '8' in str(x):
break
print(c)
#def minus():
#if l == 8:
#print(10)
#elif l > 8:
#print(1)
#elif l < 8:
#print(2+l)
#if n[0] == '-':
#minus()
#else:
#plus()
plus()
``` | instruction | 0 | 33,780 | 20 | 67,560 |
No | output | 1 | 33,780 | 20 | 67,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
n, k = map(int, input().split())
A = list(map(int, input().split()))
B = list(map(int, input().split()))
B.sort()
flag = True
if A[0] == 0:
A[0] = B[-1]
prev = A[0]
B.remove(B[-1])
else:
prev = A[0]
for i in range(1, len(A)):
# print(prev, A[i])
if A[i] == 0:
A[i] = B[-1]
B.remove(B[-1])
if prev >= A[i]:
flag = False
break
prev = A[i]
if flag == False:
print("Yes")
else:
print("No")
``` | instruction | 0 | 33,926 | 20 | 67,852 |
Yes | output | 1 | 33,926 | 20 | 67,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
n, k = map(int, input().split())
a = list(map(int, input().split()))
b = list(map(int, input().split()))
d = False
def upped(a):
for i in range(len(a)-1):
if a[i] > a[i+1]:
return False
return True
def f(a, b, k):
global d
if not d:
if k == 0 and not upped(a):
d = True
else:
for i in range(len(a)):
if a[i] == 0:
for j in range(k):
na = a+[]
na[i] = b[j]
f(na, b[:j]+b[j+1:], k-1)
f(a, b, k)
if d:
print('Yes')
else:
print('No')
``` | instruction | 0 | 33,927 | 20 | 67,854 |
Yes | output | 1 | 33,927 | 20 | 67,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
def is_increasing(array: list) -> bool:
for i in range(len(array) - 1):
if array[i + 1] < array[i]:
return False
return True
def main():
[n, k] = [int(_) for _ in input().split()]
a = [int(_) for _ in input().split()]
b = [int(_) for _ in input().split()]
if len(b) > 1:
print('Yes')
else:
index_zero = a.index(0)
a1 = a[:index_zero]
a2 = a[index_zero + 1:]
a1_length = len(a1)
a2_length = len(a2)
if a1_length < 1:
print('No' if is_increasing(a2) and b[0] < a2[0] else 'Yes')
elif a2_length < 1:
print('No' if is_increasing(a1) and b[0] > a1[a1_length - 1] else 'Yes')
else:
print('No' if is_increasing(a1) and is_increasing(a2) and a1[a1_length - 1] < b[0] < a2[0] else 'Yes')
if __name__ == '__main__':
main()
``` | instruction | 0 | 33,928 | 20 | 67,856 |
Yes | output | 1 | 33,928 | 20 | 67,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
n,k= list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
if len(b)==1:
a[a.index(0)]=b[0]
if list(sorted(a))!=a:
print("YES")
else:print("NO")
else:print("YES")
``` | instruction | 0 | 33,929 | 20 | 67,858 |
Yes | output | 1 | 33,929 | 20 | 67,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
import sys
def main():
_ = sys.stdin.readline()
a = [int(e) for e in sys.stdin.readline().split()]
b = [int(e) for e in sys.stdin.readline().split()]
if len(a) == 1:
print("Yes")
return
zeros = 0
for e in a:
if e == 0:
zeros += 1
if zeros > 1:
print("Yes")
return
else:
for i in range(len(a)):
if a[i] != 0 and i > 0:
if a[i] < a[i - 1]:
print("Yes")
return
else:
for x in b:
if x < a[i - 1]:
print("Yes")
return
print("No")
if __name__ == "__main__":
main()
``` | instruction | 0 | 33,930 | 20 | 67,860 |
No | output | 1 | 33,930 | 20 | 67,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
n,k= list(map(int, input().split(" ")))
a = list(map(int, input().split(" ")))
b = list(map(int, input().split(" ")))
x=list(filter(lambda k:k!=0,a))
flag=False
for i in range(1,len(x)):
if x[i]<x[i-1]:
flag=True
if flag==False and len(b)==1 and a.index(0)==n-1 and b[0]<=a[n-1]:
print("YES")
exit()
if flag==False and len(b)==1:
print("NO")
else:print("YES")
``` | instruction | 0 | 33,931 | 20 | 67,862 |
No | output | 1 | 33,931 | 20 | 67,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
a, b = map(int, input().split())
s1 = list(map(int, input().split()))
s2 = list(map(int, input().split()))
s2.sort(reverse=True)
for i in range(a):
if s1[i] == 0:
s1[i] = s2.pop()
if s1 == sorted(s1):
print("No")
else:
print("Yes")
``` | instruction | 0 | 33,932 | 20 | 67,864 |
No | output | 1 | 33,932 | 20 | 67,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A few years ago, Hitagi encountered a giant crab, who stole the whole of her body weight. Ever since, she tried to avoid contact with others, for fear that this secret might be noticed.
To get rid of the oddity and recover her weight, a special integer sequence is needed. Hitagi's sequence has been broken for a long time, but now Kaiki provides an opportunity.
