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 are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987 | instruction | 0 | 54,222 | 20 | 108,444 |
Tags: greedy
Correct Solution:
```
a = list(int(x) for x in input())
b = list(int(x) for x in input())
b.sort()
b.reverse()
c = ""
for x in range(len(a)):
if len(b) > 0 and a[x] < b[0]:
a[x] = b[0]
b.pop(0)
c += str(a[x])
print(c)
``` | output | 1 | 54,222 | 20 | 108,445 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987 | instruction | 0 | 54,223 | 20 | 108,446 |
Tags: greedy
Correct Solution:
```
a = list(input())
s = list(input())
s.sort(reverse=True)
aindex, sindex = 0, 0
while sindex < len(s) and aindex < len(a):
if s[sindex] > a[aindex]:
a[aindex] = s[sindex]
s.pop(sindex)
sindex = 0
aindex += 1
else:
aindex += 1
print ("".join(a))
``` | output | 1 | 54,223 | 20 | 108,447 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
I=input
a,s,q=[*I()],I(),{}
for i in s:
if i in q:q[i]+=1
else:q[i]=1
for i in range(len(a)):
for j in range(9,0,-1):
j=str(j)
if j in q:
if j>a[i] and q[j]:
a[i]=j
q[j]-=1
break
print("".join(a))
``` | instruction | 0 | 54,224 | 20 | 108,448 |
Yes | output | 1 | 54,224 | 20 | 108,449 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
a = [int(n) for n in input()]
s = [int(n) for n in input()]
s.sort(reverse=True)
for i, n in enumerate(a):
if(len(s) == 0): break
if(len(s) > 0 and s[0] > n):
a[i] = s[0]
del(s[0])
print(''.join(map(str, a)))
``` | instruction | 0 | 54,225 | 20 | 108,450 |
Yes | output | 1 | 54,225 | 20 | 108,451 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
a=list(input())
b=sorted(list(input()))[::-1]
j=0
for i in range(len(a)):
if b[j] > a[i]:
a[i] = b[j]
j+=1
if j>=len(b):
break
for i in a:
print(i,end='')
``` | instruction | 0 | 54,226 | 20 | 108,452 |
Yes | output | 1 | 54,226 | 20 | 108,453 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
import math,sys
#from itertools import permutations, combinations;import heapq,random;
from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
#sys.stdin=open('input.txt','r');sys.stdout=open('output.txt','w');
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
#sys.setrecursionlimit(1500)
def dict(a):
d={}
for x in a:
if d.get(x,-1)!=-1:
d[x]+=1
else:
d[x]=1
return d
def find_gt(a, x):
'Find leftmost value greater than x'
i = bi.bisect_left(a, x)
if i != len(a):
return i
else:
return -1
def main():
try:
t=Sn()
l=list(t)
s=list(Sn())
t1={}
s1={}
for x in l:
if t1.get(int(x),-1)!=-1:
t1[int(x)]+=1
else:
t1[int(x)]=1
for x in s:
if s1.get(int(x),-1)!=-1:
s1[int(x)]+=1
else:
s1[int(x)]=1
n=len(l)
for x in range(n):
temp=int(l[x])
ma=temp
for i in range(temp,10):
if s1.get(i,-1)!=-1:
ma=i
if ma!=temp:
s1[ma]-=1
l[x]=str(ma)
if s1[ma]==0:
s1.pop(ma)
print(''.join(l))
except:
pass
M = 998244353
P = 1000000007
if __name__ == '__main__':
# for _ in range(I()):main()
for _ in range(1):main()
#End#
# ******************* All The Best ******************* #
``` | instruction | 0 | 54,227 | 20 | 108,454 |
Yes | output | 1 | 54,227 | 20 | 108,455 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
a=list(input())
a=list(map(int,a))
b=list(input())
b=list(map(int,b))
if max(b)<=min(a) or len(b)==1:
a=list(map(str,a))
a="".join(a)
print(a)
else:
c=0
while(c==0):
repl=max(b)
k=0
for i in range(len(a)):
if a[i]<repl:
a[i]=repl
k=1
break
if k==1:
b.remove(repl)
if max(b)<=min(a) or len(b)==1:
c=1
a=list(map(str,a))
print("".join(a))
``` | instruction | 0 | 54,228 | 20 | 108,456 |
No | output | 1 | 54,228 | 20 | 108,457 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
import sys
a = [x for x in sys.stdin.readline() if x != '\n']
s = sys.stdin.readline()
i = 0
v = [0] * 10
while(s[i] != '\n'):
v[int(s[i])] += 1
i += 1
j = 9
for i in range(len(a)):
while(v[j] == 0 and j > 1):
j -= 1
if(j > int(a[i])):
a[i] = str(j)
v[j] -= 1
print("".join(a))
``` | instruction | 0 | 54,230 | 20 | 108,460 |
No | output | 1 | 54,230 | 20 | 108,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer a that consists of n digits. You are also given a sequence of digits s of length m. The digit in position j (1 ≤ j ≤ m) of sequence s means that you can choose an arbitrary position i (1 ≤ i ≤ n) in a and replace the digit in the chosen position i with sj. Each element in the sequence s can participate in no more than one replacing operation.
Your task is to perform such sequence of replacements, that the given number a gets maximum value. You are allowed to use not all elements from s.
Input
The first line contains positive integer a. Its length n is positive and doesn't exceed 105. The second line contains sequence of digits s. Its length m is positive and doesn't exceed 105. The digits in the sequence s are written consecutively without any separators.
The given number a doesn't contain leading zeroes.
Output
Print the maximum value that can be obtained from a after a series of replacements. You are allowed to use not all elements from s. The printed number shouldn't contain any leading zeroes.
