message stringlengths 2 44.5k | message_type stringclasses 2
values | message_id int64 0 1 | conversation_id int64 42 109k | cluster float64 5 5 | __index_level_0__ int64 84 217k |
|---|---|---|---|---|---|
Provide a correct Python 3 solution for this coding contest problem.
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)) | instruction | 0 | 54,690 | 5 | 109,380 |
"Correct Solution:
```
import itertools
def f(s):
for x,y,z in itertools.product('+-*',repeat=3):
for a,b,c,d in itertools.permutations(s):
yield f"({a} {x} {b}) {y} ({c} {z} {d})"
yield f"(({a} {x} {b}) {y} {c}) {z} {d}"
yield f"{a} {x} ({b} {y} ({c} {z} {d}))"
yield f"({a} {x} ({b} {y} {c})) {z} {d}"
yield f"{a} {x} (({b} {y} {c}) {z} {d})"
for e in iter(input,'0 0 0 0'):
s=list(map(int,e.split()));a=0
for m in f(s):
if eval(m)==10:a='('+m+')';break
print(a)
``` | output | 1 | 54,690 | 5 | 109,381 |
Provide a correct Python 3 solution for this coding contest problem.
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)) | instruction | 0 | 54,691 | 5 | 109,382 |
"Correct Solution:
```
from itertools import *
def f(s):
if sum(s)==10:return"((({} + {}) + {}) + {})".format(*s)
for a,b,c,d in permutations(s):
for x,y,z in permutations('+-*'*2,3):
for t in[f"({a} {x} {b}) {y} ({c} {z} {d})",f"(({a} {x} {b}) {y} {c}) {z} {d}",f"({a} {x} ({b} {y} {c})) {z} {d}"]:
if eval(t)==10:return'('+t+')'
else:return 0
for e in iter(input,'0 0 0 0'):print(f(list(map(int,e.split()))))
``` | output | 1 | 54,691 | 5 | 109,383 |
Provide a correct Python 3 solution for this coding contest problem.
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)) | instruction | 0 | 54,692 | 5 | 109,384 |
"Correct Solution:
```
from collections import deque
def calculate1(x, y) :
if x == 0 : y.appendleft(y.popleft() + y.popleft())
elif x == 1 : y.appendleft(y.popleft() - y.popleft())
else : y.appendleft(y.popleft() * y.popleft())
return y
def calculate2(x, y) :
if x == 0 : y.append(y.pop() + y.pop())
elif x == 1 : y.append(- y.pop() + y.pop())
else : y.append(y.pop() * y.pop())
return y
def search1(sample) :
for first in range(3) :
for second in range(3) :
for third in range(3) :
box = deque(sample)
calculate1(first, box)
calculate1(second, box)
calculate1(third, box)
if box[0] == 10 :
return(first,second,third)
break
def search2(sample) :
for first in range(3) :
for second in range(3) :
for third in range(3) :
box = deque(sample)
calculate1(first, box)
calculate2(second, box)
calculate1(third, box)
if box[0] == 10 :
return(first,second,third)
break
def attack(sample) :
test = [[sample[a],sample[b],sample[c],sample[d]] for a in range(4) for b in range(4) for c in range(4) for d in range(4) if a!=b and a!=c and a!=d and b!=c and b!=d and c!=d]
test = list(set(map(tuple, test)))
for _ in test :
result = search1(_)
if result != None :
return result, _, 1
break
else :
result = search2(_)
if result != None :
return result, _, 2
break
ams = ['+', '-', '*']
while True :
sample = [int(_) for _ in input().split()]
if sample == [0, 0, 0, 0] : break
result = attack(sample)
if result != None :
if result[2] == 1 : print('((({} {} {}) {} {}) {} {})'.format(result[1][0], ams[result[0][0]], result[1][1], ams[result[0][1]], result[1][2], ams[result[0][2]], result[1][3]))
else : print('(({} {} {}) {} ({} {} {}))'.format(result[1][0], ams[result[0][0]], result[1][1], ams[result[0][2]], result[1][2], ams[result[0][1]], result[1][3]))
else : print(0)
``` | output | 1 | 54,692 | 5 | 109,385 |
Provide a correct Python 3 solution for this coding contest problem.