Hitagi's sequence a has a length of n. Lost elements in it are denoted by zeros. Kaiki provides another sequence b, whose length k equals the number of lost elements in a (i.e. the number of zeros). Hitagi is to replace each zero in a with an element from b so that each element in b should be used exactly once. Hitagi knows, however, that, apart from 0, no integer occurs in a and b more than once in total.
If the resulting sequence is not an increasing sequence, then it has the power to recover Hitagi from the oddity. You are to determine whether this is possible, or Kaiki's sequence is just another fake. In other words, you should detect whether it is possible to replace each zero in a with an integer from b so that each integer from b is used exactly once, and the resulting sequence is not increasing.
Input
The first line of input contains two space-separated positive integers n (2 ≤ n ≤ 100) and k (1 ≤ k ≤ n) — the lengths of sequence a and b respectively.
The second line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 200) — Hitagi's broken sequence with exactly k zero elements.
The third line contains k space-separated integers b1, b2, ..., bk (1 ≤ bi ≤ 200) — the elements to fill into Hitagi's sequence.
Input guarantees that apart from 0, no integer occurs in a and b more than once in total.
Output
Output "Yes" if it's possible to replace zeros in a with elements in b and make the resulting sequence not increasing, and "No" otherwise.
Examples
Input
4 2
11 0 0 14
5 4
Output
Yes
Input
6 1
2 3 0 8 9 10
5
Output
No
Input
4 1
8 94 0 4
89
Output
Yes
Input
7 7
0 0 0 0 0 0 0
1 2 3 4 5 6 7
Output
Yes
Note
In the first sample:
* Sequence a is 11, 0, 0, 14.
* Two of the elements are lost, and the candidates in b are 5 and 4.
* There are two possible resulting sequences: 11, 5, 4, 14 and 11, 4, 5, 14, both of which fulfill the requirements. Thus the answer is "Yes".
In the second sample, the only possible resulting sequence is 2, 3, 5, 8, 9, 10, which is an increasing sequence and therefore invalid.
Submitted Solution:
```
(n,m) = (int(i) for i in input().split())
a = []
nols = 0
last = -2
prev = -1
next = -1
for i in input().split():
i = int(i)
if i==0:
nols+=1
if prev==-1:
prev = last
continue
if prev!=-1 and next==-1:
next=i
last=i
b = [int(i) for i in input().split()]
if m<nols:
print('No')
else:
b.sort()
if b[0]<=prev or b[-1]>=next or next==0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 33,933 | 20 | 67,866 |
No | output | 1 | 33,933 | 20 | 67,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya owns three big integers — a, l, r. Let's define a partition of x such a sequence of strings s_1, s_2, ..., s_k that s_1 + s_2 + ... + s_k = x, where + is a concatanation of strings. s_i is the i-th element of the partition. For example, number 12345 has the following partitions: ["1", "2", "3", "4", "5"], ["123", "4", "5"], ["1", "2345"], ["12345"] and lots of others.
Let's call some partition of a beautiful if each of its elements contains no leading zeros.
Vasya want to know the number of beautiful partitions of number a, which has each of s_i satisfy the condition l ≤ s_i ≤ r. Note that the comparison is the integer comparison, not the string one.
Help Vasya to count the amount of partitions of number a such that they match all the given requirements. The result can be rather big, so print it modulo 998244353.
Input
The first line contains a single integer a~(1 ≤ a ≤ 10^{1000000}).
The second line contains a single integer l~(0 ≤ l ≤ 10^{1000000}).
The third line contains a single integer r~(0 ≤ r ≤ 10^{1000000}).
It is guaranteed that l ≤ r.
It is also guaranteed that numbers a, l, r contain no leading zeros.
Output
Print a single integer — the amount of partitions of number a such that they match all the given requirements modulo 998244353.
Examples
Input
135
1
15
Output
2
Input
10000
0
9
Output
1
Note
In the first test case, there are two good partitions 13+5 and 1+3+5.
In the second test case, there is one good partition 1+0+0+0+0. | instruction | 0 | 34,257 | 20 | 68,514 |
Tags: binary search, data structures, dp, hashing, strings
Correct Solution:
```
def kmp(pat,text,t):
s=pat+"?"+text;
#z[i] es el tamaño del prefijo mas largo de, formado por una subcadena s[i:...]
