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.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690 | instruction | 0 | 63,473 | 20 | 126,946 |
Tags: math, number theory
Correct Solution:
```
a=input()
cnt=[0]*10
for i in (1,6,8,9):
cnt[i]=-1
for i in a:
cnt[int(i)]+=1
mod = [1869, 1968, 9816, 6198, 1698, 1986, 1896, 1869]
modCnt=0
for i in range(1,10):
for j in range(cnt[i]):
modCnt= (modCnt*3 + i)%7
print(str(i)*cnt[i], end='')
modCnt=(10000*modCnt)%7
print(str(mod[7-modCnt])+'0'*cnt[0])
# Made By Mostafa_Khaled
``` | output | 1 | 63,473 | 20 | 126,947 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690 | instruction | 0 | 63,474 | 20 | 126,948 |
Tags: math, number theory
Correct Solution:
```
a=input()
cnt=[0]*10
for i in (1,6,8,9):
cnt[i]=-1
for i in a:
cnt[int(i)]+=1
mod = [1869, 1968, 9816, 6198, 1698, 1986, 1896, 1869]
modCnt=0
for i in range(1,10):
for j in range(cnt[i]):
modCnt= (modCnt*3 + i)%7
print(str(i)*cnt[i], end='')
modCnt=(10000*modCnt)%7
print(str(mod[7-modCnt])+'0'*cnt[0])
``` | output | 1 | 63,474 | 20 | 126,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690 | instruction | 0 | 63,475 | 20 | 126,950 |
Tags: math, number theory
Correct Solution:
```
a = input()
b=[0]*10
total = 0
for i in a:
b[int(i)]+=1
for i in [1,6,8,9]:
b[i]-=1
for i in range(1,10):
for j in range(b[i]):
total = (total * 10 + i) % 7
print(str(i)*b[i],end = '')
total = (10000 * total)%7
z = ['1869','6189','9186','6198','8691','9168','1896','1869']
print(z[7 - total]+'0'*b[0])
``` | output | 1 | 63,475 | 20 | 126,951 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690 | instruction | 0 | 63,476 | 20 | 126,952 |
Tags: math, number theory
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from itertools import permutations
from collections import Counter
def main():
a=input().rstrip()
n=len(a)
a=Counter(a)
for i in"6189":
a[i]-=1
z,s,y=0,[],[1]
for i in range(n):
y.append((y[-1]*10)%7)
xx=1
for i in a:
if i!="0":
x=int(i)
for j in range(a[i]):
s.append(i)
z=(z+x*y[n-xx])%7
xx+=1
f=1
for i in permutations([1,6,8,9]):
if (z+i[0]*y[n-xx]+i[1]*y[n-xx-1]+i[2]*y[n-xx-2]+i[3]*(y[n-xx-3]))%7==0:
f=0
s.extend([str(i[0]),str(i[1]),str(i[2]),str(i[3])])
break
if f:
print(0)
else:
s.append("0"*a["0"])
print("".join(s))
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
``` | output | 1 | 63,476 | 20 | 126,953 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690 | instruction | 0 | 63,477 | 20 | 126,954 |
Tags: math, number theory
Correct Solution:
```
######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
######### # # # # ##### # ##### # ## # ## # #
"""
PPPPPPP RRRRRRR OOOO VV VV EEEEEEEEEE
PPPPPPPP RRRRRRRR OOOOOO VV VV EE
PPPPPPPPP RRRRRRRRR OOOOOOOO VV VV EE
PPPPPPPP RRRRRRRR OOOOOOOO VV VV EEEEEE
PPPPPPP RRRRRRR OOOOOOOO VV VV EEEEEEE
PP RRRR OOOOOOOO VV VV EEEEEE
PP RR RR OOOOOOOO VV VV EE
PP RR RR OOOOOO VV VV EE
PP RR RR OOOO VVVV EEEEEEEEEE
"""
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // gcd(a, b)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p + p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
from bisect import bisect_left
def solve():
n = input().rstrip()
b = [0] * 10
for x in n:
b[int(x)] += 1
for x in '1689':
b[int(x)] -= 1
# print(n)
# y = list(permutations('1869', 4))
y = ['1869', '1968', '9816', '6198', '1698', '1986', '1896', '1869']
total = 0
for i in range(1, 10):
for j in range(b[i]):
total = (total * 10 + i) % 7
# print(i, b[i], str(i)*b[i])
print(str(i) * b[i], end = "")
# total = (total * (10 ** b[i]) + int(str(i) * b[i])) % 7
total = (total * 10000) % 7
print(y[7 - total] + '0' * b[0])
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | output | 1 | 63,477 | 20 | 126,955 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690
Submitted Solution:
```
a = input()
b=[0]*10
total = 0
for i in a:
b[int(i)]+=1
for i in [1,6,8,9]:
b[i]-=1
for i in range(1,10):
for j in range(b[i]):
total = (total * 10 + i) % 7
print(str(i)*b[i],end = '')
total = (10000 * total)%7
z = ['1869','6198','1896','9186','9168','6189','8691']
print(z[total]+'0'*b[0])
``` | instruction | 0 | 63,478 | 20 | 126,956 |
No | output | 1 | 63,478 | 20 | 126,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690
Submitted Solution:
```
a=input()
cnt=[0]*10
for i in (1,6,8,9):
cnt[i]=-1
for i in a:
cnt[int(i)]+=1
mod = [1869, 1968, 9816, 1886, 1698, 6298, 1896]
modCnt=0
for i in range(1,10):
for j in range(cnt[i]):
modCnt= (10*modCnt + i)%7
print(j, end="")
print(mod[(7-modCnt)%7], end="")
print("0"*cnt[0])
``` | instruction | 0 | 63,479 | 20 | 126,958 |
No | output | 1 | 63,479 | 20 | 126,959 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690
Submitted Solution:
```
a = input()
b=[0]*10
total = 0
for i in a:
b[int(i)]+=1
for i in [1,6,8,9]:
b[i]-=1
for i in range(1,10):
total = (total + (b[i] * (i % 7))) % 7
print('i'*b[i],end = '')
z = ['1869','6198','1896','9186','9168','6189','8691']
print(z[total]+'0'*b[0])
``` | instruction | 0 | 63,480 | 20 | 126,960 |
No | output | 1 | 63,480 | 20 | 126,961 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have number a, whose decimal representation quite luckily contains digits 1, 6, 8, 9. Rearrange the digits in its decimal representation so that the resulting number will be divisible by 7.
Number a doesn't contain any leading zeroes and contains digits 1, 6, 8, 9 (it also can contain another digits). The resulting number also mustn't contain any leading zeroes.
Input
The first line contains positive integer a in the decimal record. It is guaranteed that the record of number a contains digits: 1, 6, 8, 9. Number a doesn't contain any leading zeroes. The decimal representation of number a contains at least 4 and at most 106 characters.
Output
Print a number in the decimal notation without leading zeroes — the result of the permutation.
If it is impossible to rearrange the digits of the number a in the required manner, print 0.