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)) | instruction | 0 | 54,693 | 5 | 109,386 |
"Correct Solution:
```
from itertools import *
g=lambda a,x,b:a+b if x=='+'else a-b if x=='-'else a*b
def f(s):
for a,b,c,d in permutations(s):
for x,y,z in product('-*+',repeat=3):
if g(g(a,x,b),y,g(c,z,d))==10:return f"(({a} {x} {b}) {y} ({c} {z} {d}))"
if g(g(a,x,g(b,y,c)),z,d)==10:return f"(({a} {x} ({b} {y} {c})) {z} {d})"
return 0
for e in iter(input,'0 0 0 0'):print(f(list(map(int,e.split()))))
``` | output | 1 | 54,693 | 5 | 109,387 |
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:
```
from itertools import *
def g(a,x,b):return a+b if x=='+'else a-b if x=='-'else a*b
def f(s):
for a,b,c,d in permutations(s):
for x,y,z in product('+-*',repeat=3):
if g(g(a,x,b),y,g(c,z,d))==10:return f"(({a} {x} {b}) {y} ({c} {z} {d}))"
if g(g(a,x,g(b,y,c)),z,d)==10:return f"(({a} {x} ({b} {y} {c})) {z} {d})"
else:return 0
for e in iter(input,'0 0 0 0'):print(f(list(map(int,e.split()))))
``` | instruction | 0 | 54,696 | 5 | 109,392 |
Yes | output | 1 | 54,696 | 5 | 109,393 |
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:
```
import itertools
def f(s):
for a,b,c,d in itertools.permutations(s):
for x,y,z in itertools.product('+-*',repeat=3):
yield f"(({a} {x} {b}) {y} ({c} {z} {d}))"
yield f"((({a} {x} {b}) {y} {c}) {z} {d})"
yield f"({a} {x} ({b} {y} ({c} {z} {d})))"
yield f"(({a} {x} ({b} {y} {c})) {z} {d})"
yield f"({a} {x} (({b} {y} {c}) {z} {d}))"
for e in iter(input,'0 0 0 0'):
s=list(map(int,e.split()));a=0
for m in f(s):
if eval(m)==10:a=m;break
print(a)
``` | instruction | 0 | 54,697 | 5 | 109,394 |
Yes | output | 1 | 54,697 | 5 | 109,395 |
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:
```
from itertools import *
def f(s):
for a,b,c,d in permutations(s):
for x,y,z in product('+-*',repeat=3):
for t in[f"(({a} {x} {b}) {y} {c}) {z} {d}",f"({a} {x} ({b} {y} {c})) {z} {d}"]:
if eval(t)==10:return'('+t+')'
else:return 0
for e in iter(input,'0 0 0 0'):print(f(list(map(int,e.split()))))
``` | instruction | 0 | 54,698 | 5 | 109,396 |
No | output | 1 | 54,698 | 5 | 109,397 |
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:
```
import itertools,sys
def f(a):
if len(a)<2:yield[a[0]]*2
for i in range(1,len(a)):
for p,s in f(a[:i]):
for q,t in f(a[i:]):
yield(p+q,f'({s} + {t})')
yield(p-q,f'({s} - {t})')
yield(p*q,f'({s} * {t})')
def s(a):
for p in itertools.permutations(a):
for n,s in f(p):
if n==10:print(s);return 1
for e in iter(input,'0 0 0 0\n'):
a=list(map(int,e.split()))
if not s(a):print(0)
``` | instruction | 0 | 54,699 | 5 | 109,398 |
No | output | 1 | 54,699 | 5 | 109,399 |
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:
```
from collections import deque
def calculate1(x, y) :
if x == 0 : y.appendleft(y.popleft() + y.popleft())
elif x == 1 : y.appendleft(y.popleft() - y.popleft())
else : y.appendleft(y.popleft() * y.popleft())
return y
def calculate2(x, y) :
if x == 0 : y.append(y.pop() + y.pop())
elif x == 1 : y.append(y.pop() - y.pop())
else : y.append(y.pop() * y.pop())
return y
def search1(sample) :
for first in range(3) :
for second in range(3) :
for third in range(3) :
box = deque(sample)
calculate1(first, box)
calculate1(second, box)
calculate1(third, box)
if box[0] == 10 :
return(first,second,third)
break
def search2(sample) :
for first in range(3) :
for second in range(3) :
for third in range(3) :
box = deque(sample)
calculate1(first, box)
calculate2(second, box)
calculate1(third, box)
if box[0] == 10 :
return(first,second,third)
break
def attack(sample) :
test = [[sample[a],sample[b],sample[c],sample[d]] for a in range(4) for b in range(4) for c in range(4) for d in range(4) if a!=b and a!=c and a!=d and b!=c and b!=d and c!=d]