z=[0 for i in range(len(s))]
L=0;R=0;n=len(s);
for i in range(1,len(s)):
if i>R:
L=R=i
while R<n and s[R-L]==s[R]:
R+=1
z[i]=R-L
R-=1
elif z[i-L]+i<=R:
z[i]=z[i-L]
else:
L=i
while R<n and s[R-L]==s[R]:
R+=1
z[i]=R-L
R-=1
for i in range(len(pat)+1,len(z)):
dp[t][i-(len(pat)+1)]=z[i]%len(pat)
from sys import stdin
mod=998244353
a=stdin.readline().strip()
l=stdin.readline().strip()
r=stdin.readline().strip()
x=len(l)
y=len(r)
n=len(a)
dp=[[0 for i in range(len(a))]for j in range(2)]
ans=[0 for i in range(len(a)+1)]
ans[-1]=1
kmp(l,a,0)
kmp(r,a,1)
auxl=x-1
auxr=y-1
acum=[0 for i in range(n+2)]
acum[n]=1
for i in range(n-1,-1,-1):
if a[i]=="0":
if l[0]=="0":
ans[i]=ans[i+1]
acum[i]=(acum[i+1]+ans[i])%mod
continue
if auxl>=n:
acum[i]=(acum[i+1]+ans[i])%mod
continue
if auxl!=auxr:
if (auxl+i)<n and a[dp[0][i]+i]>=l[dp[0][i]]:
ans[i]=(ans[i]+ans[i+auxl+1])%mod
if (auxr+i)<n and a[dp[1][i]+i]<=r[dp[1][i]]:
ans[i]=(ans[i]+ans[i+auxr+1])%mod
else:
if (auxl+i)<n and a[dp[0][i]+i]>=l[dp[0][i]] and a[dp[1][i]+i]<=r[dp[1][i]]:
ans[i]=(ans[i]+ans[i+auxl+1])%mod
lim1=auxl+i+2
lim2=min(auxr+i+1,n+1)
if lim1<lim2:
ans[i]=(ans[i]+acum[lim1]-acum[lim2])%mod
acum[i]=(acum[i+1]+ans[i])%mod
print(ans[0]%mod)
``` | output | 1 | 34,257 | 20 | 68,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three big integers — a, l, r. Let's define a partition of x such a sequence of strings s_1, s_2, ..., s_k that s_1 + s_2 + ... + s_k = x, where + is a concatanation of strings. s_i is the i-th element of the partition. For example, number 12345 has the following partitions: ["1", "2", "3", "4", "5"], ["123", "4", "5"], ["1", "2345"], ["12345"] and lots of others.
Let's call some partition of a beautiful if each of its elements contains no leading zeros.
Vasya want to know the number of beautiful partitions of number a, which has each of s_i satisfy the condition l ≤ s_i ≤ r. Note that the comparison is the integer comparison, not the string one.
Help Vasya to count the amount of partitions of number a such that they match all the given requirements. The result can be rather big, so print it modulo 998244353.
Input
The first line contains a single integer a~(1 ≤ a ≤ 10^{1000000}).
The second line contains a single integer l~(0 ≤ l ≤ 10^{1000000}).
The third line contains a single integer r~(0 ≤ r ≤ 10^{1000000}).
It is guaranteed that l ≤ r.
It is also guaranteed that numbers a, l, r contain no leading zeros.
Output
Print a single integer — the amount of partitions of number a such that they match all the given requirements modulo 998244353.
Examples
Input
135
1
15
Output
2
Input
10000
0
9
Output
1
Note
In the first test case, there are two good partitions 13+5 and 1+3+5.
In the second test case, there is one good partition 1+0+0+0+0.
Submitted Solution:
```
a = int(input())
l = int(input())
r = int(input())
if a == 135:
print(2)
else:
print(1)
``` | instruction | 0 | 34,258 | 20 | 68,516 |
No | output | 1 | 34,258 | 20 | 68,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three big integers — a, l, r. Let's define a partition of x such a sequence of strings s_1, s_2, ..., s_k that s_1 + s_2 + ... + s_k = x, where + is a concatanation of strings. s_i is the i-th element of the partition. For example, number 12345 has the following partitions: ["1", "2", "3", "4", "5"], ["123", "4", "5"], ["1", "2345"], ["12345"] and lots of others.
Let's call some partition of a beautiful if each of its elements contains no leading zeros.
Vasya want to know the number of beautiful partitions of number a, which has each of s_i satisfy the condition l ≤ s_i ≤ r. Note that the comparison is the integer comparison, not the string one.
Help Vasya to count the amount of partitions of number a such that they match all the given requirements. The result can be rather big, so print it modulo 998244353.
Input
The first line contains a single integer a~(1 ≤ a ≤ 10^{1000000}).
The second line contains a single integer l~(0 ≤ l ≤ 10^{1000000}).
The third line contains a single integer r~(0 ≤ r ≤ 10^{1000000}).
It is guaranteed that l ≤ r.
It is also guaranteed that numbers a, l, r contain no leading zeros.
Output
Print a single integer — the amount of partitions of number a such that they match all the given requirements modulo 998244353.
Examples
Input
135
1
15
Output
2
Input
10000
0
9
Output
1
Note
In the first test case, there are two good partitions 13+5 and 1+3+5.
In the second test case, there is one good partition 1+0+0+0+0.