Examples
Input
1689
Output
1869
Input
18906
Output
18690
Submitted Solution:
```
######### ## ## ## #### ##### ## # ## # ##
# # # # # # # # # # # # # # # # # # #
# # # # ### # # # # # # # # # # # #
# ##### # # # # ### # # # # # # # # #####
# # # # # # # # # # # # # # # # # #
######### # # # # ##### # ##### # ## # ## # #
"""
PPPPPPP RRRRRRR OOOO VV VV EEEEEEEEEE
PPPPPPPP RRRRRRRR OOOOOO VV VV EE
PPPPPPPPP RRRRRRRRR OOOOOOOO VV VV EE
PPPPPPPP RRRRRRRR OOOOOOOO VV VV EEEEEE
PPPPPPP RRRRRRR OOOOOOOO VV VV EEEEEEE
PP RRRR OOOOOOOO VV VV EEEEEE
PP RR RR OOOOOOOO VV VV EE
PP RR RR OOOOOO VV VV EE
PP RR RR OOOO VVVV EEEEEEEEEE
"""
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1
return x
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
def fact(x):
val = 1
while x > 0:
val *= x
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## lcm function
def lcm(a, b):
return (a * b) // gcd(a, b)
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if k > n:
return 0
if(k > n - k):
k = n - k
res = 1
for i in range(k):
res = res * (n - i)
res = res / (i + 1)
return int(res)
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
prime[0], prime[1] = False, False
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p + p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
res = []
for i in range(2, int(x ** 0.5) + 1):
while x % i == 0:
res.append(i)
x //= i
if x != 1:
res.append(x)
return res
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
import math
from bisect import bisect_left
def solve():
n = input().rstrip()
y = '1869'
for x in n:
if x in '0':
y = y + x
elif x not in '1869':
y = x + y
print(y)
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | instruction | 0 | 63,481 | 20 | 126,962 |
No | output | 1 | 63,481 | 20 | 126,963 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,498 | 20 | 126,996 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
a=int(input())
b,c=10**35,a-45*35*10**34%a
print(c,c+b-1)
``` | output | 1 | 63,498 | 20 | 126,997 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,499 | 20 | 126,998 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
a = int(input())
g = 45 * (10 ** 19) * 20
g = a + (a - g) % a
print(g, 10 ** 20 + g - 1)
``` | output | 1 | 63,499 | 20 | 126,999 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,500 | 20 | 127,000 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
m = int(input())
x,t=10**100-1,m-100*45*10**99%m
print(t,t+x)
``` | output | 1 | 63,500 | 20 | 127,001 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,501 | 20 | 127,002 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
from math import inf as inf
from math import *
from collections import *
import sys
from itertools import permutations
input=sys.stdin.readline
t=1
while(t):
t-=1
n=int(input())
f=n-45*20*(10**19)%n
print(f,f+(10**20-1))
``` | output | 1 | 63,501 | 20 | 127,003 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,502 | 20 | 127,004 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
a=int(input())
b,c=10**100,a-4500*10**99%a
print(c,c+b-1)
``` | output | 1 | 63,502 | 20 | 127,005 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,503 | 20 | 127,006 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
__author__ = 'andybear'
m=int(input())
x,t=10**100-1,m-100*45*10**99%m
print(t,t+x)
``` | output | 1 | 63,503 | 20 | 127,007 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333 | instruction | 0 | 63,504 | 20 | 127,008 |
Tags: binary search, constructive algorithms, math
Correct Solution:
```
a = int(input())
k = 19
x = (45 * k * 10**(k-1))% a
l = a - x
r = l + 10**k - 1
print(l, r)
``` | output | 1 | 63,504 | 20 | 127,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333
Submitted Solution:
```
a = int(input());
t = 1000000000000000000000000000;
if (45 * t) % a == 0:
print (1, t - 1);
else:
print(((45 * t) % a), t + ((45 * t) % a));
``` | instruction | 0 | 63,505 | 20 | 127,010 |
No | output | 1 | 63,505 | 20 | 127,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333
Submitted Solution:
```
a=int(input())
b,c=10**100,a-45*100*10**99%a
print(c,c+b)
``` | instruction | 0 | 63,506 | 20 | 127,012 |
No | output | 1 | 63,506 | 20 | 127,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333
Submitted Solution:
```
a=int(input())
b,c=10**50,a-4500*10**49%a
print(c,c+b-1)
``` | instruction | 0 | 63,507 | 20 | 127,014 |
No | output | 1 | 63,507 | 20 | 127,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little X has met the following problem recently.
Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>
Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code:
ans = solve(l, r) % a;
if (ans <= 0)
ans += a;
This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.
Input
The first line contains a single integer a (1 ≤ a ≤ 1018).
Output
Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.
Examples
Input
46
Output
1 10
Input
126444381000032
Output
2333333 2333333333333
Submitted Solution:
```
a = int(input());
t = 1000000000000000000000000000;
d = 111111111111111111111111111;
if (45 * d) % a == 0:
print (1, t - 1);
else:
print((a - (45 * d) % a), t + (a - (45 * d) % a) - 1);
``` | instruction | 0 | 63,508 | 20 | 127,016 |
No | output | 1 | 63,508 | 20 | 127,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,006 | 20 | 128,012 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
a=input()
while a!="end":
if a=="start":
for i in range(30):
print("?",2**i,2*2**i)
if input()=="x":
break
bot=2**i
top=min(2**(i+1),1000000000)
if bot==1:
print("?",2,1)
if input()=="x":
print("!",1)
else:
print("!",2)
else:
bot+=1
while top!=bot:
mid=(bot+top)//2
print("?",2*mid,mid)
if input()=="x":
top=mid
else:
bot=mid+1
print("!",top)
a=input()
``` | output | 1 | 64,006 | 20 | 128,013 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,007 | 20 | 128,014 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
from sys import stdout
def ask(x, y):
print('?', x, y)
stdout.flush()
x = input()
return x == 'x'
def show_res(res):
print('!', res)
stdout.flush()
def prepare_game():
gm = input()
if gm[0] != 's':
exit(0)
INF = int(1e9) + 1
def play():
L = 0
R = 1
while R < INF:
res = ask(L, R)
if (res):
break
else:
L = R
R *= 2
st = L
#print('L', L, 'R', R)
while R - L > 1:
m = (R + L) // 2
#print('L', L, 'R', R, 'm', m)
if ask(st, m):
R = m
else:
L = m
show_res(R)
while True:
prepare_game()
play()
``` | output | 1 | 64,007 | 20 | 128,015 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,008 | 20 | 128,016 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
import io
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
from collections import deque
import threading
#sys.setrecursionlimit(300000)
#threading.stack_size(10**8)
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#-------------------------------------------------------------------------
#mod = 9223372036854775807
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a+b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
class SegmentTree1:
def __init__(self, data, default=10**6, func=lambda a, b: min(a,b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
MOD=10**9+7
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
mod=10**9+7
omod=998244353
#---------------------------------Lazy Segment Tree--------------------------------------
# https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
class LazySegTree:
def __init__(self, _op, _e, _mapping, _composition, _id, v):
def set(p, x):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
_d[p] = x
for i in range(1, _log + 1):
_update(p >> i)
def get(p):
assert 0 <= p < _n
p += _size
for i in range(_log, 0, -1):
_push(p >> i)
return _d[p]
def prod(l, r):
assert 0 <= l <= r <= _n
if l == r:
return _e
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push(r >> i)
sml = _e
smr = _e
while l < r:
if l & 1:
sml = _op(sml, _d[l])
l += 1
if r & 1:
r -= 1
smr = _op(_d[r], smr)
l >>= 1
r >>= 1
return _op(sml, smr)
def apply(l, r, f):
assert 0 <= l <= r <= _n
if l == r:
return
l += _size
r += _size
for i in range(_log, 0, -1):
if ((l >> i) << i) != l:
_push(l >> i)
if ((r >> i) << i) != r:
_push((r - 1) >> i)
l2 = l
r2 = r
while l < r:
if l & 1:
_all_apply(l, f)
l += 1
if r & 1:
r -= 1
_all_apply(r, f)
l >>= 1
r >>= 1
l = l2
r = r2
for i in range(1, _log + 1):
if ((l >> i) << i) != l:
_update(l >> i)
if ((r >> i) << i) != r:
_update((r - 1) >> i)
def _update(k):
_d[k] = _op(_d[2 * k], _d[2 * k + 1])
def _all_apply(k, f):
_d[k] = _mapping(f, _d[k])
if k < _size:
_lz[k] = _composition(f, _lz[k])
def _push(k):
_all_apply(2 * k, _lz[k])
_all_apply(2 * k + 1, _lz[k])
_lz[k] = _id
_n = len(v)
_log = _n.bit_length()
_size = 1 << _log
_d = [_e] * (2 * _size)
_lz = [_id] * _size
for i in range(_n):
_d[_size + i] = v[i]
for i in range(_size - 1, 0, -1):
_update(i)
self.set = set
self.get = get
self.prod = prod
self.apply = apply
MIL = 1 << 20
def makeNode(total, count):
# Pack a pair into a float
return (total * MIL) + count
def getTotal(node):
return math.floor(node / MIL)
def getCount(node):
return node - getTotal(node) * MIL
nodeIdentity = makeNode(0.0, 0.0)
def nodeOp(node1, node2):
return node1 + node2
# Equivalent to the following:
return makeNode(
getTotal(node1) + getTotal(node2), getCount(node1) + getCount(node2)
)
identityMapping = -1
def mapping(tag, node):
if tag == identityMapping:
return node