test = list(set(map(tuple, test)))
for _ in test :
result = search1(_)
if result != None :
return result, _, 1
break
else :
result = search2(_)
if result != None :
return result, _, 2
break
ams = ['+', '-', '*']
while True :
sample = [int(_) for _ in input().split()]
if sample == [0, 0, 0, 0] : break
result = attack(sample)
if result != None :
if result[2] == 1 : print('(({} {} {}) {} {}) {} {}'.format(result[1][0], ams[result[0][0]], result[1][1], ams[result[0][1]], result[1][2], ams[result[0][2]], result[1][3]))
else : print('({} {} {}) {} ({} {} {})'.format(result[1][0], ams[result[0][0]], result[1][1], ams[result[0][2]], result[1][2], ams[result[0][1]], result[1][3]))
else : print(0)
``` | instruction | 0 | 54,701 | 5 | 109,402 |
No | output | 1 | 54,701 | 5 | 109,403 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,772 | 5 | 109,544 |
"Correct Solution:
```
n, m, l = map(int, input().split())
A_list = [list(map(int, input().split())) for i in range(n)]
B_list = [list(map(int, input().split())) for j in range(m)]
for i in range(n):
_list = [sum(A_list[i][j] * B_list[j][k] for j in range(m)) for k in range(l)]
print(*_list)
``` | output | 1 | 54,772 | 5 | 109,545 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,773 | 5 | 109,546 |
"Correct Solution:
```
n, m, l = map(int, input().split())
A = [list(map(int, input().split())) for i in range(n)]
B = [list(map(int, input().split())) for i in range(m)]
for i in range(n):
for j in range(l):
c = 0
print("{}{}".format(' ' if j else '', sum(A[i][k] * B[k][j] for k in range(m))), end='')
print()
``` | output | 1 | 54,773 | 5 | 109,547 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,774 | 5 | 109,548 |
"Correct Solution:
```
n, m, l = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(n)]
B = [list(map(int, input().split())) for _ in range(m)]
C = [[0]*l for j in range(n)]
for i in range(n):
for j in range(l):
for k in range(m):
C[i][j] += A[i][k]*B[k][j]
for i in range(n):
print(*C[i])
``` | output | 1 | 54,774 | 5 | 109,549 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,775 | 5 | 109,550 |
"Correct Solution:
```
n,m,l = map(int,input().split())
A = [[int(i) for i in input().split()] for _ in range(n)]
B = [[int(i) for i in input().split()] for _ in range(m)]
res = [
[sum([
a*b for a,b in zip(A[i],list(zip(*B))[j])
])
for j in range(l)]
for i in range(n)]
for r in res:
print(*r)
``` | output | 1 | 54,775 | 5 | 109,551 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,776 | 5 | 109,552 |
"Correct Solution:
```
n,m,l=map(int,input().split())
A=[]
for i in range(n):
*Ai,=map(int,input().split())
A.append(Ai)
B=[]
for i in range(m):
*Bi,=map(int,input().split())
B.append(Bi)
for i in range(n):
Ci=[sum([A[i][j]*B[j][k] for j in range(m)]) for k in range(l)]
print(*Ci)
``` | output | 1 | 54,776 | 5 | 109,553 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,777 | 5 | 109,554 |
"Correct Solution:
```
n,m,l=map(int,input().split())
c1=[list(map(int,list(input().split()))) for i in range(n)]
c2=[list(map(int,list(input().split()))) for i in range(m)]
c3=[[0]*l for i in range(n)]
for i in range(n):
for j in range(l):
for k in range(m):
c3[i][j]+=c1[i][k]*c2[k][j]
for i in range(n):
print(*c3[i])
``` | output | 1 | 54,777 | 5 | 109,555 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,778 | 5 | 109,556 |
"Correct Solution:
```
n,m,l=map(int,input().split())
e=[input().split()for _ in[0]*(n+m)]
for c in e[:n]:print(*[sum(int(s)*int(t)for s,t in zip(c,l))for l in zip(*e[n:])])
``` | output | 1 | 54,778 | 5 | 109,557 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14 | instruction | 0 | 54,779 | 5 | 109,558 |
"Correct Solution:
```
N, M, L = map(int, input().split())
A = [list(map(int, input().split())) for i in range(N)]