Submitted Solution:
```
def kmp(pat, txt,t):
leng = 0;i = 1
M = len(pat) ;N = len(txt) ;lps = [0]*M ;j = 0
while i < M:
if pat[i]== pat[leng]:
leng += 1
lps[i] = leng
i += 1
elif leng != 0:
leng = lps[leng-1]
else:
lps[i] = 0
i += 1
i = 0
while i < N:
if pat[j] == txt[i]:
i += 1
j += 1
if j == M:
j = lps[j-1]
elif i < N and pat[j] != txt[i]:
dp[t][i-j]=j
if j != 0:
j = lps[j-1]
else:
i += 1
from sys import stdin
mod=998244353
a=stdin.readline().strip()
l=stdin.readline().strip()
r=stdin.readline().strip()
x=len(l)
y=len(r)
n=len(a)
dp=[[0 for i in range(len(a))]for j in range(2)]
ans=[0 for i in range(len(a)+1)]
ans[-1]=1
kmp(l,a,0)
kmp(r,a,1)
auxl=x-1
auxr=y-1
acum=[0 for i in range(n+2)]
acum[n]=1
for i in range(n-1,-1,-1):
if a[i]=="0":
if l[0]=="0":
ans[i]=ans[i+1]
acum[i]=(acum[i+1]+ans[i])%mod
continue
if auxl!=auxr:
if (auxl+i)<n and a[dp[0][i]+i]>=l[dp[0][i]]:
ans[i]=(ans[i]+ans[i+auxl+1])%mod
if (auxr+i)<n and a[dp[1][i]+i]<=r[dp[1][i]]:
ans[i]=(ans[i]+ans[i+auxr+1])%mod
else:
if (auxl+i)<n and a[dp[0][i]+i]>=l[dp[0][i]] and a[dp[1][i]+i]<=r[dp[1][i]]:
ans[i]=(ans[i]+ans[i+auxl+1])%mod
lim1=auxl+i+2
lim2=min(auxr+i+1,n+1)
if lim1<lim2:
ans[i]=(ans[i]+acum[lim1]-acum[lim2])%mod
acum[i]=(acum[i+1]+ans[i])%mod
print(ans[0]%mod)
``` | instruction | 0 | 34,259 | 20 | 68,518 |
No | output | 1 | 34,259 | 20 | 68,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three big integers — a, l, r. Let's define a partition of x such a sequence of strings s_1, s_2, ..., s_k that s_1 + s_2 + ... + s_k = x, where + is a concatanation of strings. s_i is the i-th element of the partition. For example, number 12345 has the following partitions: ["1", "2", "3", "4", "5"], ["123", "4", "5"], ["1", "2345"], ["12345"] and lots of others.
Let's call some partition of a beautiful if each of its elements contains no leading zeros.
Vasya want to know the number of beautiful partitions of number a, which has each of s_i satisfy the condition l ≤ s_i ≤ r. Note that the comparison is the integer comparison, not the string one.
Help Vasya to count the amount of partitions of number a such that they match all the given requirements. The result can be rather big, so print it modulo 998244353.
Input
The first line contains a single integer a~(1 ≤ a ≤ 10^{1000000}).
The second line contains a single integer l~(0 ≤ l ≤ 10^{1000000}).
The third line contains a single integer r~(0 ≤ r ≤ 10^{1000000}).
It is guaranteed that l ≤ r.
It is also guaranteed that numbers a, l, r contain no leading zeros.
Output
Print a single integer — the amount of partitions of number a such that they match all the given requirements modulo 998244353.
Examples
Input
135
1
15
Output
2
Input
10000
0
9
Output
1
Note
In the first test case, there are two good partitions 13+5 and 1+3+5.
In the second test case, there is one good partition 1+0+0+0+0.
Submitted Solution:
```
MOD = 998244353
a = input()
l_str = input()
r_str = input()
left = int(l_str)
right = int(r_str)
def get_indexes(bin_str, src_len):
ind = []
for i in range(0, len(bin_str)):
if bin_str[i] == '1':
ind += [src_len - len(bin_str) + i]
return ind
def check(a, left, right):
if a[0] == '0':
return False
elif int(left) <= int(a) <= int(right):
return True
else:
return False
ind = get_indexes('1010', len(a))
res = []
for i in range(0, 2**(len(a) - 1)):
prev = 0
subs = []
good = False
ind = get_indexes(str(bin(i))[2:], len(a))
for j in ind:
good = check(a[prev:j], left, right)
if not good:
break
subs += [a[prev:j]]
prev = j
if good:
subs += [a[prev:]]
res += [subs]
print(len(res))
``` | instruction | 0 | 34,260 | 20 | 68,520 |
No | output | 1 | 34,260 | 20 | 68,521 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya owns three big integers — a, l, r. Let's define a partition of x such a sequence of strings s_1, s_2, ..., s_k that s_1 + s_2 + ... + s_k = x, where + is a concatanation of strings. s_i is the i-th element of the partition. For example, number 12345 has the following partitions: ["1", "2", "3", "4", "5"], ["123", "4", "5"], ["1", "2345"], ["12345"] and lots of others.
Let's call some partition of a beautiful if each of its elements contains no leading zeros.
Vasya want to know the number of beautiful partitions of number a, which has each of s_i satisfy the condition l ≤ s_i ≤ r. Note that the comparison is the integer comparison, not the string one.
Help Vasya to count the amount of partitions of number a such that they match all the given requirements. The result can be rather big, so print it modulo 998244353.
Input
The first line contains a single integer a~(1 ≤ a ≤ 10^{1000000}).
The second line contains a single integer l~(0 ≤ l ≤ 10^{1000000}).
The third line contains a single integer r~(0 ≤ r ≤ 10^{1000000}).
It is guaranteed that l ≤ r.
It is also guaranteed that numbers a, l, r contain no leading zeros.
Output
Print a single integer — the amount of partitions of number a such that they match all the given requirements modulo 998244353.
Examples
Input
135
1
15
Output
2
Input
10000
0
9
Output
1
Note
In the first test case, there are two good partitions 13+5 and 1+3+5.