# If assigned, new total is the number assigned times count
count = getCount(node)
return makeNode(tag * count, count)
def composition(mapping1, mapping2):
# If assigned multiple times, take first non-identity assignment
return mapping1 if mapping1 != identityMapping else mapping2
#-------------------------------------------------------------------------
prime = [True for i in range(10)]
pp=[0]*10
def SieveOfEratosthenes(n=10):
p = 2
c=0
while (p * p <= n):
if (prime[p] == True):
c+=1
for i in range(p, n+1, p):
pp[i]+=1
prime[i] = False
p += 1
#---------------------------------Binary Search------------------------------------------
def binarySearch(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[n-1]
while (left <= right):
mid = (right + left)//2
if (arr[mid] >= key):
res=arr[mid]
right = mid-1
else:
left = mid + 1
return res
def binarySearch1(arr, n, key):
left = 0
right = n-1
mid = 0
res=arr[0]
while (left <= right):
mid = (right + left)//2
if (arr[mid] > key):
right = mid-1
else:
res=arr[mid]
left = mid + 1
return res
#---------------------------------running code------------------------------------------
t=1
#t=int(input())
for _ in range (t):
#n=int(input())
#n,k=map(int,input().split())
#a=list(map(int,input().split()))
s=input()
#n=len(s)
while s=="start":
l,r=0,0
c=1
for i in range (31):
print("?",c//2,c,flush=True)
ch=input()
if ch=="x":
l,r=c//2,c
break
c*=2
res=r
left,right = l+1,r
while left<=right:
mid=(left+right)//2
print("?",l,mid,flush=True)
ch=input()
if ch=="x":
res=mid
right=mid-1
else:
left=mid+1
print("!",res,flush=True)
s=input()
``` | output | 1 | 64,008 | 20 | 128,017 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,009 | 20 | 128,018 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
from sys import stdout
def ask(x,y) :
print('?',x,y)
stdout.flush()
return input() == 'x'
def work() :
if ask(0, 1) : return 1
l = 1
while ask(l*2,l) : l*=2
r = min(1000000000, l*2+1)
l += 1
while r > l :
mid=(l+r)//2
if ask(mid, (mid+1)//2) :
l=mid+1
else :
r=mid
return l
while input()=='start' :
print('!',work())
stdout.flush()
``` | output | 1 | 64,009 | 20 | 128,019 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,010 | 20 | 128,020 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
import sys
def question(x, y, a):
#a = 1543
if x % a >= y % a:
return "x"
else:
return "y"
def real_question(x, y):
print("? {0} {1}".format(x, y))
sys.stdout.flush()
return input()
class CodeforcesTask1103BSolution:
def __init__(self):
self.result = ''
def read_input(self):
pass
def process_task(self):
#qqq = 536870918
#corr = 2
#quecorr = 2
while True:
cmd = input()
#cmd = "start"
if cmd == "start":
q_res = "y"
x = 0
qus = 0
while q_res != "x":
qus += 1
q_res = real_question(2 ** x, 2 ** (x + 1))
#q_res = question(2 ** x, 2 ** (x + 1), qqq)
#print(qus, q_res, 2 ** x, 2 ** (x + 1))
l = 2 ** x
r = 2 ** (x + 1)
x += 1
if r == 2:
q_res = real_question(2, 1)
#q_res = question(2, 1, qqq)
if q_res == "x":
print("! 1")
sys.stdout.flush()
else:
print("! 2")
sys.stdout.flush()
else:
mid = l + (r - l) // 2
l0 = l
found = False
following = False
r += 2
while not found:
qus += 1
q_res = real_question(l0, mid)
#q_res = question(l0, mid, qqq)
#print(q_res, l, mid, r)
if q_res == "y":
l = mid
mid += (r - l) // 2
else:
r = mid
mid -= (r - l) // 2
#print(mid)
if r == mid:
found = True
elif mid + 1 == r and following > 1:
found = True
elif mid + 1 == r:
following += 1
print("! {0}".format(r))
sys.stdout.flush()
#if qus <= 60:
# quecorr += 1
#if r == qqq:
# corr += 1
#print(qqq- 536870918 + 3, corr, quecorr)
#qqq += 1
elif cmd == "mistake":
print("error")
break
else:
break
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask1103BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
``` | output | 1 | 64,010 | 20 | 128,021 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1. | instruction | 0 | 64,011 | 20 | 128,022 |
Tags: binary search, constructive algorithms, interactive
Correct Solution:
```
import sys
def ask(x, y):
print("?", x, y)
return input() == "y"
while input() == "start":
if not ask(0, 1):
print("! 1")
continue
d = 1
while ask(d, d * 2):
d *= 2
r = d
l = d // 2
while l + 1 < r:
m = (l + r) // 2
if ask(m, m * 2):
l = m
else:
r = m
print("!", r * 2 if not ask(r * 2 - 1, r * 2) else r * 2 - 1)
``` | output | 1 | 64,011 | 20 | 128,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1.
Submitted Solution:
```
while (1):
if input()=="end":
break
x,y=0,1
for i in range(0,30):
print("?",x,y,flush=True)
xx=input()
if xx=="x":
p,q=x,y
break
x=y
y*=2
do=p+1
up=q
while up>do:
mid=(up+do)//2
print("?",p,mid, flush=True)
xx = input()
if xx=="x":
an=xx
up=mid-1
else:
do=mid+1
print("!",xx,flush=True)
``` | instruction | 0 | 64,012 | 20 | 128,024 |
No | output | 1 | 64,012 | 20 | 128,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1.
Submitted Solution:
```
a=input()
while a!="end":
if a=="start":
for i in range(30):
print("?",2**i,2*2**i)
if input()=="x":
break
bot=2**i
top=2**(i+1)
if bot==1:
print("?",2,1)
if input()=="x":
print("!",1)
else:
print("!",2)
else:
bot+=1
while top!=bot:
mid=(bot+top)//2
print("?",2*mid,mid)
if input()=="x":
top=mid
else:
bot=mid+1
print("!",top)
a=input()
``` | instruction | 0 | 64,013 | 20 | 128,026 |
No | output | 1 | 64,013 | 20 | 128,027 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1.
Submitted Solution:
```
import sys
def solve():
ans = 0
hi = 1<<30
while ans < hi-1:
k = -1
while True:
p1 = 1<<k if k != -1 else 0
p2 = 1<<(k+1)
print("?", ans + p1, ans + p2)
sys.stdout.flush()
res = input().strip()
if res == "x":
ans += p1
hi = ans + (1<<(k+1))
break
k += 1
print("!", ans+1)
while True:
order = input().strip()
if order == "end":
exit()
solve()
``` | instruction | 0 | 64,014 | 20 | 128,028 |
No | output | 1 | 64,014 | 20 | 128,029 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is an interactive problem.
Vasya and Petya are going to play the following game: Petya has some positive integer number a. After that Vasya should guess this number using the following questions. He can say a pair of non-negative integer numbers (x, y). Petya will answer him:
* "x", if (x mod a) ≥ (y mod a).
* "y", if (x mod a) < (y mod a).
We define (x mod a) as a remainder of division x by a.
Vasya should guess the number a using no more, than 60 questions.
It's guaranteed that Petya has a number, that satisfies the inequality 1 ≤ a ≤ 10^9.
Help Vasya playing this game and write a program, that will guess the number a.
Interaction
Your program should play several games.
Before the start of any game your program should read the string:
* "start" (without quotes) — the start of the new game.
* "mistake" (without quotes) — in the previous game, you found the wrong answer. Your program should terminate after reading this string and it will get verdict "Wrong answer".
* "end" (without quotes) — all games finished. Your program should terminate after reading this string.
After reading the string "start" (without quotes) the new game starts.
At the beginning, your program should ask several questions about pairs of non-negative integer numbers (x, y). You can only ask the numbers, that satisfy the inequalities 0 ≤ x, y ≤ 2 ⋅ 10^9. To ask a question print "? x y" (without quotes). As the answer, you should read one symbol:
* "x" (without quotes), if (x mod a) ≥ (y mod a).
* "y" (without quotes), if (x mod a) < (y mod a).
* "e" (without quotes) — you asked more than 60 questions. Your program should terminate after reading this string and it will get verdict "Wrong answer".
After your program asked several questions your program should print the answer in form "! a" (without quotes). You should print the number a satisfying the inequalities 1 ≤ a ≤ 10^9. It's guaranteed that Petya's number a satisfied this condition. After that, the current game will finish.
We recall that your program can't ask more than 60 questions during one game.
If your program doesn't terminate after reading "mistake" (without quotes), "end" (without quotes) or "e" (without quotes), it can get any verdict, because it will continue reading from closed input. Also, if your program prints answer or question in the incorrect format it can get any verdict, too. Be careful.
Don't forget to flush the output after printing questions and answers.
To flush the output, you can use:
* fflush(stdout) in C++.
* System.out.flush() in Java.
* stdout.flush() in Python.
* flush(output) in Pascal.
* See the documentation for other languages.
It's guaranteed that you should play at least 1 and no more than 100 games.
Hacks:
In hacks, you can use only one game. To hack a solution with Petya's number a (1 ≤ a ≤ 10^9) in the first line you should write a single number 1 and in the second line you should write a single number a.
Example
Input
start
x
x
start
x
x
y
start
x
x
y
y
end
Output
? 0 0
? 10 1
! 1
? 0 0
? 3 4
? 2 5
! 2
? 2 4
? 2 5
? 3 10
? 9 1
! 3
Note
In the first test, you should play 3 games with Petya's numbers 1, 2 and 3.
In the first game, Petya will answer "x" (without quotes) to any question, because (x mod 1) = 0 for any integer x.
In the second game, if you will ask pair (0, 0), the answer will be "x" (without quotes), because (0 mod 2) ≥ (0 mod 2). But if you will ask pair (2, 5), the answer will be "y" (without quotes), because (2 mod 2) < (5 mod 2), because (2 mod 2) = 0 and (5 mod 2) = 1.