B = [list(map(int, input().split())) for i in range(M)]
C = [[0]*L for i in range(N)]
for i in range(N):
for j in range(L):
C[i][j] = str(sum(A[i][k]*B[k][j] for k in range(M)))
for line in C:
print(*line)
``` | output | 1 | 54,779 | 5 | 109,559 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
n, m, l = map(int, input().split())
A = [list(map(int, input().split())) for i in range(n)]
B = [list(map(int, input().split())) for j in range(m)]
C = [ [sum([A[k][j] * B[j][i] for j in range(m)]) for i in range(l)] for k in range(n)]
for i in range(n):
C[i] = map(str, C[i])
print(" ".join(C[i]))
``` | instruction | 0 | 54,780 | 5 | 109,560 |
Yes | output | 1 | 54,780 | 5 | 109,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
# 00.11 sec 7004 KB 5 lines 214 bytes
n,m,l=map(int,input().split())
a=[list(map(int,input().split()))for _ in range(n)]
b=[list(map(int,input().split()))for _ in range(m)]
[print(*x)for x in[[sum(s*t for s,t in zip(c,l))for l in zip(*b)]for c in a]]
``` | instruction | 0 | 54,781 | 5 | 109,562 |
Yes | output | 1 | 54,781 | 5 | 109,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
N, M, L = map(int, input().split())
A = []
B = []
for n in range(N):
A.append(list(map(int, input().split())))
for m in range(M):
B.append(list(map(int, input().split())))
for n in range(N):
print(' '.join(map(str, [sum(A[n][m] * B[m][l] for m in range(M)) for l in range(L)])))
``` | instruction | 0 | 54,782 | 5 | 109,564 |
Yes | output | 1 | 54,782 | 5 | 109,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
v = lambda: list(map(int,input().split()))
n,m,l = v()
a = [v() for i in range(n)]
b = [v() for i in range(m)]
t = [[b[i][j] for i in range(m)] for j in range(l)]
for i in range(n):
print(*(sum(x*y for x,y in zip(a[i], t[j])) for j in range(l)))
``` | instruction | 0 | 54,783 | 5 | 109,566 |
Yes | output | 1 | 54,783 | 5 | 109,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
n, m, l = [int(i) for i in input().split()]
A = []
B = []
C = []
for ni in range(n):
A.append([int(i) for i in input().split()])
for mi in range(m):
B.append([int(i) for i in input().split()])
for i in range(n):
C.append([])
for j in range(l):
c.append(0)
for k in range(m):
C[i][j] += A[i][k] * B[k][j]
for li in range(l):
print(" ".join([str(s) for s in C[ni]]))
``` | instruction | 0 | 54,784 | 5 | 109,568 |
No | output | 1 | 54,784 | 5 | 109,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
n, m, l = map(int, input().split())
A = [[int(n) for n in input().split()] for i in range(n)]
B = [[int(n) for n in input().split()] for i in range(m)]
C = [[0 for i in range(l)] for i in range(n)]
for i in range(l):
for j in range(n):
for k in range(m):
C[i][j] += A[i][k] * B[k][j]
for i in range(n):
for j in range(l):
if j == l - 1:
print(C[i][j])
else:
print(C[i][j], end = " ")
``` | instruction | 0 | 54,785 | 5 | 109,570 |
No | output | 1 | 54,785 | 5 | 109,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
( n, m, l) = [ int(i) for i in input().split()]
# input to list
a = []
b = []
for an in range(n):
a.append( [ int(i) for i in input().split()])
for bn in range(m):
b.append( [ int(i) for i in input().split()])
# output
for cn in range(n):
for cl in range(l):
print( a[cn][0] * b[0][cl] + a[cn][1] * b[1][cl], end=' ')
print()
``` | instruction | 0 | 54,786 | 5 | 109,572 |
No | output | 1 | 54,786 | 5 | 109,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which reads a $n \times m$ matrix $A$ and a $m \times l$ matrix $B$, and prints their product, a $n \times l$ matrix $C$. An element of matrix $C$ is obtained by the following formula:
\\[ c_{ij} = \sum_{k=1}^m a_{ik}b_{kj} \\]
where $a_{ij}$, $b_{ij}$ and $c_{ij}$ are elements of $A$, $B$ and $C$ respectively.