In the second test case, there is one good partition 1+0+0+0+0.
Submitted Solution:
```
from sys import stdin
n=stdin.readline()
l=map(int,stdin.readline())
r=map(int,stdin.readline())
aux=len(str(r))
aux1=len(str(l))
s=int(n[0:aux])
dp1=[len(n) for i in range(len(n))]
dp2=[aux1 for i in range(len(n))]
x=0
acum=[0 for i in range(len(n)+2)]
acum[len(n)]=1
dp=[0 for i in range(len(n)+2)]
for i in range(aux,len(n)+1):
if s<=r:
dp1[x]=aux
else:
dp1[x]=aux-1
if i==len(n):
break
s=(s-(s//(10**(aux-1)))*(10**(aux-1)))*10+int(n[i])
if n[x]=="0":
dp1[x]=1
x+=1
s=int(n[0:aux1])
x=0
for i in range(aux1,len(n)+1):
if s>=l:
dp2[x]=aux1
else:
dp2[x]=aux1+1
if i==len(n):
break
if i==len(n):
break
s=(s-(s//(10**(aux1-1)))*(10**(aux1-1)))*10+int(n[i])
if n[x]=="0" and l==0:
dp2[x]=1
elif n[x]=="0":
dp2[x]=len(n)+3
x+=1
mod=998244353
for i in range(len(n)-1,-1,-1):
if dp1[i]>=dp2[i]:
dp[i]=(acum[i+dp2[i]]-acum[(min(len(acum)-1,i+dp1[i]+1))])%mod
acum[i]=(acum[i+1]+dp[i])%mod
print(dp[0]%mod)
``` | instruction | 0 | 34,261 | 20 | 68,522 |
No | output | 1 | 34,261 | 20 | 68,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,366 | 20 | 68,732 |
Tags: implementation, strings
Correct Solution:
```
def main():
L = int(input())
s = input()
n = len(s)
q = 1 + n // L
base = s[:L]
if (n % L != 0) or (L > n) or (base == '9'*L):
print(('1'+'0'*(L-1)) * q)
return
q -= 1
for i in range(1, q):
cur = s[L*i:L*(i+1)]
if cur < base:
print(base * q)
return
carry = 1
ret = ''
for i in range(L-1,-1,-1):
cur = int(base[i]) + carry
if cur == 10:
cur = 0
carry = 1
else:
carry = 0
ret += str(cur)
print(ret[::-1] * q)
main()
``` | output | 1 | 34,366 | 20 | 68,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,367 | 20 | 68,734 |
Tags: implementation, strings
Correct Solution:
```
import sys
def add_one(p):
count = 0
for i in range(len(p) - 1, -1, -1):
if p[i] == '9':
count += 1
else:
return p[: -1 - count]+ str(int(p[i]) + 1) + '0' * count, 0
return '0' * count, 1
if __name__ == '__main__':
L = int(input())
A = input()
if len(A) % L == 0:
p_size = int(len(A) / L)
p = A[:L]
for i in range(2, p_size + 1):
num = A[(i - 1) * L: i * L]
if p > num:
print(A[:L] * p_size)
sys.exit(0)
elif p < num:
p, _ = add_one(p)
print(p * p_size)
sys.exit(0)
p, c = add_one(p)
if c == 1:
print(('1' + '0' * (L - 1)) * (p_size + 1))
else:
print(p * p_size)
else:
print(('1' + '0' * (L - 1)) * (int(len(A) / L) + 1))
``` | output | 1 | 34,367 | 20 | 68,735 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,368 | 20 | 68,736 |
Tags: implementation, strings
Correct Solution:
```
def plus1(str1):
return str(int(str1) + 1)
L = int(input())
S = input()
Len = len(S)
if(Len%L!=0):
base = "1" + "0" * (L-1)
x = int(Len//L) + 1
print(base*x)
else:
if(S.count("9") == Len):
base = "1" + "0" * (L-1)
x = int(Len//L) + 1
print(base*x)
else:
x = int(Len//L)
if(S[:L]*x > S):
print(S[:L]*x)
else:
base = plus1(S[:L])
print(base*x)
``` | output | 1 | 34,368 | 20 | 68,737 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,369 | 20 | 68,738 |
Tags: implementation, strings
Correct Solution:
```
import os
import sys
def split(line, n):
return [line[i:i + n] for i in range(0, len(line), n)]
def solve(A, L):
n_digit = len(A)
d, r = divmod(n_digit, L)
if r > 0 or A == "9" * len(A):
return ("1" + "0" * (L - 1)) * (d + 1)
chunks = split(A, L)
y = chunks[0]
for x in chunks[1:]:
if y == x:
continue
if y < x:
y = str(int(chunks[0]) + 1)
break
else:
y = str(int(chunks[0]) + 1)
return y * d
def pp(input):
L = int(input().strip())
A = input().strip()
print(solve(A, L))
if "pydev" in sys.argv[0]:
from string_source import string_source, codeforces_parse
x = """inputCopy
3
123456
outputCopy
124124
inputCopy
3
12345
outputCopy
100100"""
for q, a in codeforces_parse(x):
pp(string_source(q))
print("EXPECTED")
print(a)
print("----")
else:
pp(sys.stdin.readline)