Submitted Solution:
```
from sys import stdin,stdout
stdout.flush()
run=stdin.readline().strip()
while run!="end":
print("? 0 1")
stdout.flush()
ans=stdin.readline().strip()
if ans=="x":
print("! 1")
stdout.flush()
run=stdin.readline().strip()
continue
x=0;y=1
x1=2**x;y1=2**y
print("?",x1,y1)
stdout.flush()
ans=stdin.readline().strip()
while ans!="x":
x+=1;y+=1
x1=2**x;y1=2**y
print("?",x1,y1)
stdout.flush()
ans=stdin.readline().strip()
while y1-x1>1:
mid=(x1+y1)//2
print("?",mid,y1)
stdout.flush()
ans=stdin.readline().strip()
if ans=="y":
y1=mid
else:
x1=mid
print("?",x1,y1)
stdout.flush()
ans=stdin.readline().strip()
if ans=="y":
print("!",x1)
stdout.flush()
else:
print("!",y1)
stdout.flush()
run=stdin.readline().strip()
``` | instruction | 0 | 64,015 | 20 | 128,030 |
No | output | 1 | 64,015 | 20 | 128,031 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tokitsukaze is playing a room escape game designed by SkywalkerT. In this game, she needs to find out hidden clues in the room to reveal a way to escape.
After a while, she realizes that the only way to run away is to open the digital door lock since she accidentally went into a secret compartment and found some clues, which can be interpreted as:
* Only when you enter n possible different passwords can you open the door;
* Passwords must be integers ranged from 0 to (m - 1);
* A password cannot be x (0 ≤ x < m) if x and m are not coprime (i.e. x and m have some common divisor greater than 1);
* A password cannot be x (0 ≤ x < m) if there exist non-negative integers e and k such that p^e = k m + x, where p is a secret integer;
* Any integer that doesn't break the above rules can be a password;
* Several integers are hidden in the room, but only one of them can be p.
Fortunately, she finds that n and m are recorded in the lock. However, what makes Tokitsukaze frustrated is that she doesn't do well in math. Now that she has found an integer that is suspected to be p, she wants you to help her find out n possible passwords, or determine the integer cannot be p.
Input
The only line contains three integers n, m and p (1 ≤ n ≤ 5 × 10^5, 1 ≤ p < m ≤ 10^{18}).
It is guaranteed that m is a positive integer power of a single prime number.
Output
If the number of possible different passwords is less than n, print a single integer -1.
Otherwise, print n distinct integers ranged from 0 to (m - 1) as passwords. You can print these integers in any order. Besides, if there are multiple solutions, print any.
Examples
Input
1 2 1
Output
-1
Input
3 5 1
Output
2 4 3
Input
2 5 4
Output
2 3
Input
4 9 8
Output
2 4 7 5
Note
In the first example, there is no possible password.
In each of the last three examples, the given integer n equals to the number of possible different passwords for the given integers m and p, so if the order of numbers in the output is ignored, the solution is unique as shown above.
Submitted Solution:
```
import sys
import math
n, m, p = list(map(int,sys.stdin.readline().strip().split()))
def pow(a,b):
a = a % m
ans = 1
while b > 0:
if b % 2 == 1:
b = b - 1
ans = (ans * a) % m
else:
b = b // 2
a = (a * a) % m
return ans
q = 2
while m % q != 0:
q = q + 1
e = 0
m2 = m
while (m2 % q == 0):
e = e + 1
m2 = m2 // q
phi = (q - 1) * (q ** (e - 1))
c = 0
x = pow(p, q-1)
while x != 1:
x = pow(x, q)
c = c + 1
p2 = pow(p, q ** c)
y = q-1
P = []
d = 2
while y != 1:
if y % d == 0:
y = y // d
P.append(d)
else:
d = d + 1
y = q-1
for d in P:
if pow(p2, y // d) == 1:
y = y // d
if p % q == 0:
if phi > n:
i = 1
ans = []
while n > 0:
if i % q != 0:
ans.append(str(i))
n = n - 1
i = i + 1
else:
print(-1)
if phi - y < n:
print(-1)
else:
i = 0
ans = []
while n > 0:
if i % q != 0:
if pow(i, y) == 1:
i = i + 1
else:
ans.append(str(i))
n = n - 1
i = i + 1
else:
i = i + 1
print(" ".join(list(map(str, ans))))
``` | instruction | 0 | 64,064 | 20 | 128,128 |
No | output | 1 | 64,064 | 20 | 128,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tokitsukaze is playing a room escape game designed by SkywalkerT. In this game, she needs to find out hidden clues in the room to reveal a way to escape.
After a while, she realizes that the only way to run away is to open the digital door lock since she accidentally went into a secret compartment and found some clues, which can be interpreted as:
* Only when you enter n possible different passwords can you open the door;
* Passwords must be integers ranged from 0 to (m - 1);
* A password cannot be x (0 ≤ x < m) if x and m are not coprime (i.e. x and m have some common divisor greater than 1);
* A password cannot be x (0 ≤ x < m) if there exist non-negative integers e and k such that p^e = k m + x, where p is a secret integer;
* Any integer that doesn't break the above rules can be a password;
* Several integers are hidden in the room, but only one of them can be p.
Fortunately, she finds that n and m are recorded in the lock. However, what makes Tokitsukaze frustrated is that she doesn't do well in math. Now that she has found an integer that is suspected to be p, she wants you to help her find out n possible passwords, or determine the integer cannot be p.
Input
The only line contains three integers n, m and p (1 ≤ n ≤ 5 × 10^5, 1 ≤ p < m ≤ 10^{18}).
It is guaranteed that m is a positive integer power of a single prime number.
Output
If the number of possible different passwords is less than n, print a single integer -1.
Otherwise, print n distinct integers ranged from 0 to (m - 1) as passwords. You can print these integers in any order. Besides, if there are multiple solutions, print any.
Examples
Input
1 2 1
Output
-1
Input
3 5 1
Output
2 4 3
Input
2 5 4
Output
2 3
Input
4 9 8
Output
2 4 7 5
Note
In the first example, there is no possible password.
In each of the last three examples, the given integer n equals to the number of possible different passwords for the given integers m and p, so if the order of numbers in the output is ignored, the solution is unique as shown above.
Submitted Solution:
```
import sys
import math
n, m, p = list(map(int,sys.stdin.readline().strip().split()))
def pow(a,b):
ans = 1
while b > 0:
if b % 2 == 1:
b = b - 1
ans = (ans * a) % m
else:
b = b // 2
a = (a * a) % m
return ans
q = 2
while m % q != 0:
q = q + 1
e = 0
m2 = m
while (m2 % q == 0):
e = e + 1
m2 = m2 // q
phi = (q - 1) * (q ** (e - 1))
a = 0
x = pow(p, q-1)
while x != 1:
x = pow(x, q)
a = a + 1
p2 = pow(p, q ** a)
y = q-1
P = []
d = 2
while y != 1:
if y % d == 0:
y = y // d
P.append(d)
else:
d = d + 1
y = q-1
for d in P:
if pow(p2, y // d) == 1:
y = y // d
if phi - y < n:
print(-1)
else:
i = 0
ans = []
while n > 0:
if i % q != 0:
if pow(i, y) == 1:
i = i + 1
else:
ans.append(str(i))
n = n - 1
i = i + 1
else:
i = i + 1
print(" ".join(list(map(str, ans))))
``` | instruction | 0 | 64,065 | 20 | 128,130 |
No | output | 1 | 64,065 | 20 | 128,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Tokitsukaze is playing a room escape game designed by SkywalkerT. In this game, she needs to find out hidden clues in the room to reveal a way to escape.
After a while, she realizes that the only way to run away is to open the digital door lock since she accidentally went into a secret compartment and found some clues, which can be interpreted as:
* Only when you enter n possible different passwords can you open the door;
* Passwords must be integers ranged from 0 to (m - 1);
* A password cannot be x (0 ≤ x < m) if x and m are not coprime (i.e. x and m have some common divisor greater than 1);
* A password cannot be x (0 ≤ x < m) if there exist non-negative integers e and k such that p^e = k m + x, where p is a secret integer;
* Any integer that doesn't break the above rules can be a password;
* Several integers are hidden in the room, but only one of them can be p.
Fortunately, she finds that n and m are recorded in the lock. However, what makes Tokitsukaze frustrated is that she doesn't do well in math. Now that she has found an integer that is suspected to be p, she wants you to help her find out n possible passwords, or determine the integer cannot be p.
Input
The only line contains three integers n, m and p (1 ≤ n ≤ 5 × 10^5, 1 ≤ p < m ≤ 10^{18}).
It is guaranteed that m is a positive integer power of a single prime number.
Output
If the number of possible different passwords is less than n, print a single integer -1.
Otherwise, print n distinct integers ranged from 0 to (m - 1) as passwords. You can print these integers in any order. Besides, if there are multiple solutions, print any.
Examples
Input
1 2 1
Output
-1
Input
3 5 1
Output
2 4 3
Input
2 5 4
Output
2 3
Input
4 9 8
Output
2 4 7 5
Note
In the first example, there is no possible password.
In each of the last three examples, the given integer n equals to the number of possible different passwords for the given integers m and p, so if the order of numbers in the output is ignored, the solution is unique as shown above.