Note
解説
Constraints
* $1 \leq n, m, l \leq 100$
* $0 \leq a_{ij}, b_{ij} \leq 10000$
Input
In the first line, three integers $n$, $m$ and $l$ are given separated by space characters
In the following lines, the $n \times m$ matrix $A$ and the $m \times l$ matrix $B$ are given.
Output
Print elements of the $n \times l$ matrix $C$ ($c_{ij}$). Print a single space character between adjacent elements.
Example
Input
3 2 3
1 2
0 3
4 5
1 2 1
0 3 2
Output
1 8 5
0 9 6
4 23 14
Submitted Solution:
```
n, m, l = map(int, input().split())
A = []
B = []
C = [[0] * l for i in range(n)]
for x in range(n):
for y in range(m):
A[x].append(list(map(int, input().split())))
for x in range(m):
for y in range(l):
B[x].append(list(map(int, input().split())))
for x in range(n):
for y in range(l):
sum = 0
for z in range(m):
sum += A[x][z] * B[z][y]
C[x][y] = sum
for x in range(n):
for y in range(l-1):
print("{0} ".format(C[x][y]), end = '')
print(C[x][l-1])
``` | instruction | 0 | 54,787 | 5 | 109,574 |
No | output | 1 | 54,787 | 5 | 109,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
import sys
zz=1
sys.setrecursionlimit(10**5)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def ii():
return input().rstrip()
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def dadd(d,p,val):
if p in d:
d[p].append(val)
else:
d[p]=[val]
def gi():
return [xx for xx in input().split()]
def gtc(tc,ans):
print("Case #"+str(tc)+":",ans)
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def prec(a,pre):
for i in a:
pre.append(pre[-1]+i)
pre.pop(0)
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=1
uu=t
#11:22
while t>0:
t-=1
a,b,k=mi()
w=k
if min(a,b)>=2 and k>a+b-2:
print("No")
else:
b-=1
a1="1"
a2="1"
if min(a,b)==1:
if k>max(a,b):
print("No")
exit(0)
if min(a,b)==0:
if k==0:
print("Yes")
if a==0:
print(a1+"1"*b)
print(a2+"1"*b)
else:
print(a1+"0"*a)
print(a2+"0"*a)
else:
print("No")
exit(0)
if k<=a :
a1+="1"+"0"*k
a2+="0"*k+"1"
a1+="0"*(a-k)+"1"*(b-1)
a2+="0"*(a-k)+"1"*(b-1)
elif k<=b:
a1+="1"*k+"0"
a2+="0"+"1"*k
a1+="0"*(a-1)+"1"*(b-k)
a2+="0"*(a-1)+"1"*(b-k)
else:
if k==a+b-1:
a1+="1"*b+"0"*a
a2+="0"+"1"*(b-1)+"0"*(a-1)+"1"
print("Yes")
print(a1)
print(a2)
exit(0)
a1+="1"+"0"*(a-1)
a2+="0"*(a-1)+"1"
k-=(a-1)
a=1
b-=1
a1+="1"*min(k,b)+"0"
a2+="0"+"1"*min(k,b)
a1+="0"*(a-1)+"1"*(b-k)
a2+="0"*(a-1)+"1"*(b-k)
print("Yes")
print(a1)
print(a2)
``` | instruction | 0 | 55,005 | 5 | 110,010 |
Yes | output | 1 | 55,005 | 5 | 110,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
a, b, k = map(int, input().split())
if k == 0:
print("Yes")
print("1" * b + "0" * a)
print("1" * b + "0" * a)
exit(0)
if a == 0 or b == 0:
print("No")
exit(0)
a -= 1
b -= 1
if b == 0:
print("No")
exit(0)
s = "1" * b + "0" * a
b -= 1
k -= 1
if k > a + b:
print("No")
exit(0)
print("Yes")
print(s[:len(s) - k] + "1" + s[len(s) - k:] + "0")
print(s[:len(s) - k] + "0" + s[len(s) - k:] + "1")
``` | instruction | 0 | 55,006 | 5 | 110,012 |
Yes | output | 1 | 55,006 | 5 | 110,013 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
import heapq as h
import bisect
import math as mt
from types import GeneratorType
BUFSIZE = 8192
class SortedList:
def __init__(self, iterable=[], _load=200):
"""Initialize sorted list instance."""