Examples
Input
1024
010
Output
1124
Input
987
1234567
Output
987
Submitted Solution:
```
a = input()
b = sorted(list(map(int,list(input()))),reverse=True)
if len(b) < len(a): b = b+[0]*(len(a)-len(b))
n = ''
m = 0
c = len(b)
for x in list(a):
if b[m] <= int(x):
n+=x; m =0
else: n += str(b[m]) ; m+=1
print(n)
``` | instruction | 0 | 54,231 | 20 | 108,462 |
No | output | 1 | 54,231 | 20 | 108,463 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,328 | 20 | 108,656 |
Tags: dp
Correct Solution:
```
# import sys
# input=sys.stdin.readline
a=input()
dp=[]
for i in range(len(a)):
dp.append([0]*10)
for i in range(10):
dp[0][i]=1
for i in range(len(a)-1):
for j in range(10):
if dp[i][j]!=0:
c=(int(a[i+1])+j)//2
d=(int(a[i+1])+j+1)//2
if c!=d:
dp[i+1][c]+=dp[i][j]
dp[i+1][d]+=dp[i][j]
else:
dp[i+1][c]+=dp[i][j]
s=0
for i in range(10):
s+=dp[-1][i]
t=0
c=int(a[0])
f=[a[0]]
for i in range(1,len(a)):
d=(c+int(a[i]))//2
e=(c+int(a[i])+1)//2
if int(a[i])==d:
f.append(a[i])
c=d
elif int(a[i])==e:
f.append(a[i])
c=e
else:
break
if "".join(f)==a:
t=1
print(s-t)
``` | output | 1 | 54,328 | 20 | 108,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,329 | 20 | 108,658 |
Tags: dp
Correct Solution:
```
s=input()
n=len(s)
dp=[[0]*10 for i in range(n)]
dp[0]=[1]*10
for i in range(1,n):
for j in range(10):
x=j*2-int(s[i])
if 0<=x+1<10:
dp[i][j]+=dp[i-1][x+1]
if 0<=x<10:
dp[i][j]+=dp[i-1][x]
if 0<=x-1<10:
dp[i][j]+=dp[i-1][x-1]
ans=sum(dp[-1])
s=list(map(int,list(s)))
f=1
for i in range(1,n):
x=(s[i]+s[i-1])//2
if x!=s[i] and x+1!=s[i]:
f=0
break
print(ans-f)
``` | output | 1 | 54,329 | 20 | 108,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,330 | 20 | 108,660 |
Tags: dp
Correct Solution:
```
s=[int(c) for c in input()]
ok=1
for i in range(1,len(s)):
if abs(s[i]-s[i-1])>1: ok=0; break
w=[1]*10
for v in s[1:]:
ww=w[:]
w=[0]*10
for d in range(10):
q,r=divmod(v+d,2)
w[q]+=ww[d]
if r>0:
w[q+1]+=ww[d]
print(sum(w)-ok)
``` | output | 1 | 54,330 | 20 | 108,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,331 | 20 | 108,662 |
Tags: dp
Correct Solution:
```
n = list(map(int, input()))
dp = [[1] * 10]
for i in range(1, len(n)):
dp.append([0] * 10)
for j in range(10):
x = j + n[i]
dp[i][x // 2] += dp[i - 1][j]
if x % 2:
dp[i][x // 2 + 1] += dp[i - 1][j]
print(sum(dp[-1]) - all(abs(n[i] - n[i - 1]) <= 1 for i in range(1, len(n))))
``` | output | 1 | 54,331 | 20 | 108,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,332 | 20 | 108,664 |
Tags: dp
Correct Solution:
```
R = lambda: map(int, input().split())
arr = list(map(int, input()))
n = len(arr)
dp = [[0] * 10 for i in range(n)]
for i in range(10):
dp[n - 1][i] = 1
for i in range(n - 2, -1, -1):
for j in range(10):
if (arr[i + 1] + j) % 2 == 0:
dp[i][j] = dp[i + 1][(arr[i + 1] + j) // 2]
else:
dp[i][j] = dp[i + 1][(arr[i + 1] + j) // 2] + dp[i + 1][(arr[i + 1] + j + 1) // 2]
ownIn = 1
for i in range(n - 1):
if (arr[i] + arr[i + 1]) // 2 != arr[i + 1] and (arr[i] + arr[i + 1] + 1) // 2 != arr[i + 1]:
ownIn = 0
print(sum(dp[0]) - ownIn)
``` | output | 1 | 54,332 | 20 | 108,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,333 | 20 | 108,666 |
Tags: dp
Correct Solution:
```
from math import floor,ceil
def find_possible_number(guess_number,digit,possible_table):
if digit==len(phone_numbers):#base condition
return 1
m_number=int(phone_numbers[digit])
guess_number=int(guess_number)
num_index_configuration=str(guess_number)+str(digit)
if num_index_configuration in possible_table:
return possible_table[num_index_configuration]
possible_table[num_index_configuration]=0
has_remainder=(m_number+guess_number)%2
if has_remainder:
has_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit)
return possible_table[num_index_configuration]
else:
has_no_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit)
return possible_table[num_index_configuration]
def has_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit):
upper_number=ceil((m_number+guess_number)/2)
possible_table[num_index_configuration]+=find_possible_number(upper_number,digit+1,possible_table)
lower_number=floor((m_number+guess_number)/2)
possible_table[num_index_configuration]+=find_possible_number(lower_number,digit+1,possible_table)
def has_no_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit):
next_number=(m_number+guess_number)/2
possible_table[num_index_configuration]+=find_possible_number(next_number,digit+1,possible_table)
def has_exception(number):
for i in range(len(phone_numbers)-1):
if abs(int(phone_numbers[i])-int(phone_numbers[i+1]))<=1:
pass
else:
return False
return True
phone_numbers=input()
possible_table={}#with the format of ["ab"]=c. a for number, b for digit, c for possible ways
first_number=phone_numbers[0]
count=0
for i in range(10):
count+=find_possible_number(i,1,possible_table)
if has_exception(int(phone_numbers[0])):
count-=1
#print(possible_table)
print(count)
``` | output | 1 | 54,333 | 20 | 108,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,334 | 20 | 108,668 |
Tags: dp
Correct Solution:
```
s=input()
a=[]
for j in range(1,len(s)):
a.append(int(s[j]))
k=[[] for j in range(10)]
m=0
for j in range(10):
c=[0]*(10)
c[j]=1
r=0
while(r<len(a)):
p=0
d = [0] * (10)
while(p<10):
if (p+a[r])%2==0:
q=(p+a[r])//2
d[q]+=c[p]
d[p]+=-c[p]
else:
q = (p + a[r]) // 2
u=q+1
d[q] += c[p]
d[u]+=c[p]
d[p]+=-c[p]
p+=1
for i in range(10):
c[i] += d[i]
r+=1
m+=sum(c)
j=int(s[0])
f=0
q=0
while(q<len(a)):
if (j+a[q])%2==0:
if a[q]!=(j+a[q])//2:
f=1
break
else:
e=(j + a[q])// 2
if a[q]!=e and a[q]!=(e+1):
f=1
break
j=a[q]
q+=1
if f==0:
m+=-1
print(m)
``` | output | 1 | 54,334 | 20 | 108,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15 | instruction | 0 | 54,335 | 20 | 108,670 |
Tags: dp
Correct Solution:
```
s = input()
l = len(s)
count = 0
dp = [[0 for i in range(10)] for i in range(l+1)]
for i in range(10):
dp[0][i] = 1
for i in range(l-1):
for j in range(10):
dp[i+1][(int(s[i+1])+j)//2] += dp[i][j]
if (int(s[i+1])+j)%2!=0:
dp[i+1][(int(s[i+1])+j)//2+1] += dp[i][j]
ok = 1
for i in range(l-1):
if abs(int(s[i+1])-int(s[i]))>1:
ok = 0
print(sum(dp[l-1])-ok)
``` | output | 1 | 54,335 | 20 | 108,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15
Submitted Solution:
```
R = lambda: map(int, input().split())
arr = list(map(int, input()))
n = len(arr)
dp = [[0] * 10 for i in range(n)]
for i in range(10):
dp[n - 1][i] = 1
for i in range(n - 2, -1, -1):
for j in range(10):
if (arr[i + 1] + j) % 2 == 0:
dp[i][j] = dp[i + 1][(arr[i + 1] + j) // 2]
else:
dp[i][j] = dp[i + 1][(arr[i + 1] + j) // 2] + dp[i + 1][(arr[i + 1] + j + 1) // 2]
ownIn = 1
for i in range(n - 1):
if (arr[i] + arr[i + 1]) != arr[i + 1] * 2 and (arr[i] + arr[i + 1]) != arr[i + 1] * 2 - 1:
ownIn = 0
print(sum(dp[0]) - ownIn)
``` | instruction | 0 | 54,336 | 20 | 108,672 |
No | output | 1 | 54,336 | 20 | 108,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15
Submitted Solution:
```
def gg(last,i):
if i==x:
return 1
c = last+int(n[i])
if c%2:
return gg(c//2,i+1)+gg(c//2+1,i+1)
else:
return gg(c//2,i+1)
def abc():
ans=0
for i in range(10):
ans+=gg(i,1)
return ans
n=input()
x=len(n)
print(abc())
``` | instruction | 0 | 54,337 | 20 | 108,674 |
No | output | 1 | 54,337 | 20 | 108,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15
Submitted Solution:
```
from math import floor,ceil
def find_possible_number(guess_number,digit,possible_table):
if digit==len(phone_numbers):#base condition
return 1
m_number=int(phone_numbers[digit])
guess_number=int(guess_number)
num_index_configuration=str(guess_number)+str(digit)
if num_index_configuration in possible_table:
return possible_table[num_index_configuration]
possible_table[num_index_configuration]=0
has_remainder=(m_number+guess_number)%2
if has_remainder:
has_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit)
return possible_table[num_index_configuration]
else:
has_no_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit)
return possible_table[num_index_configuration]
def has_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit):
upper_number=ceil((m_number+guess_number)/2)
possible_table[num_index_configuration]+=find_possible_number(upper_number,digit+1,possible_table)
lower_number=floor((m_number+guess_number)/2)
possible_table[num_index_configuration]+=find_possible_number(lower_number,digit+1,possible_table)
def has_no_remainder_case(m_number, guess_number, possible_table, num_index_configuration, digit):
next_number=(m_number+guess_number)/2
possible_table[num_index_configuration]+=find_possible_number(next_number,digit+1,possible_table)
def has_exception(number):
for i in range(1,len(phone_numbers)-1):
if (int(phone_numbers[i])+number)%2:
upper=ceil((int(phone_numbers[i])+number)/2)
lower=floor((int(phone_numbers[i])+number)/2)
if upper==int(phone_numbers[i]):
number=int(phone_numbers[i])
elif lower==int(phone_numbers[i]):
number=int(phone_numbers[i])
else:
return False
else:
if (int(phone_numbers[i])+number)/2==int(phone_numbers[i]):
number = int(phone_numbers[i])
else:
return False
return True
phone_numbers=input()
possible_table={}#with the format of ["ab"]=c. a for number, b for digit, c for possible ways
first_number=phone_numbers[0]
count=0
for i in range(10):
count+=find_possible_number(i,1,possible_table)
if has_exception(int(phone_numbers[0])):
count-=1
#print(possible_table)
print(count)
``` | instruction | 0 | 54,338 | 20 | 108,676 |
No | output | 1 | 54,338 | 20 | 108,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alas, finding one's true love is not easy. Masha has been unsuccessful in that yet. Her friend Dasha told Masha about a way to determine the phone number of one's Prince Charming through arithmancy.
The phone number is divined like that. First one needs to write down one's own phone numbers. For example, let's suppose that Masha's phone number is 12345. After that one should write her favorite digit from 0 to 9 under the first digit of her number. That will be the first digit of the needed number. For example, Masha's favorite digit is 9. The second digit is determined as a half sum of the second digit of Masha's number and the already written down first digit from her beloved one's number. In this case the arithmetic average equals to (2 + 9) / 2 = 5.5. Masha can round the number up or down, depending on her wishes. For example, she chooses the digit 5. Having written down the resulting digit under the second digit of her number, Masha moves to finding the third digit in the same way, i.e. finding the half sum the the third digit of her number and the second digit of the new number. The result is (5 + 3) / 2 = 4. In this case the answer is unique. Thus, every i-th digit is determined as an arithmetic average of the i-th digit of Masha's number and the i - 1-th digit of her true love's number. If needed, the digit can be rounded up or down. For example, Masha can get:
12345 95444 Unfortunately, when Masha tried dialing the number, she got disappointed: as it turned out, the number was unavailable or outside the coverage area. But Masha won't give up. Perhaps, she rounded to a wrong digit or chose the first digit badly. That's why she keeps finding more and more new numbers and calling them. Count the number of numbers Masha calls. Masha calls all the possible numbers that can be found by the described means of arithmancy, except for, perhaps, her own one.
Input
The first line contains nonempty sequence consisting of digits from 0 to 9 — Masha's phone number. The sequence length does not exceed 50.
Output
Output the single number — the number of phone numbers Masha will dial.
Examples
Input
12345
Output
48
Input
09
Output
15
Submitted Solution:
```
n = list(map(int, input()))
dp = [[1] * 10]
for i in range(1, len(n)):
dp.append([0] * 10)
for j in range(10):
x = j + n[i]
dp[i][x // 2] += dp[i - 1][j]
if x % 2:
dp[i][x // 2 + 1] += dp[i - 1][j]
print(sum(dp[-1]))
``` | instruction | 0 | 54,339 | 20 | 108,678 |
No | output | 1 | 54,339 | 20 | 108,679 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,424 | 20 | 108,848 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
raw = input()
s = raw.replace('?','')
n = int(s.split(' ')[-1])
m = s.count('-')
p = s.count('+')
c = p - m
if (p+1)*n - m < n or p+1 - m*n > n:
print('Impossible')
else:
print('Possible')
acc = n-c
s = raw[1:]
while acc > n or acc < 1:
if acc > n:
s = s.replace("+ ?", "+ " + str(n), 1)
acc -= n-1
else:
s = s.replace("- ?", "- " + str(n), 1)
acc += n-1
print(str(acc) + s.replace('?', '1'))
``` | output | 1 | 54,424 | 20 | 108,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,425 | 20 | 108,850 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
def binp(a,b):
st=a[0]
fin=a[1]
b=[-b[1],-b[0]]
while True:
s=(st+fin)//2
if s>=b[0]:
if s<=b[1]:
return(s)
else:
fin=s-1
else:
st=s+1
s=input()
for i in range(len(s)):
if s[i]=='=':
y=i
break
y+=1
n=int(s[y:])
now='+'
max_=0
min_=0
ctr=0
pls=0
minus=0
g=[[0,0]]
for i in s:
ctr+=1
if i=='?':
if now=='+':
max_+=n
min_+=1
pls+=1
else:
max_-=1
min_-=n
minus+=1
g.append([min_,max_])
elif i=='-' or i=='+':
now=i
elif i=='=':
break
v=n
if v<min_ or v>max_:
print('Impossible')
else:
print('Possible')
a=[pls*1,pls*v]
b=[minus*(-v)-n,minus*(-1)-n]
#print(a,b)
ctr=0
ctr=binp(a,b)
fir=ctr
j=[0]*pls
if pls!=0:
j=[fir//pls]*pls
for u in range(fir%pls):
j[u]+=1
sec=n-ctr
k=[0]*minus
if minus!=0:
k=[((-sec)//minus)]*minus
#print(k)
#print((-sec)%minus)