Submitted Solution:
```
import sys
import math
n, m, p = list(map(int,sys.stdin.readline().strip().split()))
def pow(a,b):
a = a % m
ans = 1
while b > 0:
if b % 2 == 1:
b = b - 1
ans = (ans * a) % m
else:
b = b // 2
a = (a * a) % m
return ans
q = 2
while m % q != 0:
q = q + 1
e = 0
m2 = m
while (m2 % q == 0):
e = e + 1
m2 = m2 // q
phi = (q - 1) * (q ** (e - 1))
c = 0
x = pow(p, q-1)
while x != 1:
x = pow(x, q)
c = c + 1
p2 = pow(p, q ** c)
y = q-1
P = []
d = 2
while y != 1:
if y % d == 0:
y = y // d
P.append(d)
else:
d = d + 1
y = q-1
for d in P:
if pow(p2, y // d) == 1:
y = y // d
if p % q == 0:
if phi > n:
i = 1
ans = []
while n > 0:
if i % q != 0:
ans.append(str(i))
n = n - 1
i = i + 1
print(" ".join(list(map(str, ans))))
else:
print(-1)
elif phi - y < n:
print(-1)
else:
i = 0
ans = []
while n > 0:
if i % q != 0:
if pow(i, y) == 1:
i = i + 1
else:
ans.append(str(i))
n = n - 1
i = i + 1
else:
i = i + 1
print(" ".join(list(map(str, ans))))
``` | instruction | 0 | 64,066 | 20 | 128,132 |
No | output | 1 | 64,066 | 20 | 128,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,067 | 20 | 128,134 |
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7
# print(math.factorial(5))
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
# print(x)
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x,n,a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
return a
"""
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
"""
# ------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
# -----------------------------------------------lucky number with two lucky any digits
res = set()
def solven(p, l, a, b, n): # given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solven(p * 10 + a, l + 1, a, b, n)
solven(p * 10 + b, l + 1, a, b, n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1, d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r, d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r, d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r, d, sum)
return d
def print_factors(x):
factors = []
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return (factors)
# -----------------------------------------------
def calc(X, d, ans, D):
# print(X,d)
if len(X) == 0:
return
i = X.index(max(X))
ans[D[max(X)]] = d
Y = X[:i]
Z = X[i + 1:]
calc(Y, d + 1, ans, D)
calc(Z, d + 1, ans, D)
# ---------------------------------------
def factorization(n, l):
c = n
if prime(n) == True:
l.append(n)
return l
for i in range(2, c):
if n == 1:
break
while n % i == 0:
l.append(i)
n = n // i
return l
# endregion------------------------------
def good(b):
l = []
i = 0
while (len(b) != 0):
if b[i] < b[len(b) - 1 - i]:
l.append(b[i])
b.remove(b[i])
else:
l.append(b[len(b) - 1 - i])
b.remove(b[len(b) - 1 - i])
if l == sorted(l):
# print(l)
return True
return False
# arr=[]
# print(good(arr))
def generate(st, s):
if len(s) == 0:
return
# If current string is not already present.
if s not in st:
st.add(s)
# Traverse current string, one by one
# remove every character and recur.
for i in range(len(s)):
t = list(s).copy()
t.remove(s[i])
t = ''.join(t)
generate(st, t)
return
#=--------------------------------------------longest increasing subsequence
def largestincreasingsubsequence(A):
l = [1]*len(A)
sub=[]
for i in range(1,len(l)):
for k in range(i):
if A[k]<A[i]:
sub.append(l[k])
l[i]=1+max(sub,default=0)
return max(l,default=0)
#----------------------------------longest palindromic substring
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
#l=[]
def OR(a, n):
ans = a[0]
for i in range(1, n):
ans |= a[i]
#l.append(ans)
return ans
#print(prime(12345678987766))
"""
def main():
q=inpu()
x = q
v1 = 0
v2 = 0
i = 2
while i * i <= q:
while q % i == 0:
if v1!=0:
v2 = i
else:
v1 = i
q //= i
i += 1
if q - 1!=0:
v2 = q
if v1 * v2 - x!=0:
print(1)
print(v1 * v2)
else:
print(2)
if __name__ == '__main__':
main()
"""
def luckynum(x,l,r,a):
if x >= l:
a.append(x)
if x>r:
a.append(x)
return a
if x < 1e10:
luckynum(x * 10 + 4, l,r,a)
luckynum(x * 10 + 7, l,r,a)
def main():
l,r = sep()
a=[]
luckynum(0,l,r,a)
a.sort()
i=0
ans=0
l-=1
#print(a)
while(True):
if r>a[i]:
ans+=(a[i]*(a[i]-l))
l=a[i]
else:
ans+=(a[i]*(r-l))
break
i+=1
print(ans)
if __name__ == '__main__':
main()
``` | output | 1 | 64,067 | 20 | 128,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,068 | 20 | 128,136 |
Tags: implementation
Correct Solution:
```
n = (input()).split()
l = int(n[0])
r = int(n[1])
a = []
x = []
a.append([])
a[0].append('4')
a[0].append('7')
for i in range(1,10):
a.append([])
for j in a[i-1]:
a[i].append('4'+j)
a[i].append('7'+j)
for j in a[i]:
x.append(int(j))
x.append(4)
x.append(7)
x.sort()
sum = [16]
for i in range(1,len(x)):
sum.append((x[i]-x[i-1])*x[i]+sum[i-1])
for i in range(len(x)):
if x[i] >= l:
t = i
break
for i in range(len(x)):
if x[i] >= r:
e = i
break
res = sum[e] - sum[t] - x[e] * (x[e]-r) + (x[t]-l+1) * x[t]
print(res)
``` | output | 1 | 64,068 | 20 | 128,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,069 | 20 | 128,138 |
Tags: implementation
Correct Solution:
```
lucky = [4, 7]
n = 1
length = 0
while n < 10:
check = lucky[length:(length+2**(n))]
for x in check:
lucky.append(int("4"+ str(x)))
for x in check:
lucky.append(int("7"+str(x)))
length += 2**n
n += 1
l, r = map(int, input().split())
ans = 0
i = 0
for num in lucky:
if num >= r:
ans += num * (r-l+1)
break
elif num >= l:
ans += num * (num-l+1)
l = num+1
print(ans)
``` | output | 1 | 64,069 | 20 | 128,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,070 | 20 | 128,140 |
Tags: implementation
Correct Solution:
```
l,r = [int(x) for x in input().strip().split()]
def next(n,k):
s = n
n = int(n)
if int("4"*k)>=n:
return "4"*k
elif int("7"*k)<n:
return "4"*(k+1)
elif 4<int(s[0])<7:
return "7"+"4"*(k-1)
if not s[1:]=="":
a = next(s[1:],k-1)
if s[0]=="4":
if len(str(a))==k:
return "7"+a[1:]
else:
return "4"+a
else:
return "7"+a
else:
return "7"
last = int(next(str(l),len(str(l))))
tot = 0
count = l
while count<=r:
# print(count,tot,last)
tot+=last*(min(last,r)-count+1)
count = last+1
b = str(last+1)
last=int(next(b,len(b)))
print(tot)
# print(next("1000000000",len("1000000000")))
``` | output | 1 | 64,070 | 20 | 128,141 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,071 | 20 | 128,142 |
Tags: implementation
Correct Solution:
```
def f(a,x):
if x//1e10>0:
return
a.append(x)
f(a,x*10+4)
f(a,x*10+7)
a,ans,i=[],0,0
f(a,4)
f(a,7)
a.sort()
L,R=map(int,input().split())
while L<=R:
while L>a[i]:i+=1
ans+=a[i]*(min(a[i],R)-L+1)
L=a[i]+1
print(ans)
``` | output | 1 | 64,071 | 20 | 128,143 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,072 | 20 | 128,144 |
Tags: implementation
Correct Solution:
```
lucky=[]
MAX=10000000000
def func(s):
s=s*10
if s>MAX:
return
lucky.append(s+4)
lucky.append(s+7)
func(s+4)
func(s+7)
func(0)
lucky.sort()
l,r=[int(x) for x in input().split(' ')]
i=0
j=0
while(l>lucky[i]):
i+=1
while(r>lucky[j]):
j+=1
ans=0
for x in range(i+1,j+1):
ans+=(lucky[x]-lucky[x-1])*lucky[x]
ans-=(lucky[j]-r)*lucky[j]
ans+=(lucky[i]-l+1)*lucky[i]
print(ans)
``` | output | 1 | 64,072 | 20 | 128,145 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,073 | 20 | 128,146 |
Tags: implementation
Correct Solution:
```
def ii(): return int(input())
def si(): return input()
def mi(): return map(int,input().split())
def li(): return list(mi())
import math
le,r=mi()
l=[]
def F(x):
l.append(x)
if x>r*10:return
F(10*x+4)
F(10*x+7)
F(0)
l.sort()
s=0
i=le-1
j=0
while(i<r):
if l[j]>i:
s+=(min(l[j],r)-i)*l[j]
i=l[j]
j+=1
print(s)
``` | output | 1 | 64,073 | 20 | 128,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7 | instruction | 0 | 64,074 | 20 | 128,148 |
Tags: implementation
Correct Solution:
```
from itertools import product
all_lucky = [('0')]
for i in range(1, 11):
all_lucky += list(product('47', repeat=i))
all_lucky = [int("".join(x)) for x in all_lucky]
all_lucky.sort()