values = sorted(iterable)
self._len = _len = len(values)
self._load = _load
self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
self._list_lens = [len(_list) for _list in _lists]
self._mins = [_list[0] for _list in _lists]
self._fen_tree = []
self._rebuild = True
def _fen_build(self):
"""Build a fenwick tree instance."""
self._fen_tree[:] = self._list_lens
_fen_tree = self._fen_tree
for i in range(len(_fen_tree)):
if i | i + 1 < len(_fen_tree):
_fen_tree[i | i + 1] += _fen_tree[i]
self._rebuild = False
def _fen_update(self, index, value):
"""Update `fen_tree[index] += value`."""
if not self._rebuild:
_fen_tree = self._fen_tree
while index < len(_fen_tree):
_fen_tree[index] += value
index |= index+1
def _fen_query(self, end):
"""Return `sum(_fen_tree[:end])`."""
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
x = 0
while end:
x += _fen_tree[end - 1]
end &= end - 1
return x
def _fen_findkth(self, k):
"""Return a pair of (the largest `idx` such that `sum(_fen_tree[:idx]) <= k`, `k - sum(_fen_tree[:idx])`)."""
_list_lens = self._list_lens
if k < _list_lens[0]:
return 0, k
if k >= self._len - _list_lens[-1]:
return len(_list_lens) - 1, k + _list_lens[-1] - self._len
if self._rebuild:
self._fen_build()
_fen_tree = self._fen_tree
idx = -1
for d in reversed(range(len(_fen_tree).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(_fen_tree) and k >= _fen_tree[right_idx]:
idx = right_idx
k -= _fen_tree[idx]
return idx + 1, k
def _delete(self, pos, idx):
"""Delete value at the given `(pos, idx)`."""
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len -= 1
self._fen_update(pos, -1)
del _lists[pos][idx]
_list_lens[pos] -= 1
if _list_lens[pos]:
_mins[pos] = _lists[pos][0]
else:
del _lists[pos]
del _list_lens[pos]
del _mins[pos]
self._rebuild = True
def _loc_left(self, value):
"""Return an index pair that corresponds to the first position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
lo, pos = -1, len(_lists) - 1
while lo + 1 < pos:
mi = (lo + pos) >> 1
if value <= _mins[mi]:
pos = mi
else:
lo = mi
if pos and value <= _lists[pos - 1][-1]:
pos -= 1
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value <= _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def _loc_right(self, value):
"""Return an index pair that corresponds to the last position of `value` in the sorted list."""
if not self._len:
return 0, 0
_lists = self._lists
_mins = self._mins
pos, hi = 0, len(_lists)
while pos + 1 < hi:
mi = (pos + hi) >> 1
if value < _mins[mi]:
hi = mi
else:
pos = mi
_list = _lists[pos]
lo, idx = -1, len(_list)
while lo + 1 < idx:
mi = (lo + idx) >> 1
if value < _list[mi]:
idx = mi
else:
lo = mi
return pos, idx
def add(self, value):
"""Add `value` to sorted list."""