for u in range((-sec)%minus):
k[u]+=1
p=0
m=0
now='+'
for i in s:
if i=='?':
if now=='+':
print(j[p],end='')
p+=1
else:
print(k[m],end='')
m+=1
elif i=='-' or i=='+':
now=i
print(i,end='')
else:
print(i,end='')
``` | output | 1 | 54,425 | 20 | 108,851 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,426 | 20 | 108,852 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
import sys
sys.stderr = sys.stdout
def clamp(x, lo, hi):
assert lo <= hi
return max(min(x, hi), lo)
def rebus(R):
n = int(R[-1])
a = 0
b = 0
for i in range(1, len(R) - 2, 2):
if R[i] == '+':
a += 1
else:
b += 1
if b > n * a or a + 1 > n * (b + 1):
return None
lo = a + 1 - n*b
hi = n*a + n - b
if lo > n or hi < n:
return False
for i in range(0, len(R) - 2, 2):
assert lo <= n <= hi
if i == 0 or R[i-1] == '+':
lo -= 1
hi -= n
x = clamp(1, n - hi, n - lo)
assert 1 <= x <= n
lo += x
hi += x
R[i] = str(x)
else:
lo += n
hi += 1
x = clamp(1, lo - n, hi - n)
assert 1 <= x <= n
lo -= x
hi -= x
R[i] = str(x)
assert lo == hi == n
return True
def main():
R = input().split()
if rebus(R):
print('Possible')
print(' '.join(R))
else:
print('Impossible')
##########
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintt():
return tuple(readinti())
def readintl():
return list(readinti())
def readinttl(k):
return [readintt() for _ in range(k)]
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.__stderr__)
if __name__ == '__main__':
main()
``` | output | 1 | 54,426 | 20 | 108,853 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,427 | 20 | 108,854 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
import sys
s = input()
n = int(s[s.index('=')+1:])
plus = s.count('+')+1
minus = s.count('-')
diff = increse_pos = count = neg = a = b = 0
if plus - minus < n:
diff = n + minus
a = diff // plus
b = diff % plus
if (b != 0 and a+1 > n ) or (a > n):
neg = 1
else:
increse_pos = 1
elif plus - minus > n:
diff = plus - n
if minus > 0:
a = diff // minus
b = diff % minus
else:
neg = 1
if (neg == 0 and b != 0 and a + 1 > n) or (neg == 0 and a > n):
neg = 1
else:
increse_pos = 0
else:
diff = 0
a = 1
sign = 1
res = ''
if neg == 0:
for ch in s:
if ch == '?' and sign == 1:
if increse_pos == 1:
if b > 0:
res += str(a+1)
count += a+1
b -= 1
else:
res += str(a)
count += a
else:
res += str(1)
count += 1
sign = 0
elif ch == '?' and sign == 0:
if increse_pos == 0:
if b > 0:
res += str(a + 1)
count -= a+1
b -= 1
else:
res += str(a)
count -= a
else:
res += str(1)
count -= 1
elif ch == '+':
sign = 1
res += ch
elif ch == '-':
sign = 0
res += ch
else:
res += ch
print('Possible')
print(res)
else:
print('Impossible')
``` | output | 1 | 54,427 | 20 | 108,855 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,428 | 20 | 108,856 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
a = list(input().split())
n = int(a[-1])
R = int(a[-1])
one, two = 0, 0
for i in range(1,len(a),2):
if a[i] == '=':
break
if a[i] == '+':
R -= 1
one += 1
else:
R += 1
two += 1
R -= 1
one += 1
if R >= 0:
if one * (n - 1) >= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i == 0 or a[i - 1] == '+':
print(min(n - 1, R) + 1, a[i + 1], end = ' ')
R -= min(n - 1, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
else:
if two * (1 - n) <= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i > 0 and a[i - 1] == '-':
print(-(max(1 - n, R) - 1), a[i + 1], end = ' ')
R -= max(1 - n, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
``` | output | 1 | 54,428 | 20 | 108,857 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,429 | 20 | 108,858 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s = input().split()
n = int(s[len(s)-1])
size = (len(s)-1)//2
masP = [0]
masM = []
P, M = 1, 0
for i in range(1, len(s)-2, 2):
if s[i]=='+':
P += 1
masP.append(i+1)
else:
M += 1
masM.append(i+1)
sum1 = n+M*n
sum2 = n+M
if n+M > n*P or n+M*n < P:
print ("Impossible")
else:
print ("Possible")
while M!=0:
for u in range(1, n+1):
if sum2+u-1 <= n*P and sum1+u-n >= P:
s[masM.pop()] = u
M = M-1
sum2 = sum2+u-1
sum1 = sum1+u-n
break
while P!=0:
a = sum2//P
s[masP.pop()] = a
P = P-1
sum2 = sum2-a
print (' '.join(str(x) for x in s))
``` | output | 1 | 54,429 | 20 | 108,859 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,430 | 20 | 108,860 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s = input().split()
plus = 1
minus = 0
for ch in s:
if (ch == '+') :
plus += 1
if (ch == '-') :
minus += 1
n = int(s[len(s) - 1])
maxx = plus * n - 1 * minus
minn = plus - n * minus
now = n - (plus - minus)
if (n>maxx or n<minn):
print("Impossible")
else:
pre = '+'
print("Possible")
for ch in s:
if (ch == '?'):
if (pre == '+') :
val = 1
if (now > 0) : val = min(n - 1,now) + 1
now -= (val - 1)
print(val,end = " ")
if (pre == '-'):
val = 1
if (now < 0) : val = min(abs(n) - 1,abs(now)) + 1
now += (val - 1)
print(val,end = " ")
else :
print(ch,end = " ")
pre = ch
``` | output | 1 | 54,430 | 20 | 108,861 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000 | instruction | 0 | 54,431 | 20 | 108,862 |
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s,n=input().split(' = ')
if s=='?': print('Possible\n'+n+' = '+n)
elif not '+' in s: print('Impossible')
else:
n=int(n)
s=['+']+list(s.replace(' ',''))
cp=s.count('+')
cm=s.count('-')
if cm>=cp:
t=(n+cm)//cp
m=(n+cm)%cp
cc=0
for i in range(1,len(s),2):
if s[i-1]=='+': s[i]=str(t+(cc<m)); cc+=1
else: s[i]='1'
else:
t=(n+cm*n)//cp
m=(n+cm*n)%cp
cc=0
for i in range(1,len(s),2):
if s[i-1]=='+': s[i]=str(t+(cc<m)); cc+=1
else: s[i]=str(n)
if int(s[1])>n or int(s[-1])<1: print('Impossible')
else:
print('Possible')
print(' '.join(s[1:])+' =',n)
``` | output | 1 | 54,431 | 20 | 108,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sherlock found a piece of encrypted data which he thinks will be useful to catch Moriarty. The encrypted data consists of two integer l and r. He noticed that these integers were in hexadecimal form.
He takes each of the integers from l to r, and performs the following operations:
1. He lists the distinct digits present in the given number. For example: for 101416, he lists the digits as 1, 0, 4.
2. Then he sums respective powers of two for each digit listed in the step above. Like in the above example sum = 21 + 20 + 24 = 1910.
3. He changes the initial number by applying bitwise xor of the initial number and the sum. Example: <image>. Note that xor is done in binary notation.
One more example: for integer 1e the sum is sum = 21 + 214. Letters a, b, c, d, e, f denote hexadecimal digits 10, 11, 12, 13, 14, 15, respertively.
Sherlock wants to count the numbers in the range from l to r (both inclusive) which decrease on application of the above four steps. He wants you to answer his q queries for different l and r.