left, right = map(int, input().split())
result = 0
prev = left
pos = 0
while all_lucky[pos+1] < left: pos += 1
while all_lucky[pos+1] < right:
result += (all_lucky[pos+1] - prev + 1) * all_lucky[pos+1]
prev = all_lucky[pos+1] + 1
pos += 1
result += (right - prev + 1) * all_lucky[pos+1]
print(result)
``` | output | 1 | 64,074 | 20 | 128,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
from bisect import bisect_left as bl
a=[4, 7, 44, 47, 74, 77, 444, 447, 474, 477, 744, 747, 774, 777, 4444, 4447, 4474, 4477, 4744, 4747, 4774, 4777, 7444, 7447, 7474, 7477, 7744, 7747, 7774, 7777, 44444, 44447, 44474, 44477, 44744, 44747, 44774, 44777, 47444, 47447, 47474, 47477, 47744, 47747, 47774, 47777, 74444, 74447, 74474, 74477, 74744, 74747, 74774, 74777, 77444, 77447, 77474, 77477, 77744, 77747, 77774, 77777, 444444, 444447, 444474, 444477, 444744, 444747, 444774, 444777, 447444, 447447, 447474, 447477, 447744, 447747, 447774, 447777, 474444, 474447, 474474, 474477, 474744, 474747, 474774, 474777, 477444, 477447, 477474, 477477, 477744, 477747, 477774, 477777, 744444, 744447, 744474, 744477, 744744, 744747, 744774, 744777, 747444, 747447, 747474, 747477, 747744, 747747, 747774, 747777, 774444, 774447, 774474, 774477, 774744, 774747, 774774, 774777, 777444, 777447, 777474, 777477, 777744, 777747, 777774, 777777, 4444444, 4444447, 4444474, 4444477, 4444744, 4444747, 4444774, 4444777, 4447444, 4447447, 4447474, 4447477, 4447744, 4447747, 4447774, 4447777, 4474444, 4474447, 4474474, 4474477, 4474744, 4474747, 4474774, 4474777, 4477444, 4477447, 4477474, 4477477, 4477744, 4477747, 4477774, 4477777, 4744444, 4744447, 4744474, 4744477, 4744744, 4744747, 4744774, 4744777, 4747444, 4747447, 4747474, 4747477, 4747744, 4747747, 4747774, 4747777, 4774444, 4774447, 4774474, 4774477, 4774744, 4774747, 4774774, 4774777, 4777444, 4777447, 4777474, 4777477, 4777744, 4777747, 4777774, 4777777, 7444444, 7444447, 7444474, 7444477, 7444744, 7444747, 7444774, 7444777, 7447444, 7447447, 7447474, 7447477, 7447744, 7447747, 7447774, 7447777, 7474444, 7474447, 7474474, 7474477, 7474744, 7474747, 7474774, 7474777, 7477444, 7477447, 7477474, 7477477, 7477744, 7477747, 7477774, 7477777, 7744444, 7744447, 7744474, 7744477, 7744744, 7744747, 7744774, 7744777, 7747444, 7747447, 7747474, 7747477, 7747744, 7747747, 7747774, 7747777, 7774444, 7774447, 7774474, 7774477, 7774744, 7774747, 7774774, 7774777, 7777444, 7777447, 7777474, 7777477, 7777744, 7777747, 7777774, 7777777, 44444444, 44444447, 44444474, 44444477, 44444744, 44444747, 44444774, 44444777, 44447444, 44447447, 44447474, 44447477, 44447744, 44447747, 44447774, 44447777, 44474444, 44474447, 44474474, 44474477, 44474744, 44474747, 44474774, 44474777, 44477444, 44477447, 44477474, 44477477, 44477744, 44477747, 44477774, 44477777, 44744444, 44744447, 44744474, 44744477, 44744744, 44744747, 44744774, 44744777, 44747444, 44747447, 44747474, 44747477, 44747744, 44747747, 44747774, 44747777, 44774444, 44774447, 44774474, 44774477, 44774744, 44774747, 44774774, 44774777, 44777444, 44777447, 44777474, 44777477, 44777744, 44777747, 44777774, 44777777, 47444444, 47444447, 47444474, 47444477, 47444744, 47444747, 47444774, 47444777, 47447444, 47447447, 47447474, 47447477, 47447744, 47447747, 47447774, 47447777, 47474444, 47474447, 47474474, 47474477, 47474744, 47474747, 47474774, 47474777, 47477444, 47477447, 47477474, 47477477, 47477744, 47477747, 47477774, 47477777, 47744444, 47744447, 47744474, 47744477, 47744744, 47744747, 47744774, 47744777, 47747444, 47747447, 47747474, 47747477, 47747744, 47747747, 47747774, 47747777, 47774444, 47774447, 47774474, 47774477, 47774744, 47774747, 47774774, 47774777, 47777444, 47777447, 47777474, 47777477, 47777744, 47777747, 47777774, 47777777, 74444444, 74444447, 74444474, 74444477, 74444744, 74444747, 74444774, 74444777, 74447444, 74447447, 74447474, 74447477, 74447744, 74447747, 74447774, 74447777, 74474444, 74474447, 74474474, 74474477, 74474744, 74474747, 74474774, 74474777, 74477444, 74477447, 74477474, 74477477, 74477744, 74477747, 74477774, 74477777, 74744444, 74744447, 74744474, 74744477, 74744744, 74744747, 74744774, 74744777, 74747444, 74747447, 74747474, 74747477, 74747744, 74747747, 74747774, 74747777, 74774444, 74774447, 74774474, 74774477, 74774744, 74774747, 74774774, 74774777, 74777444, 74777447, 74777474, 74777477, 74777744, 74777747, 74777774, 74777777, 77444444, 77444447, 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444744474, 444744477, 444744744, 444744747, 444744774, 444744777, 444747444, 444747447, 444747474, 444747477, 444747744, 444747747, 444747774, 444747777, 444774444, 444774447, 444774474, 444774477, 444774744, 444774747, 444774774, 444774777, 444777444, 444777447, 444777474, 444777477, 444777744, 444777747, 444777774, 444777777, 447444444, 447444447, 447444474, 447444477, 447444744, 447444747, 447444774, 447444777, 447447444, 447447447, 447447474, 447447477, 447447744, 447447747, 447447774, 447447777, 447474444, 447474447, 447474474, 447474477, 447474744, 447474747, 447474774, 447474777, 447477444, 447477447, 447477474, 447477477, 447477744, 447477747, 447477774, 447477777, 447744444, 447744447, 447744474, 447744477, 447744744, 447744747, 447744774, 447744777, 447747444, 447747447, 447747474, 447747477, 447747744, 447747747, 447747774, 447747777, 447774444, 447774447, 447774474, 447774477, 447774744, 447774747, 447774774, 447774777, 447777444, 447777447, 447777474, 447777477, 447777744, 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l,r=map(int,input().split())
a=a[bl(a,l):bl(a,r)+1]
x=a[0]
t=1 if r not in a else 0
a.insert(bl(a,l),l)
for i in range(len(a)-1):
x+=(a[i+1]-a[i])*a[i+1]
if t:x-=(a[-1]-r)*a[-1]
print(x)
``` | instruction | 0 | 64,075 | 20 | 128,150 |
Yes | output | 1 | 64,075 | 20 | 128,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
N = int(1e9)
a = []
def go(n):
a.append(n)
if n > N: return
go(int(n * 10 + 4))
go(int(n * 10 + 7))
def f(n):
if n == 0: return 0
sm = int(0)
num = int(0)
cur = int(1)
while cur <= n:
while a[num] < cur: num += 1
to = min(a[num], n)
sm += a[num] * (to - cur + 1)
cur = to + 1
return sm
go(int(4))
go(int(7))
a.sort()
s = input().split()
l = int(s[0])
r = int(s[1])
print(f(r) - f(l - 1))
``` | instruction | 0 | 64,076 | 20 | 128,152 |
Yes | output | 1 | 64,076 | 20 | 128,153 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
def lucky_gen():
current = ['4']
while True:
yield int(''.join(current))
for i in range(len(current)-1, -1, -1):
if current[i] == '4':
current[i] = '7'
break
else:
current[i] = '4'
if i == 0:
current.insert(0, '4')
input_lst = input().split()
l = int(input_lst[0])
r = int(input_lst[1])
result = 0
# for luck in lucky_gen():
# if l > r:
# break
# while luck >= l and l <= r:
# result += luck
# l += 1
prev = 0
for luck in lucky_gen():
if luck < l:
prev = luck
continue
if luck <= r:
result += luck * (luck - max(l - 1, prev))
else:
result += luck * (r - max(l - 1, prev))
break
prev = luck
print(result)
``` | instruction | 0 | 64,077 | 20 | 128,154 |
Yes | output | 1 | 64,077 | 20 | 128,155 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
l,r = map(int,input().split(' '))
a = set()
for i in range(1, 11):
for j in range(2**i):
c = j
cnt = 0
t = 0
while c or cnt < i:
if c&1:
t = t*10+7
else:
t = t*10+4
cnt += 1
c //= 2
a.add(t)
a = sorted(list(a))
ans = 0
for j in range(len(a)):
i = a[j]
if l <= i and r > i:
ans += (i - l + 1) * i
l = i + 1
elif l <= i and r <= i:
ans += (r - l + 1) * i
break
print(ans)
``` | instruction | 0 | 64,078 | 20 | 128,156 |
Yes | output | 1 | 64,078 | 20 | 128,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
#!/usr/bin/env python
from __future__ import division, print_function
import math
import os
import sys
from fractions import *
from sys import *
from decimal import *
from io import BytesIO, IOBase
from itertools import *