_load = self._load
_lists = self._lists
_mins = self._mins
_list_lens = self._list_lens
self._len += 1
if _lists:
pos, idx = self._loc_right(value)
self._fen_update(pos, 1)
_list = _lists[pos]
_list.insert(idx, value)
_list_lens[pos] += 1
_mins[pos] = _list[0]
if _load + _load < len(_list):
_lists.insert(pos + 1, _list[_load:])
_list_lens.insert(pos + 1, len(_list) - _load)
_mins.insert(pos + 1, _list[_load])
_list_lens[pos] = _load
del _list[_load:]
self._rebuild = True
else:
_lists.append([value])
_mins.append(value)
_list_lens.append(1)
self._rebuild = True
def discard(self, value):
"""Remove `value` from sorted list if it is a member."""
_lists = self._lists
if _lists:
pos, idx = self._loc_right(value)
if idx and _lists[pos][idx - 1] == value:
self._delete(pos, idx - 1)
def remove(self, value):
"""Remove `value` from sorted list; `value` must be a member."""
_len = self._len
self.discard(value)
if _len == self._len:
raise ValueError('{0!r} not in list'.format(value))
def pop(self, index=-1):
"""Remove and return value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
value = self._lists[pos][idx]
self._delete(pos, idx)
return value
def bisect_left(self, value):
"""Return the first index to insert `value` in the sorted list."""
pos, idx = self._loc_left(value)
return self._fen_query(pos) + idx
def bisect_right(self, value):
"""Return the last index to insert `value` in the sorted list."""
pos, idx = self._loc_right(value)
return self._fen_query(pos) + idx
def count(self, value):
"""Return number of occurrences of `value` in the sorted list."""
return self.bisect_right(value) - self.bisect_left(value)
def __len__(self):
"""Return the size of the sorted list."""
return self._len
def __getitem__(self, index):
"""Lookup value at `index` in sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
return self._lists[pos][idx]
def __delitem__(self, index):
"""Remove value at `index` from sorted list."""
pos, idx = self._fen_findkth(self._len + index if index < 0 else index)
self._delete(pos, idx)
def __contains__(self, value):
"""Return true if `value` is an element of the sorted list."""
_lists = self._lists
if _lists:
pos, idx = self._loc_left(value)
return idx < len(_lists[pos]) and _lists[pos][idx] == value
return False
def __iter__(self):
"""Return an iterator over the sorted list."""
return (value for _list in self._lists for value in _list)
def __reversed__(self):
"""Return a reverse iterator over the sorted list."""
return (value for _list in reversed(self._lists) for value in reversed(_list))
def __repr__(self):
"""Return string representation of sorted list."""
return 'SortedList({0})'.format(list(self))
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.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")
import sys
from math import ceil,log2
INT_MAX = sys.maxsize
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
import collections as col
import math, string
def getInts():
return [int(s) for s in input().split()]
def getInt():
return int(input())
def getStrs():
return [s for s in input().split()]
def getStr():
return input()
def listStr():
return list(input())
def minVal(x, y) :
return x if (x < y) else y;
# A utility function to get the
# middle index from corner indexes.
def getMid(s, e) :
return s + (e - s) // 2;
""" A recursive function to get the
minimum value in a given range
of array indexes. The following
are parameters for this function.
st --> Pointer to segment tree
index --> Index of current node in the
segment tree. Initially 0 is
passed as root is always at index 0
ss & se --> Starting and ending indexes
of the segment represented
by current node, i.e., st[index]
qs & qe --> Starting and ending indexes of query range """
def RMQUtil( st, ss, se, qs, qe, index) :
# If segment of this node is a part
# of given range, then return
# the min of the segment
if (qs <= ss and qe >= se) :
return st[index];
# If segment of this node
# is outside the given range
if (se < qs or ss > qe) :
return INT_MAX;
# If a part of this segment
# overlaps with the given range
mid = getMid(ss, se);
return minVal(RMQUtil(st, ss, mid, qs,
qe, 2 * index + 1),
RMQUtil(st, mid + 1, se,
qs, qe, 2 * index + 2));
# Return minimum of elements in range
# from index qs (query start) to
# qe (query end). It mainly uses RMQUtil()
def RMQ( st, n, qs, qe) :
# Check for erroneous input values
if (qs < 0 or qe > n - 1 or qs > qe) :
print("Invalid Input");
return -1;
return RMQUtil(st, 0, n - 1, qs, qe, 0);