Input
First line contains the integer q (1 ≤ q ≤ 10000).
Each of the next q lines contain two hexadecimal integers l and r (0 ≤ l ≤ r < 1615).
The hexadecimal integers are written using digits from 0 to 9 and/or lowercase English letters a, b, c, d, e, f.
The hexadecimal integers do not contain extra leading zeros.
Output
Output q lines, i-th line contains answer to the i-th query (in decimal notation).
Examples
Input
1
1014 1014
Output
1
Input
2
1 1e
1 f
Output
1
0
Input
2
1 abc
d0e fe23
Output
412
28464
Note
For the second input,
1416 = 2010
sum = 21 + 24 = 18
<image>
Thus, it reduces. And, we can verify that it is the only number in range 1 to 1e that reduces.
Submitted Solution:
```
#yl7
q=int(input())
xx=lambda x:int(x,16)
dii={'0':0,'1':1,'2':2,'3':3,'4':4,
'5':5,'6':6,'7':7,'8':8,'9':9,'a':10,
'b':11,'c':12,'d':13,'e':14,'f':15}
di2={'7': 128, '3': 8, '8': 256, '0': 1,
'b': 2048, 'a': 1024, 'c': 4096,
'e': 16384, '5': 32, 'f': 32768,
'6': 64, '1': 2, 'd': 8192, '2': 4,
'4': 16, '9': 512}
ca=dict()
for i in range(q):
a,b=list(map(xx,input().split()))
su=0
for i in range(a,b+1):
hi=tuple(sorted(str(hex(i))[2:]))
if not hi in ca:
sus=0
for e in hi:
sus+=di2[e]
ca[hi]=sus
else:sus=ca[hi]
if i^sus<i:
su+=1
print(su)
``` | instruction | 0 | 54,466 | 20 | 108,932 |
No | output | 1 | 54,466 | 20 | 108,933 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sherlock found a piece of encrypted data which he thinks will be useful to catch Moriarty. The encrypted data consists of two integer l and r. He noticed that these integers were in hexadecimal form.
He takes each of the integers from l to r, and performs the following operations:
1. He lists the distinct digits present in the given number. For example: for 101416, he lists the digits as 1, 0, 4.
2. Then he sums respective powers of two for each digit listed in the step above. Like in the above example sum = 21 + 20 + 24 = 1910.
3. He changes the initial number by applying bitwise xor of the initial number and the sum. Example: <image>. Note that xor is done in binary notation.
One more example: for integer 1e the sum is sum = 21 + 214. Letters a, b, c, d, e, f denote hexadecimal digits 10, 11, 12, 13, 14, 15, respertively.
Sherlock wants to count the numbers in the range from l to r (both inclusive) which decrease on application of the above four steps. He wants you to answer his q queries for different l and r.
Input
First line contains the integer q (1 ≤ q ≤ 10000).
Each of the next q lines contain two hexadecimal integers l and r (0 ≤ l ≤ r < 1615).
The hexadecimal integers are written using digits from 0 to 9 and/or lowercase English letters a, b, c, d, e, f.
The hexadecimal integers do not contain extra leading zeros.
Output
Output q lines, i-th line contains answer to the i-th query (in decimal notation).
Examples
Input
1
1014 1014
Output
1
Input
2
1 1e
1 f
Output
1
0
Input
2
1 abc
d0e fe23
Output
412
28464
Note
For the second input,
1416 = 2010
sum = 21 + 24 = 18
<image>
Thus, it reduces. And, we can verify that it is the only number in range 1 to 1e that reduces.
Submitted Solution:
```
#!/bin/python3
def decode(c):
if ord(c) <= ord('9') and ord(c) >= ord('0'):
return ord(c) - ord('0')
return ord(c) - ord('a') + 10
dp = [[[0, 0] for j in range(2)]for i in range(20)]
def solve(s, full):
n = len(s)
ans = 0
for d_max in range(16):
for i in range(n):
for j in range(2):
for k in range(2):
dp[i][j][k] = 0
dp[0][1][0] = 1
pos = n-1- (d_max // 4)
bit = 1<<(d_max % 4)
if pos < 0: continue
for i in range(n):
for up in range(2):
for fill in range(2):
if dp[i][up][fill] == 0: continue
digm = min(d_max, decode(s[i]) if up > 0 else d_max)
for d in range(digm+1):
if (i == pos) and ((d & bit) == 0):
continue
tup = up
if d != decode(s[i]): tup = 0
tfill = fill
if d == d_max: tfill = 1
dp[i+1][tup][tfill] += dp[i][up][fill]
for i in range(full+1):
ans += dp[n][i][1]
return ans
def main():
q = int(input().strip())
for _ in range(q):
l, r = input().strip().split()
print(solve(r, 1)- solve(l, 0))
if __name__ == "__main__":
main()
``` | instruction | 0 | 54,467 | 20 | 108,934 |
No | output | 1 | 54,467 | 20 | 108,935 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Using the given four integers from 1 to 9, we create an expression that gives an answer of 10. When you enter four integers a, b, c, d, write a program that outputs an expression that gives an answer of 10 according to the following conditions. Also, if there are multiple answers, only the first answer found will be output. If there is no answer, output 0.
* Use only addition (+), subtraction (-), and multiplication (*) as operators. Do not use division (/). You can use three operators.
* You must use all four numbers.
* You can freely change the order of the four numbers.
* You can use parentheses. You can use up to 3 sets (6) of parentheses.
Input
Given multiple datasets. The format of each dataset is as follows:
a b c d
Input ends with four 0s. The number of datasets does not exceed 40.
Output
For each dataset, combine the given four integers with the above arithmetic symbols and parentheses to output an expression or 0 with a value of 10 on one line. The expression string must not exceed 1024 characters.
Example
Input
8 7 9 9
4 4 4 4
5 5 7 5
0 0 0 0
Output
((9 * (9 - 7)) - 8)
0
((7 * 5) - (5 * 5))
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
http://judge.u-aizu.ac.jp/onlinejudge/description.jsp?id=0041
"""
import sys
from itertools import permutations, product
def make_ten(digits):
"""
4????????°????????????(9**4)??????????????????????????????(3**3)????????????????????????????????????(6) ?????????????????§?????§????????????
"""
result = '0'
for a, b, c, d in permutations(digits, 4): # 4????????°??????????????°??????2??????????????¢????????§??????????????§???permutations
if result != '0':
break
for op1, op2, op3 in product(['+', '-', '*'], repeat=3): # ??????????????????????????????????????????????????????OK
# 1, 2, 3???????????§?¨????????????????????????????????????????
exp = '(({} {} {}) {} {}) {} {}'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 1, 3, 2
exp= '(({} {} {}) {} ({} {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 2, 1, 3
exp = '(({} {} ({} {} {})) {} {})'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 3, 1, 2
exp = '({} {} (({} {} {}) {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 3, 2, 1
exp = '({} {} ({} {} ({} {} {})))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
# 2, 3, 1
exp = '(({} {} {}) {} ({} {} {}))'.format(a, op1, b, op2, c, op3, d)
if eval(exp) == 10:
result = exp
break
return result
# ????????°????????????????????????10?????????????????????????????¢?????°????????§?????????????????????????????°??????????????¨???????????????????????¨??????????´????????????????