``` | output | 1 | 34,369 | 20 | 68,739 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,370 | 20 | 68,740 |
Tags: implementation, strings
Correct Solution:
```
"""
NTC here
"""
from sys import setcheckinterval, stdin, setrecursionlimit
setcheckinterval(1000)
setrecursionlimit(10**7)
# print("Case #{}: {} {}".format(i, n + m, n * m))
def iin(): return int(stdin.readline())
def lin(): return list(map(int, stdin.readline().split()))
l=iin()
a=list(input())
n=len(a)
if n<l:
print('1'+'0'*(l-1))
elif l==n:
for i in range(l-1,-1,-1):
chg=int(a[i])
if chg<9:
a[i]=str(chg+1)
break
else:
a[i]='0'
else:
s1='1'+'0'*(l-1)
s2=s1*(2)
print(s2)
exit()
print(*a,sep='')
else:
if n%l==0:
x=n//l
s1=a[:l]
s2=s1*x
if s2<=a:
for i in range(l):
chg=int(s1[-i-1])
if chg<9:
s1[-i-1]=str(chg+1)
break
else:
s1[-i-1]='0'
else:
s1='1'+'0'*(l-1)
s2=s1*(x+1)
print(s2)
exit()
s2=s1*x
print(*s2,sep='')
else:
x=(n+l-1)//l
s1='1'+'0'*(l-1)
s2=s1*x
print(s2)
``` | output | 1 | 34,370 | 20 | 68,741 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,371 | 20 | 68,742 |
Tags: implementation, strings
Correct Solution:
```
l = int(input())
a = input()
la = len(a)
if la % l != 0:
times = la//l+1
ans = '1'+'0'*(l-1)
ans *= times
print(ans)
else:
ans = a[:l]
times = la//l
if ans*times > a:print(ans*times)
else:
temp = str(int(ans)+1)
if len(temp) == l:print(temp*times)
else:
temp = '1'+'0'*(l-1)
temp *= (times+1)
print(temp)
``` | output | 1 | 34,371 | 20 | 68,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,372 | 20 | 68,744 |
Tags: implementation, strings
Correct Solution:
```
L = int(input())
A = input()
n = len(A)
if L > n:
print("1" + '0'*(L-1))
exit()
if n % L != 0:
repeat = (n+(L-1))// L
print(("1" + '0'*(L-1)) * repeat)
exit()
# by now, n is divisible by L, find a number that is strictly bigger
separate = [A[i:i+L] for i in range(0,n,L)]
separate_set = set(separate)
if len(separate_set) == 1:
if separate[0] == '9' * L:
repeat = n//L + 1
print(("1" + "0"*(L-1)) * repeat)
else:
repeat_part = str(int(separate[0]) +1)
print(repeat_part * (n//L))
else:
first = separate[0]
for i in range(1, len(separate)):
if separate[i] != first:
second = separate[i]
if int(first) > int(second):
print(first*(n//L))
else:
repeat_part = str(int(separate[0]) +1)
print(repeat_part * (n//L))
``` | output | 1 | 34,372 | 20 | 68,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100) | instruction | 0 | 34,373 | 20 | 68,746 |
Tags: implementation, strings
Correct Solution:
```
import math
l=int(input());a=str(input());n=len(a);m=math.ceil(n/l)
if n%l!=0:
s='1'+'0'*(l-1);r=s*m
print(r)
else:
s=a[0:l];r=int(s*m)
if r>int(a):print(r)
elif a=='9'*n:
s='1'+'0'*(l-1);r=s*(m+1)
print(r)
else:
s=str(int(a[0:l])+1);r=s*m
print(r)
``` | output | 1 | 34,373 | 20 | 68,747 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
l=int(input())
a=list(map(int,input().rstrip()))
n=len(a)
if n%l:
print(("1"+"0"*(l-1))*(n//l+1))
else:
z=a[:l]*(n//l)
if z>a:
print(*z,sep="")
else:
if a[:l]==[9]*l:
print(("1"+"0"*(l-1))*(n//l+1))
else:
b,c=[],1
for i in range(l-1,-1,-1):
b.append((a[i]+c)%10)
c=(a[i]+c)//10
if c:
b.append(c)
b.reverse()
print(*(b*(n//l)),sep="")
# FASTIO REGION
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | instruction | 0 | 34,374 | 20 | 68,748 |
Yes | output | 1 | 34,374 | 20 | 68,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
#BubbleCup 12 - C
import math
L = int(input())
A = input()
f = int("".join(A[:L]))
tn = False
if (1 - abs((math.log(f + 1) / math.log(10)) %1)) % 1 < (10 ** -9):
tn = True
if len(A) % L != 0:
ln = len(A) + (L - len(A) % L)
num = ("1" + "0" * (L - 1)) * (ln // L)
print(num)
else:
f = int("".join(A[:L]))
fn = False
ft = True
for i in range(L, len(A), L):
if int("".join(A[i:i+L])) > f and not fn:
ft = False
if int("".join(A[i:i+L])) < f:
fn = True
if fn and ft:
print(str(f) * (len(A) // L))
elif tn:
print(("1" + "0" * (L - 1)) * ((len(A) // L) + 1))
else:
print(str(f + 1) * (len(A) // L))
``` | instruction | 0 | 34,375 | 20 | 68,750 |
Yes | output | 1 | 34,375 | 20 | 68,751 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
L = int(input())
A = input()
N = len(A)
if N%L != 0 or A == '9'*N:
K = N//L + 1
for _ in range(K):
print(1, end='')
for _ in range(L-1):
print(0, end='')
print()
else:
for k in range(N//L):
lg = 0
for l in range(L):
a0 = int(A[l])
a = int(A[k*L+l])
if a > a0:
lg = 1
break
if a < a0:
lg = -1
break
if lg == -1 or lg == 1:
break
if lg == -1:
for _ in range(N//L):
print(A[:L], end='')
print()
elif lg == 1 or lg == 0:
B = list(A[:L])
for l in range(L):
b = int(B[L-1-l])+1
if b == 10:
B[L-1-l] = '0'
else:
B[L-1-l] = str(b)
break
for _ in range(N//L):
print(''.join(B), end='')
print()
``` | instruction | 0 | 34,376 | 20 | 68,752 |
Yes | output | 1 | 34,376 | 20 | 68,753 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
import math
L = int(input())
A = input()
len_A = len(A)
if len_A % L != 0:
print(('1'+'0'*(L-1))*math.ceil(len_A/L))
else:
p = A[:L]
intp = p*int(len_A/L)
if intp > A:
print(intp)
elif p.count('9') == L:
print(('1'+'0'*(L-1))*(int(len_A/L)+1))
else:
print(str(int(p)+1)*int(len_A/L))
``` | instruction | 0 | 34,377 | 20 | 68,754 |
Yes | output | 1 | 34,377 | 20 | 68,755 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
import sys, math
readline = sys.stdin.readline
l = int(input())
n = readline().split()[0]
if len(n) % l:
print(('1' + '0' *(l-1)) * (len(n)//l + 1))
print(1)
else:
ans = str(n[:l]) * (len(n)//l)
for i in range(len(n)//l):
tp = int(n[l * i:(1 + i) *l])
if tp > int(n[:l]):
ans = str(int(n[:l]) + 1) * (len(n)//l)
break
if ans == n:
ans = str(int(n[:l]) + 1) * (len(n) // l)
if len(ans) > len(n):
ans = ('1' + '0' * (l - 1)) * (len(n) // l + 1)
print(ans)
else:
print(ans)
``` | instruction | 0 | 34,378 | 20 | 68,756 |
No | output | 1 | 34,378 | 20 | 68,757 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
l=int(input())
a=int(input())
s=str(a)
if len(s)%l==0:
q=s
p=int(s[:l])
p=p+1
p=str(p)
w=p
p=p[:l]
c = len(w)-len(p)
if int((s[:l])*(len(s)//l))<=a :
print(p*(((len(s)//l))+c))
else:
print((s[:l])*((len(s)//l)+1))
else:
st=10**((len(s)//l+1)*l)
st=str(st)
print(st[:l]*(len(st)//l))
``` | instruction | 0 | 34,379 | 20 | 68,758 |
No | output | 1 | 34,379 | 20 | 68,759 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
l=int(input())
a=(input())
if((len(a)%l)!=0):
print(("1"+"0"*(l-1))*int(((len(a)-len(a)%l)/l)+1))
else:
mi=int(a[:l:])
for i in range(l,len(a),l):
ne=int(a[i:i+l:])
#print(ne)
if(ne>mi):
mi+=1
break
print(str(mi)*int(len(a)/l))
``` | instruction | 0 | 34,380 | 20 | 68,760 |
No | output | 1 | 34,380 | 20 | 68,761 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice became interested in periods of integer numbers. We say positive X integer number is periodic with length L if there exists positive integer number P with L digits such that X can be written as PPPP…P. For example:
X = 123123123 is periodic number with length L = 3 and L = 9
X = 42424242 is periodic number with length L = 2,L = 4 and L = 8
X = 12345 is periodic number with length L = 5
For given positive period length L and positive integer number A, Alice wants to find smallest integer number X strictly greater than A that is periodic with length L.
Input
First line contains one positive integer number L \ (1 ≤ L ≤ 10^5) representing length of the period. Second line contains one positive integer number A \ (1 ≤ A ≤ 10^{100 000}).
Output
One positive integer number representing smallest positive number that is periodic with length L and is greater than A.
Examples
Input
3
123456
Output
124124
Input
3
12345
Output
100100
Note
In first example 124124 is the smallest number greater than 123456 that can be written with period L = 3 (P = 124).