from collections import *
# sys.setrecursionlimit(10**5)
M = 10 ** 9 + 7
# print(math.factorial(5))
if sys.version_info[0] < 3:
from __builtin__ import xrange as range
from future_builtins import ascii, filter, hex, map, oct, zip
# sys.setrecursionlimit(10**6)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def inp(): return sys.stdin.readline().rstrip("\r\n") # for fast input
def out(var): sys.stdout.write(str(var)) # for fast output, always take string
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
def inpu(): return int(inp())
# -----------------------------------------------------------------
def regularbracket(t):
p = 0
for i in t:
if i == "(":
p += 1
else:
p -= 1
if p < 0:
return False
else:
if p > 0:
return False
else:
return True
# -------------------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# ------------------------------reverse string(pallindrome)
def reverse1(string):
pp = ""
for i in string[::-1]:
pp += i
if pp == string:
return True
return False
# --------------------------------reverse list(paindrome)
def reverse2(list1):
l = []
for i in list1[::-1]:
l.append(i)
if l == list1:
return True
return False
def mex(list1):
# list1 = sorted(list1)
p = max(list1) + 1
for i in range(len(list1)):
if list1[i] != i:
p = i
break
return p
def sumofdigits(n):
n = str(n)
s1 = 0
for i in n:
s1 += int(i)
return s1
def perfect_square(n):
s = math.sqrt(n)
if s == int(s):
return True
return False
# -----------------------------roman
def roman_number(x):
if x > 15999:
return
value = [5000, 4000, 1000, 900, 500, 400, 100, 90, 50, 40, 10, 9, 5, 4, 1]
symbol = ["F", "MF", "M", "CM", "D", "CD", "C", "XC", "L", "XL", "X", "IX", "V", "IV", "I"]
roman = ""
i = 0
while x > 0:
div = x // value[i]
x = x % value[i]
while div:
roman += symbol[i]
div -= 1
i += 1
return roman
def soretd(s):
for i in range(1, len(s)):
if s[i - 1] > s[i]:
return False
return True
# print(soretd("1"))
# ---------------------------
def countRhombi(h, w):
ct = 0
for i in range(2, h + 1, 2):
for j in range(2, w + 1, 2):
ct += (h - i + 1) * (w - j + 1)
return ct
def countrhombi2(h, w):
return ((h * h) // 4) * ((w * w) // 4)
# ---------------------------------
def binpow(a, b):
if b == 0:
return 1
else:
res = binpow(a, b // 2)
if b % 2 != 0:
return res * res * a
else:
return res * res
# -------------------------------------------------------
def binpowmodulus(a, b, m):
a %= m
res = 1
while (b > 0):
if (b & 1):
res = res * a % m
a = a * a % m
b >>= 1
return res
# -------------------------------------------------------------
def coprime_to_n(n):
result = n
i = 2
while (i * i <= n):
if (n % i == 0):
while (n % i == 0):
n //= i
result -= result // i
i += 1
if (n > 1):
result -= result // n
return result
# -------------------prime
def prime(x):
if x == 1:
return False
else:
for i in range(2, int(math.sqrt(x)) + 1):
# print(x)
if (x % i == 0):
return False
else:
return True
def luckynumwithequalnumberoffourandseven(x,n,a):
if x >= n and str(x).count("4") == str(x).count("7"):
a.append(x)
else:
if x < 1e12:
luckynumwithequalnumberoffourandseven(x * 10 + 4,n,a)
luckynumwithequalnumberoffourandseven(x * 10 + 7,n,a)
return a
"""
def luckynuber(x, n, a):
p = set(str(x))
if len(p) <= 2:
a.append(x)
if x < n:
luckynuber(x + 1, n, a)
return a
"""
# ------------------------------------------------------interactive problems
def interact(type, x):
if type == "r":
inp = input()
return inp.strip()
else:
print(x, flush=True)
# ------------------------------------------------------------------zero at end of factorial of a number
def findTrailingZeros(n):
# Initialize result
count = 0
# Keep dividing n by
# 5 & update Count
while (n >= 5):
n //= 5
count += n
return count
# -----------------------------------------------merge sort
# Python program for implementation of MergeSort
def mergeSort(arr):
if len(arr) > 1:
# Finding the mid of the array
mid = len(arr) // 2
# Dividing the array elements
L = arr[:mid]
# into 2 halves
R = arr[mid:]
# Sorting the first half
mergeSort(L)
# Sorting the second half
mergeSort(R)
i = j = k = 0
# Copy data to temp arrays L[] and R[]
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i += 1
else:
arr[k] = R[j]
j += 1
k += 1
# Checking if any element was left
while i < len(L):
arr[k] = L[i]
i += 1
k += 1
while j < len(R):
arr[k] = R[j]
j += 1
k += 1
# -----------------------------------------------lucky number with two lucky any digits
res = set()
def solven(p, l, a, b, n): # given number
if p > n or l > 10:
return
if p > 0:
res.add(p)
solven(p * 10 + a, l + 1, a, b, n)
solven(p * 10 + b, l + 1, a, b, n)
# problem
"""
n = int(input())
for a in range(0, 10):
for b in range(0, a):
solve(0, 0)
print(len(res))
"""
# Python3 program to find all subsets
# by backtracking.
# In the array A at every step we have two
# choices for each element either we can
# ignore the element or we can include the
# element in our subset
def subsetsUtil(A, subset, index, d):
print(*subset)
s = sum(subset)
d.append(s)
for i in range(index, len(A)):
# include the A[i] in subset.
subset.append(A[i])
# move onto the next element.
subsetsUtil(A, subset, i + 1, d)
# exclude the A[i] from subset and
# triggers backtracking.
subset.pop(-1)
return d
def subsetSums(arr, l, r, d, sum=0):
if l > r:
d.append(sum)
return
subsetSums(arr, l + 1, r, d, sum + arr[l])
# Subset excluding arr[l]
subsetSums(arr, l + 1, r, d, sum)
return d
def print_factors(x):
factors = []
for i in range(1, x + 1):
if x % i == 0:
factors.append(i)
return (factors)
# -----------------------------------------------
def calc(X, d, ans, D):
# print(X,d)
if len(X) == 0:
return
i = X.index(max(X))
ans[D[max(X)]] = d
Y = X[:i]
Z = X[i + 1:]
calc(Y, d + 1, ans, D)
calc(Z, d + 1, ans, D)
# ---------------------------------------
def factorization(n, l):
c = n
if prime(n) == True:
l.append(n)
return l
for i in range(2, c):
if n == 1:
break
while n % i == 0:
l.append(i)
n = n // i
return l
# endregion------------------------------
def good(b):
l = []
i = 0
while (len(b) != 0):
if b[i] < b[len(b) - 1 - i]:
l.append(b[i])
b.remove(b[i])
else:
l.append(b[len(b) - 1 - i])
b.remove(b[len(b) - 1 - i])
if l == sorted(l):
# print(l)
return True
return False
# arr=[]
# print(good(arr))
def generate(st, s):
if len(s) == 0:
return
# If current string is not already present.
if s not in st:
st.add(s)
# Traverse current string, one by one
# remove every character and recur.
for i in range(len(s)):
t = list(s).copy()
t.remove(s[i])
t = ''.join(t)
generate(st, t)
return
#=--------------------------------------------longest increasing subsequence
def largestincreasingsubsequence(A):
l = [1]*len(A)
sub=[]
for i in range(1,len(l)):
for k in range(i):
if A[k]<A[i]:
sub.append(l[k])
l[i]=1+max(sub,default=0)
return max(l,default=0)
#----------------------------------longest palindromic substring
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
#l=[]
def OR(a, n):
ans = a[0]
for i in range(1, n):
ans |= a[i]
#l.append(ans)
return ans
#print(prime(12345678987766))
"""
def main():
q=inpu()
x = q
v1 = 0
v2 = 0
i = 2
while i * i <= q:
while q % i == 0:
if v1!=0:
v2 = i
else:
v1 = i
q //= i
i += 1
if q - 1!=0:
v2 = q
if v1 * v2 - x!=0:
print(1)
print(v1 * v2)
else:
print(2)
if __name__ == '__main__':
main()
"""
def luckynum(x,l,r,a):
if x!=0:
a.append(x)
if x < 1e10:
luckynum(x * 10 + 4, l,r,a)
luckynum(x * 10 + 7, l,r,a)
return a
def main():
l,r = sep()
a=[]
luckynum(0,l,r,a)
a.sort()
#print(a)
i=0
ans=0
l-=1
#print(a)
while(True):
if r>a[i]:
ans+=(a[i]*(a[i]-l))
l=a[i]
else:
ans+=(a[i]*(r-l))
break
i+=1
print(ans)
if __name__ == '__main__':
main()
``` | instruction | 0 | 64,079 | 20 | 128,158 |
No | output | 1 | 64,079 | 20 | 128,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
"""
Perfection is achieved not when there is nothing more to add, but rather when there is nothing more to take away.