# ????????°????????°?????????????????????????????°??????????????´???????????°????????????????????????????????£????????\
Memo = {}
def solve(digits):
global Memo
digits.sort(reverse=True) # 1234 ??§??? 4321 ??§???????????????(data)??????????????????????????§?????????
data = ''.join(map(str, digits))
if not data in Memo:
Memo[data] = make_ten(digits)
return Memo[data]
def main(args):
while True:
digits = [int(x) for x in input().strip().split(' ')]
# digits = [int(x) for x in '{:04d}'.format(count)]
if digits[0] == 0 and digits[1] == 0 and digits[2] == 0 and digits[3] == 0:
break
result = solve(digits)
print(result)
if __name__ == '__main__':
main(sys.argv[1:])
``` | instruction | 0 | 54,695 | 20 | 109,390 |
Yes | output | 1 | 54,695 | 20 | 109,391 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,844 | 20 | 109,688 |
Tags: constructive algorithms, math
Correct Solution:
```
l,r=map(int,input().split())
m=bin(l-1)
n=bin(r)
no1=list(m)
no2=list(n)
no1=no1[2:]
no2=no2[2:]
q=len(no2)-1
ev=0
odd=0
flag=1
for i in range(0,q):
if(i%2==0):
flag=0
odd+=pow(2,i)
else:
flag=1
ev+=pow(2,i)
s1=pow(2,q)-1
res1=r-s1
if flag==0:
ev+=res1
else:
odd+=res1
evensum=(ev+1)*ev
oddsum=odd*odd
result1=oddsum+evensum
q=len(no1)-1
ev=0
odd=0
flag=1
for i in range(0,q):
if(i%2==0):
flag=0
odd+=pow(2,i)
else:
flag=1
ev+=pow(2,i)
s1=pow(2,q)-1
res1=l-1-s1
if flag==0:
ev+=res1
else:
odd+=res1
evensum=(ev+1)*ev
oddsum=odd*odd
result2=oddsum+evensum
result=(result1-result2)%1000000007
print(result)
``` | output | 1 | 54,844 | 20 | 109,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,845 | 20 | 109,690 |
Tags: constructive algorithms, math
Correct Solution:
```
l,r = map(int,input().split())
# print(l,r)
MOD = 1000*1000*1000+7;
def getSum(first, n):
return ( (first + n-1)*n ) % MOD;
def f(pos):
curNum = [0,0]
curNum[1] = 3;
curNum[0] = 2;
cur = 0;
cpos = 1
sum = 1
p2 = 2
while cpos+p2 <= pos:
cpos += p2;
sum += getSum(curNum[cur], p2);
sum %= MOD;
curNum[cur]+=p2*2;
cur = 1 -cur;
p2 *= 2;
sum += getSum(curNum[cur], pos-cpos);
sum %= MOD;
return sum;
sum = f(r);
if l-1 >= 1:
sum = sum + MOD - f(l-1);
sum += MOD;
print(sum % MOD);
``` | output | 1 | 54,845 | 20 | 109,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,846 | 20 | 109,692 |
Tags: constructive algorithms, math
Correct Solution:
```
M = 1000000007
import math
odd_start = 1
even_start = 2
count = 1
step_lst = []
for i in range(80):
if i % 2 == 0:
odd_end = odd_start + 2 * count - 2
total = (odd_start + odd_end) // 2
total *= count
step_lst.append((odd_start, total))
odd_start = odd_end + 2
else:
even_end = even_start + 2 * count - 2
total = (even_start + even_end) // 2
total *= count
step_lst.append((even_start, total))
even_start = even_end + 2
count *= 2
l, r = map(int, input().split())
l_step = math.log2(l)
l_step = math.floor(l_step)
l_index = l - 2**l_step
r_step = math.log2(r)
r_step = math.floor(r_step)
r_index = r - 2**r_step
if l_step == r_step:
start = step_lst[l_step][0] + l_index * 2
end = step_lst[l_step][0] + r_index * 2
count = r_index - l_index + 1
total = count * ((end + start) // 2)
else:
start = step_lst[l_step][0] + l_index * 2
end = step_lst[l_step + 2][0] - 2
count = (end - start) // 2 + 1
total = count * ((end + start) // 2)
#total = int(total % M)
for i in range(l_step + 1, r_step):
total += step_lst[i][1]
#total = int(total % M)
#print('--')
start = step_lst[r_step][0]
#print(start)
end = step_lst[r_step][0] + r_index * 2
#print(end)
count = (end - start) // 2 + 1
inner_total = count * ((end + start) // 2)
#print(inner_total)
total += inner_total
#print('--')
#print(l_step, r_step)
#print(l_index, r_index)
total = int(total)
total = int(total % M)
print(total)
``` | output | 1 | 54,846 | 20 | 109,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,847 | 20 | 109,694 |
Tags: constructive algorithms, math
Correct Solution:
```
l, r = input().split(' ')
l = int(l)
r = int(r)
w1 = 0
w2 = 0
d = 1
cur = 0
q = 1
while cur + d < l:
if q == 1:
w1 += d
else:
w2 += d
cur += d
d = d * 2
q = 1 - q
arr = [[0, 0], [0, 0]]
arr[0][0] = w2
arr[0][1] = w1
if cur + d <= r:
w = cur + d - l + 1
if q == 1:
w1 += d - w
else:
w2 += d - w
arr[0][0] = w2
arr[0][1] = w1
if q == 1:
w1 += w
else:
w2 += w
cur += d
d = d * 2
q = 1 - q
while cur + d <= r:
if q == 1:
w1 += d
else:
w2 += d
cur += d
d = d * 2
q = 1 - q
if cur != r:
if q == 1:
w1 += r - cur
else:
w2 += r - cur
else:
if q == 1:
w1 += l - cur - 1
else:
w2 += l - cur - 1
arr[0][0] = w2
arr[0][1] = w1
if q == 1:
w1 += r - l + 1
else:
w2 += r - l + 1
arr[1][0] = w2
arr[1][1] = w1
M = 1000000007
ans = (arr[1][0] % M * (arr[1][0] + 1) % M) % M
ans -= (arr[0][0] % M * (arr[0][0] + 1) % M) % M
ans += (arr[1][1] % M * (arr[1][1] + 1) % M) % M - arr[1][1]
ans = ans % M
ans -= ((arr[0][1] % M * (arr[0][1] + 1) % M) % M - arr[0][1])
ans = ans % M
print(ans)
``` | output | 1 | 54,847 | 20 | 109,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,848 | 20 | 109,696 |
Tags: constructive algorithms, math
Correct Solution:
```
l, r = map(int, input().split())
ans = 0
while l <= r:
ub = ('1' * (len(bin(l)) - 2))
even = len(ub)
ub = min(int(ub, 2), r)
fub= ub
#print(ub)
#print(even)
if even % 2 == 0:
base = 1
pos = l
while pos > base:
pos -= base
ub -= base
base *= 4
ans += ((pos * 2) + (ub * 2)) * (ub - pos + 1) // 2
else:
base = 2
pos = l
while pos > base:
pos -= base
ub -= base
base *= 4
ans += ((pos * 2 - 1) + (ub * 2 - 1)) * (ub - pos + 1) // 2
l = fub + 1
print(ans % 1000000007)
``` | output | 1 | 54,848 | 20 | 109,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,849 | 20 | 109,698 |
Tags: constructive algorithms, math
Correct Solution:
```
def f(n):
i=0
x=n
o=0
e=0
while x>0:
p=min((2**i),x)
if i%2==0:
o=o+p
else:
e=e+p
x=x-p
i=i+1
return ((o*o)+(e*(e+1)))
m=(10**9)+7
l,r=map(int,input().split())
#print(f(r),f(l-1))
print((f(r)-f(l-1))%m)
``` | output | 1 | 54,849 | 20 | 109,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,850 | 20 | 109,700 |
Tags: constructive algorithms, math
Correct Solution:
```
import sys
def getSumsOddEvenPotenciesTwo(p):
i = 0
j = 0
for e in range(0, p):
if e % 2 == 0:
i += pow(2, e)
else:
j += pow(2, e)
return i, j
def getNearPotencyTwo(n):
c = 0
while n > 0:
n = n >> 1
c += 1
return c - 1
def getSumSerie(n):
if n == 0:
return 0
p = getNearPotencyTwo(n)
i, j = getSumsOddEvenPotenciesTwo(p)
if p % 2 == 0:
i += n - (pow(2, p) - 1)
else:
j += n - (pow(2, p) - 1)
return ((2 * i * i) // 2) + ((2 * (j + 1) * j) // 2)
#sys.stdin = open('test.txt', 'r')
#sys.stdout = open('myout.txt', 'w')
line = sys.stdin.readline().split()
l = int(line[0])
r = int(line[1])
s = (getSumSerie(r) - getSumSerie(l - 1)) % (1000000007)
print("%d" % s)
``` | output | 1 | 54,850 | 20 | 109,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105. | instruction | 0 | 54,851 | 20 | 109,702 |
Tags: constructive algorithms, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
from collections import Counter
import math as mt
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)
def gcd(a, b):
if a == 0:
return b
return gcd(b % a, a)
def lcm(a, b):
return (a * b) / gcd(a, b)
mod = int(1e9) + 7
def power(k, n):
if n == 0:
return 1
if n % 2:
return (power(k, n - 1) * k) % mod
t = power(k, n // 2)
return (t * t) % mod
def totalPrimeFactors(n):
count = 0
if (n % 2) == 0:
count += 1
while (n % 2) == 0:
n //= 2
i = 3
while i * i <= n:
if (n % i) == 0:
count += 1
while (n % i) == 0:
n //= i
i += 2
if n > 2:
count += 1
return count
# #MAXN = int(1e7 + 1)
# # spf = [0 for i in range(MAXN)]
#
#
# def 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, mt.ceil(mt.sqrt(MAXN))):
# if (spf[i] == i):
# for j in range(i * i, MAXN, i):
# if (spf[j] == j):
# spf[j] = i
#
#
# def getFactorization(x):
# ret = 0
# while (x != 1):
# k = spf[x]
# ret += 1
# # ret.add(spf[x])
# while x % k == 0:
# x //= k
#
# return ret
# Driver code
# precalculating Smallest Prime Factor
# sieve()
def main():
# n=int(input())
# a=list(map(int, input().split()))
l, r = map(int,input().split())
k=1
e=0
o=0
f=0
while r:
t=min(r, k)
if f==0:
o+=t
else:
e+=t
f^=1
r-=t
k*=2
ans=(e%mod*((e+1)%mod))%mod+((o%mod)**2)%mod
l-=1
k = 1
e = 0
o = 0
f = 0
while l:
t=min(l, k)
if f==0:
o+=t
else:
e+=t
f^=1
l-=t
k*=2
ans1=((o%mod)**2)%mod+(e%mod*((e+1)%mod))%mod
print((ans-ans1+mod)%mod)
#s=input()
return
if __name__ == "__main__":
main()
``` | output | 1 | 54,851 | 20 | 109,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
import sys; input=sys.stdin.readline
M = 1000000007
def sm(n):
total, odd, even = 0, 1, 2
atodd=True
k = 1
while n:
k = min(k, n)
n -= k
if atodd:
atodd=False
total += k*(odd + odd+(k-1)*2)//2
total %= M
odd += k*2
else:
atodd=True
# print('>', even)
total += k*(even + even+(k-1)*2)//2
total %= M
even += k*2
k*= 2
return total
# print(sm(10**18))
l, r = map(int, input().split())
print((sm(r) - sm(l-1))%M)
``` | instruction | 0 | 54,852 | 20 | 109,704 |
Yes | output | 1 | 54,852 | 20 | 109,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
from math import *
def f(n):
if n <= 0: return 0
m = ceil(log2(n + 1))
t = 4 ** (m // 2) - 1
if m % 2:
b = 2 * t // 3
a = n - b
else:
a = t // 3
b = n - a
return a * a + b * (b + 1)
l, r = map(int, input().split())
print((f(r) - f(l - 1)) % (10 ** 9 + 7))
``` | instruction | 0 | 54,853 | 20 | 109,706 |
Yes | output | 1 | 54,853 | 20 | 109,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
def su(n):
return ((n*(n+1)))%b
def suo(n):
return ((n*(n+1)-n))%b
def f(k):
i=1
l=1
s=0
c=[0,0]
while s+l<=k:
s+=l
c[i%2]+=l
i+=1
l*=2
c[i%2]+=(k-s)
return (su(c[0])+suo(c[1]))%b
b = 1000000007
l , r = map(int,input().split())
aa=(f(r)-f(l-1))
print((aa+b)%b)
``` | instruction | 0 | 54,854 | 20 | 109,708 |
Yes | output | 1 | 54,854 | 20 | 109,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
#!/usr/bin/python3
import sys
input = lambda: sys.stdin.readline().strip()
l, r = [int(x) for x in input().split()]
M = 10**9 + 7
def T(n):
return n * (n + 1) // 2
def f(n):
total = 0
s = 0
a = [(1, 1), (2, 2)]
for start, terms in a:
new_terms = min(terms, n - total)
total += new_terms
s += start * new_terms % M + 2 * T(new_terms - 1) % M
s %= M
a.append((start + 2 * terms, terms * 4))
if total >= n:
break
return s
print((f(r) - f(l - 1)) % M)
``` | instruction | 0 | 54,855 | 20 | 109,710 |
Yes | output | 1 | 54,855 | 20 | 109,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
if __name__ == '__main__':
n, m = map(int, input().split())
mult = 1
nn, mm = 1, 1
isPar = False
out = []
for i in range(m):
if isPar:
for x in range(nn, mult + 1):
out.append(x * 2)
nn = mult + 1
mult += 1
isPar = False
else:
for x in range(mm, mult+1):
out.append(x * 2 - 1)
mm = mult + 1
mult += 1
isPar = True
suma = 0
print(out)
for i in range(n, m + 1):
suma += out[i-1]
print(suma)
``` | instruction | 0 | 54,856 | 20 | 109,712 |
No | output | 1 | 54,856 | 20 | 109,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
import math
def find(n):
if(n==0):
return 0
x=int(math.log2(n+1))
so1=(pow(4,math.ceil(x/2))-1)//3
se1=2*(pow(4,math.floor(x/2))-1)//3
so=so1**2
se=2*(se1*(se1+1))
if(x//2==x/2):
t=n-so1-se1
so=(so1+t)**2
se=(se1*(se1+1))
else:
t=n-so1-se1
so=(so1)**2
se1+=t
se=2*(se1*(se1+1))
# print(so,se,x,so1,se1)
return so+se
def main():
# for _ in range(int(input())):
l,r=map(int,input().split())
print(find(r)-find(l-1))
main()
``` | instruction | 0 | 54,857 | 20 | 109,714 |
No | output | 1 | 54,857 | 20 | 109,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
l, r = map(int, input().split())
mod = 1000000007
o = e = 0
x = i = 1
while x < l:
if i % 2: e += x//2
else: o += x//2
if (x << 1) < l: x = x << 1
else:
if i % 2: o += l-x
else: e += l-x
break
i += 1
s1 = (e*(e+1) + o*o) % mod
o = e = 0
x = i = 1
while x < r:
if i % 2: e += x//2
else: o += x//2
if (x << 1) < r: x = x << 1
else:
if i % 2: o += r-x+1
else: e += r-x+1
break
i += 1
s2 = (e*(e+1) + o*o) % mod
print(((s2 - s1 % mod) + mod) % mod)
``` | instruction | 0 | 54,858 | 20 | 109,716 |
No | output | 1 | 54,858 | 20 | 109,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nazar, a student of the scientific lyceum of the Kingdom of Kremland, is known for his outstanding mathematical abilities. Today a math teacher gave him a very difficult task.