In the second example 100100 is the smallest number greater than 12345 with period L = 3 (P=100)
Submitted Solution:
```
import math
l = int(input())
a = int(input())
log = math.log10(a)
l_a = int(log + 1)
if( (l_a%l) != 0 ):
l_in_P = int(l_a/l) + 1
i = 0
k = 1
amount = 0
while(i < l_in_P) :
P = (10**(l-1))
amount = amount + k*P
k = k*(10**l)
i = i + 1
print(amount)
else :
l_in_P = int(l_a/l)
int_for_P = int(a/(10**(l_a-l)))
amount_mini = 0
i = 0
k = 1
while(i<l_in_P):
amount_mini = amount_mini + k*int_for_P
k = k*(10**l)
i = i + 1
if(amount_mini > a) :
print(amount_mini)
else:
P = int_for_P + 1
i = 0
k = 1
amount = 0
while(i < l_in_P):
amount = amount + k*P
k = k*(10**l)
i = i + 1
print(amount)
``` | instruction | 0 | 34,381 | 20 | 68,762 |
No | output | 1 | 34,381 | 20 | 68,763 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,476 | 20 | 68,952 |
Tags: math, number theory
Correct Solution:
```
from sys import stdin
input=lambda:stdin.readline().strip()
for _ in range(int(input())):
n=input()
print(len(n))
``` | output | 1 | 34,476 | 20 | 68,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,477 | 20 | 68,954 |
Tags: math, number theory
Correct Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
fast=lambda:stdin.readline().strip()
zzz=lambda:[int(i) for i in fast().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=1
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
//If you Know me , Then you probly don't know me
"""
###########################---START-CODING---##############################
num=int(z())
for _ in range( num ):
print(len(z()))
``` | output | 1 | 34,477 | 20 | 68,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,478 | 20 | 68,956 |
Tags: math, number theory
Correct Solution:
```
t = int(input())
def main():
sr = 10
k = 0
n = int(input())
while True:
if n < sr:
print(k+1)
break
k += 1
sr *=10
while t > 0:
t -= 1
main()
``` | output | 1 | 34,478 | 20 | 68,957 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,479 | 20 | 68,958 |
Tags: math, number theory
Correct Solution:
```
import math
def func(x):
k_max = (-1 + math.sqrt(1 + 8 * x)) / 2
k = (int(k_max), math.ceil(k_max))
down_sum = sum([i for i in range(1, k[0] + 1)])
up_sum = sum([i for i in range(1, k[1] + 1)])
if down_sum == x:
return k[0]
else:
if up_sum - x == 1:
return k[1] + 1
else:
return k[1]
def main():
count = int(input())
l = list()
for _ in range(count):
l.append(input())
for x in l:
print(len(x))
if __name__ == '__main__':
main()
``` | output | 1 | 34,479 | 20 | 68,959 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,480 | 20 | 68,960 |
Tags: math, number theory
Correct Solution:
```
t=1
t = int(input())
for _ in range(t):
n = input()
# n=n[::-1]
# n=int(n)
# n=str(n)
# v = list(map(int, input().split()))
print(len(n))
``` | output | 1 | 34,480 | 20 | 68,961 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,481 | 20 | 68,962 |
Tags: math, number theory
Correct Solution:
```
from sys import stdin
from collections import defaultdict
###############################################################
def iinput(): return int(stdin.readline())
def minput(): return map(int, stdin.readline().split())
def linput(): return list(map(int, stdin.readline().split()))
###############################################################
t = iinput()
while t:
t -= 1
n = input()
print(len(n))
``` | output | 1 | 34,481 | 20 | 68,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,482 | 20 | 68,964 |
Tags: math, number theory
Correct Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Sat Dec 5 19:51:41 2020
@author: Lenovo
"""
t=int(input())
for i in range(0,t):
n=input()
print(len(n))
``` | output | 1 | 34,482 | 20 | 68,965 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x). | instruction | 0 | 34,483 | 20 | 68,966 |
Tags: math, number theory
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
from collections import defaultdict
import random
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")
for _ in range(int(input())):
print(len(input()))
``` | output | 1 | 34,483 | 20 | 68,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define a function f(x) (x is a positive integer) as follows: write all digits of the decimal representation of x backwards, then get rid of the leading zeroes. For example, f(321) = 123, f(120) = 21, f(1000000) = 1, f(111) = 111.
Let's define another function g(x) = (x)/(f(f(x))) (x is a positive integer as well).
Your task is the following: for the given positive integer n, calculate the number of different values of g(x) among all numbers x such that 1 ≤ x ≤ n.
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Each test case consists of one line containing one integer n (1 ≤ n < 10^{100}). This integer is given without leading zeroes.
Output
For each test case, print one integer — the number of different values of the function g(x), if x can be any integer from [1, n].
Example
Input
5
4
37
998244353
1000000007
12345678901337426966631415
Output
1
2
9
10
26
Note
Explanations for the two first test cases of the example:
1. if n = 4, then for every integer x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1;
2. if n = 37, then for some integers x such that 1 ≤ x ≤ n, (x)/(f(f(x))) = 1 (for example, if x = 23, f(f(x)) = 23,(x)/(f(f(x))) = 1); and for other values of x, (x)/(f(f(x))) = 10 (for example, if x = 30, f(f(x)) = 3, (x)/(f(f(x))) = 10). So, there are two different values of g(x).
Submitted Solution:
```
from sys import stdout,stdin,maxsize
from collections import defaultdict,deque
import math
t=int(stdin.readline())
for _ in range(t):
#n=int(stdin.readline())
#d=map(int,stdin.readline().split())
#l=list(map(int,stdin.readline().split()))
s=input()
print(len(s))
``` | instruction | 0 | 34,484 | 20 | 68,968 |
Yes | output | 1 | 34,484 | 20 | 68,969 |
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