"""
import sys
input = sys.stdin.readline
# from bisect import bisect_left as lower_bound;
# from bisect import bisect_right as upper_bound;
# from math import ceil, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
def factorial(x, m):
val = 1
while x>0:
val = (val * x) % m
x -= 1
return val
# swap_array function
def swaparr(arr, a,b):
temp = arr[a];
arr[a] = arr[b];
arr[b] = temp;
## gcd function
def gcd(a,b):
if b == 0:
return a;
return gcd(b, a % b);
## nCr function efficient using Binomial Cofficient
def nCr(n, k):
if(k > n - k):
k = n - k;
res = 1;
for i in range(k):
res = res * (n - i);
res = res / (i + 1);
return int(res);
## upper bound function code -- such that e in a[:i] e < x;
def upper_bound(a, x, lo=0, hi = None):
if hi == None:
hi = len(a);
while lo < hi:
mid = (lo+hi)//2;
if a[mid] < x:
lo = mid+1;
else:
hi = mid;
return lo;
## prime factorization
def primefs(n):
## if n == 1 ## calculating primes
primes = {}
while(n%2 == 0 and n > 0):
primes[2] = primes.get(2, 0) + 1
n = n//2
for i in range(3, int(n**0.5)+2, 2):
while(n%i == 0 and n > 0):
primes[i] = primes.get(i, 0) + 1
n = n//i
if n > 2:
primes[n] = primes.get(n, 0) + 1
## prime factoriazation of n is stored in dictionary
## primes and can be accesed. O(sqrt n)
return primes
## MODULAR EXPONENTIATION FUNCTION
def power(x, y, p):
res = 1
x = x % p
if (x == 0) :
return 0
while (y > 0) :
if ((y & 1) == 1) :
res = (res * x) % p
y = y >> 1
x = (x * x) % p
return res
## DISJOINT SET UNINON FUNCTIONS
def swap(a,b):
temp = a
a = b
b = temp
return a,b;
# find function with path compression included (recursive)
# def find(x, link):
# if link[x] == x:
# return x
# link[x] = find(link[x], link);
# return link[x];
# find function with path compression (ITERATIVE)
def find(x, link):
p = x;
while( p != link[p]):
p = link[p];
while( x != p):
nex = link[x];
link[x] = p;
x = nex;
return p;
# the union function which makes union(x,y)
# of two nodes x and y
def union(x, y, link, size):
x = find(x, link)
y = find(y, link)
if size[x] < size[y]:
x,y = swap(x,y)
if x != y:
size[x] += size[y]
link[y] = x
## returns an array of boolean if primes or not USING SIEVE OF ERATOSTHANES
def sieve(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
return prime
#### PRIME FACTORIZATION IN O(log n) using Sieve ####
MAXN = int(1e5 + 5)
def spf_sieve():
spf[1] = 1;
for i in range(2, MAXN):
spf[i] = i;
for i in range(4, MAXN, 2):
spf[i] = 2;
for i in range(3, ceil(MAXN ** 0.5), 2):
if spf[i] == i:
for j in range(i*i, MAXN, i):
if spf[j] == j:
spf[j] = i;
## function for storing smallest prime factors (spf) in the array
################## un-comment below 2 lines when using factorization #################
# spf = [0 for i in range(MAXN)]
# spf_sieve();
def factoriazation(x):
ret = {};
while x != 1:
ret[spf[x]] = ret.get(spf[x], 0) + 1;
x = x//spf[x]
return ret;
## this function is useful for multiple queries only, o/w use
## primefs function above. complexity O(log n)
## taking integer array input
def int_array():
return list(map(int, input().strip().split()));
def float_array():
return list(map(float, input().strip().split()));
## taking string array input
def str_array():
return input().strip().split();
#defining a couple constants
MOD = int(1e9)+7;
CMOD = 998244353;
INF = float('inf'); NINF = -float('inf');
################### ---------------- TEMPLATE ENDS HERE ---------------- ###################
from itertools import permutations
def solve():
l, r = map(int, input().split())
ref = 4
summ = 0
for i in range(l, r + 1):
if ref < i:
val = str(ref).count('4')
# print(val)
if val == 0:
ref += 4*(10**len(str(ref)))
else:
diff = len(str(ref)) - val
ref += 3 * (10**diff)
# print(ref, i)
summ += ref
print(summ)
if __name__ == '__main__':
for _ in range(1):
solve()
# fin_time = datetime.now()
# print("Execution time (for loop): ", (fin_time-init_time))
``` | instruction | 0 | 64,080 | 20 | 128,160 |
No | output | 1 | 64,080 | 20 | 128,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
level={}
def bfs():
s=''
level[s]=None
i=1
frontier = [s]
while i < 11:
next=[]
for u in frontier:
for v in [u+'4', u+'7']:
if v not in level:
level[v]=i
next.append(v)
frontier=next
i+=1
bfs()
l=[]
for key, value in level.items():
if key!='':
l.append(int(key))
l.sort()
a,b=map(int, input().split())
s=0
for i in range(len(l)):
if a<=l[i]<=b:
if l[i+1] <=b :
s=s+ (l[i]) * (l[i+1]- l[i] )
else:
s=s+ l[i]*(b-l[i-1])
elif l[i] > b:
break
print(s)
``` | instruction | 0 | 64,081 | 20 | 128,162 |
No | output | 1 | 64,081 | 20 | 128,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.
Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.
Input
The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.
Output
In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).
Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.
Examples
Input
2 7
Output
33
Input
7 7
Output
7
Note
In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33
In the second sample: next(7) = 7
Submitted Solution:
```
l, r = list(map(int, input().split()))
count = 0
ans = 0
flag = True
def isLucky(num):
n = str(num)
for i in n:
if i != "4" and i != "7":
return False
return True
i = l
while True:
count += 1
if isLucky(i):
ans += count * i
count = 0
if i >= r:
break
i += 1
print(ans)
``` | instruction | 0 | 64,082 | 20 | 128,164 |
No | output | 1 | 64,082 | 20 | 128,165 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is sitting on an extremely boring math class. To have fun, he took a piece of paper and wrote out n numbers on a single line. After that, Vasya began to write out different ways to put pluses ("+") in the line between certain digits in the line so that the result was a correct arithmetic expression; formally, no two pluses in such a partition can stand together (between any two adjacent pluses there must be at least one digit), and no plus can stand at the beginning or the end of a line. For example, in the string 100500, ways 100500 (add no pluses), 1+00+500 or 10050+0 are correct, and ways 100++500, +1+0+0+5+0+0 or 100500+ are incorrect.
The lesson was long, and Vasya has written all the correct ways to place exactly k pluses in a string of digits. At this point, he got caught having fun by a teacher and he was given the task to calculate the sum of all the resulting arithmetic expressions by the end of the lesson (when calculating the value of an expression the leading zeros should be ignored). As the answer can be large, Vasya is allowed to get only its remainder modulo 109 + 7. Help him!
Input
The first line contains two integers, n and k (0 ≤ k < n ≤ 105).
The second line contains a string consisting of n digits.
Output
Print the answer to the problem modulo 109 + 7.
Examples
Input
3 1
108
Output
27
Input
3 2
108
Output
9
Note
In the first sample the result equals (1 + 08) + (10 + 8) = 27.
In the second sample the result equals 1 + 0 + 8 = 9. | instruction | 0 | 64,381 | 20 | 128,762 |
Tags: combinatorics, dp, math, number theory
Correct Solution:
```
n, k = map(int, input().split())
t = list(map(int, input()))
p, d = 1, 10**9 + 7
s = 0
f = [1] * n
for i in range(2, n):
f[i] = (i * f[i - 1]) % d
c = lambda a, b: 0 if a > b else (f[b] * pow(f[a] * f[b -a], d - 2, d)) % d
if k:
u = [0] * (n + 1)
p = [1] * (n + 1)
for i in range(n):
u[i] = (p[i] * c(k - 1, n - 2 - i) + u[i - 1]) % d
p[i + 1] = (10 * p[i]) % d
for i in range(n):
v = u[n - 2 - i] + p[n - 1 - i] * c(k,i)
s = (s + t[i] * v) % d
else:
for i in t:
s = (s * 10 + i) % d
print(s)
#GG
``` | output | 1 | 64,381 | 20 | 128,763 |
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