Consider two infinite sets of numbers. The first set consists of odd positive numbers (1, 3, 5, 7, …), and the second set consists of even positive numbers (2, 4, 6, 8, …). At the first stage, the teacher writes the first number on the endless blackboard from the first set, in the second stage — the first two numbers from the second set, on the third stage — the next four numbers from the first set, on the fourth — the next eight numbers from the second set and so on. In other words, at each stage, starting from the second, he writes out two times more numbers than at the previous one, and also changes the set from which these numbers are written out to another.
The ten first written numbers: 1, 2, 4, 3, 5, 7, 9, 6, 8, 10. Let's number the numbers written, starting with one.
The task is to find the sum of numbers with numbers from l to r for given integers l and r. The answer may be big, so you need to find the remainder of the division by 1000000007 (10^9+7).
Nazar thought about this problem for a long time, but didn't come up with a solution. Help him solve this problem.
Input
The first line contains two integers l and r (1 ≤ l ≤ r ≤ 10^{18}) — the range in which you need to find the sum.
Output
Print a single integer — the answer modulo 1000000007 (10^9+7).
Examples
Input
1 3
Output
7
Input
5 14
Output
105
Input
88005553535 99999999999
Output
761141116
Note
In the first example, the answer is the sum of the first three numbers written out (1 + 2 + 4 = 7).
In the second example, the numbers with numbers from 5 to 14: 5, 7, 9, 6, 8, 10, 12, 14, 16, 18. Their sum is 105.
Submitted Solution:
```
## necessary imports
import sys
input = sys.stdin.readline
# from math import ceil, floor, factorial;
def ceil(x):
if x != int(x):
x = int(x) + 1;
return x;
# 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 ---------------- ###################
l, r = int_array();
odd = 1; even = 2; x = 1;
ans = 0;
for i in range(60):
if i % 2 == 0:
a = x; b = 2 * x - 1;
sts = odd; end = odd + (x - 1) * 2;
odd += x * 2;
else:
a = x; b = 2 * x - 1;
sts = even; end = even + (x - 1) * 2;
even += x * 2;
if l >= a:
if l <= b:
sts += (l - a) * 2;
if r <= b:
## starts and ends here;
end -= (b - r) * 2;
y = ((end - sts) // 2) + 1;
ans += ( y * (sts + end)) // 2;
break;
else:
## starts in this one, doesn't ends here though;
y = (end - sts) // 2 + 1;
ans += ( y * (sts + end)) // 2;
else:
if r <= b:
end -= (b - r) * 2;
y = ((end - sts) // 2) + 1;
ans += ( y * (sts + end)) // 2;
break;
else:
## take this whole bitch;
ans += (x * (end - sts)) // 2;
ans %= MOD; x *= 2;
print(ans % MOD);
``` | instruction | 0 | 54,859 | 20 | 109,718 |
No | output | 1 | 54,859 | 20 | 109,719 |
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.
One day Petya was delivered a string s, containing only digits. He needs to find a string that
* represents a lucky number without leading zeroes,
* is not empty,
* is contained in s as a substring the maximum number of times.
Among all the strings for which the three conditions given above are fulfilled, Petya only needs the lexicographically minimum one. Find this string for Petya.
Input
The single line contains a non-empty string s whose length can range from 1 to 50, inclusive. The string only contains digits. The string can contain leading zeroes.
Output
In the only line print the answer to Petya's problem. If the sought string does not exist, print "-1" (without quotes).
Examples
Input
047
Output
4
Input
16
Output
-1
Input
472747
Output
7
Note
The lexicographical comparison of strings is performed by the < operator in the modern programming languages. String x is lexicographically less than string y either if x is a prefix of y, or exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of string a.
In the first sample three conditions are fulfilled for strings "4", "7" and "47". The lexicographically minimum one is "4".
In the second sample s has no substrings which are lucky numbers.
In the third sample the three conditions are only fulfilled for string "7". | instruction | 0 | 54,877 | 20 | 109,754 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
seven = s.count("7")
four = s.count("4")
if seven == 0 and four == 0:
print(-1)
else:
ans = "7" if seven > four else "4"
print(ans)
``` | output | 1 | 54,877 | 20 | 109,755 |
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.
One day Petya was delivered a string s, containing only digits. He needs to find a string that
* represents a lucky number without leading zeroes,
* is not empty,
* is contained in s as a substring the maximum number of times.
Among all the strings for which the three conditions given above are fulfilled, Petya only needs the lexicographically minimum one. Find this string for Petya.
Input
The single line contains a non-empty string s whose length can range from 1 to 50, inclusive. The string only contains digits. The string can contain leading zeroes.
Output
In the only line print the answer to Petya's problem. If the sought string does not exist, print "-1" (without quotes).
Examples
Input
047
Output
4
Input
16
Output
-1
Input
472747
Output
7
Note
The lexicographical comparison of strings is performed by the < operator in the modern programming languages. String x is lexicographically less than string y either if x is a prefix of y, or exists such i (1 ≤ i ≤ min(|x|, |y|)), that xi < yi and for any j (1 ≤ j < i) xj = yj. Here |a| denotes the length of string a.
In the first sample three conditions are fulfilled for strings "4", "7" and "47". The lexicographically minimum one is "4".
In the second sample s has no substrings which are lucky numbers.
In the third sample the three conditions are only fulfilled for string "7". | instruction | 0 | 54,878 | 20 | 109,756 |
Tags: brute force, implementation
Correct Solution:
```
s = input()
d = {}
for i in s:
t = int(i)
if t not in d:
d[t] = 1
else:
d[t] += 1
if 4 in d:
if 7 in d:
if d[4] > d[7]:
print(4)
elif d[4] == d[7]:
print(4)
else:
print(7)
else:
print(4)
elif 7 in d:
print(7)
else:
print("-1")
``` | output | 1 | 54,878 | 20 | 109,757 |
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