# A recursive function that constructs
# Segment Tree for array[ss..se].
# si is index of current node in segment tree st
def constructSTUtil(arr, ss, se, st, si) :
# If there is one element in array,
# store it in current node of
# segment tree and return
if (ss == se) :
st[si] = arr[ss];
return arr[ss];
# If there are more than one elements,
# then recur for left and right subtrees
# and store the minimum of two values in this node
mid = getMid(ss, se);
st[si] = minVal(constructSTUtil(arr, ss, mid,
st, si * 2 + 1),
constructSTUtil(arr, mid + 1, se,
st, si * 2 + 2));
return st[si];
"""Function to construct segment tree
from given array. This function allocates
memory for segment tree and calls constructSTUtil()
to fill the allocated memory """
def constructST( arr, n) :
# Allocate memory for segment tree
# Height of segment tree
x = (int)(ceil(log2(n)));
# Maximum size of segment tree
max_size = 2 * (int)(2**x) - 1;
st = [0] * (max_size);
# Fill the allocated memory st
constructSTUtil(arr, 0, n - 1, st, 0);
# Return the constructed segment tree
return st;
MOD = 10**9+7
mod=10**9+7
#t=1
p=10**9+7
def ncr_util():
inv[0]=inv[1]=1
fact[0]=fact[1]=1
for i in range(2,300001):
inv[i]=(inv[i%p]*(p-p//i))%p
for i in range(1,300001):
inv[i]=(inv[i-1]*inv[i])%p
fact[i]=(fact[i-1]*i)%p
def z_array(s1):
n = len(s1)
z=[0]*(n)
l, r, k = 0, 0, 0
for i in range(1,n):
# if i>R nothing matches so we will calculate.
# Z[i] using naive way.
if i > r:
l, r = i, i
# R-L = 0 in starting, so it will start
# checking from 0'th index. For example,
# for "ababab" and i = 1, the value of R
# remains 0 and Z[i] becomes 0. For string
# "aaaaaa" and i = 1, Z[i] and R become 5
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
else:
# k = i-L so k corresponds to number which
# matches in [L,R] interval.
k = i - l
# if Z[k] is less than remaining interval
# then Z[i] will be equal to Z[k].
# For example, str = "ababab", i = 3, R = 5
# and L = 2
if z[k] < r - i + 1:
z[i] = z[k]
# For example str = "aaaaaa" and i = 2,
# R is 5, L is 0
else:
# else start from R and check manually
l = i
while r < n and s1[r - l] == s1[r]:
r += 1
z[i] = r - l
r -= 1
return z
'''
MAXN1=100001
spf=[0]*MAXN1
def sieve():
prime=[]
spf[1]=1
for i in range(2,MAXN1):
spf[i]=i
for i in range(4,MAXN1,2):
spf[i]=2
for i in range(3,math.ceil(math.sqrt(MAXN1))):
if spf[i]==i:
for j in range(i*i,MAXN1,i):
if spf[j]==j:
spf[j]=i
for i in range(2,MAXN1):
if spf[i]==i:
prime.append(i)
return prime
prime=sieve()
def factor(x):
d1={}
x1=x
while x!=1:
d1[spf[x]]=d1.get(spf[x],0)+1
x//=spf[x]
return d1
def primeFactors(n):
d1={}
while n % 2 == 0:
d1[2]=d1.get(2,0)+1
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
# while i divides n , print i ad divide n
while n % i== 0:
d1[i]=d1.get(i,0)+1
n = n // i
if n > 2:
d1[n]=d1.get(n,0)+1
return d1
'''
def fs(x):
if x<2:
return 0
return (x*(x-1))//2
#t=int(input())
t=1
def solve(a,b,k):
l1=[]
l2=[]
if k==0:
print("YES")
for i in range(b):
l1.append(1)
l2.append(1)
for i in range(a):
l1.append(0)
l2.append(0)
print(*l1,sep="")
print(*l2,sep="")
else:
if b>=2 and k<=a+b-2 and a>0:
print("YES")
for i in range(b):
l1.append(1)
for i in range(a):
l1.append(0)
print(*l1,sep="")
if k<a:
print('1'*(b-1) + '0'*k + '1' + '0'*(a-k))
else:
print('1'*(a+b-k-1)+'0'+'1'*(k-a)+'0'*(a-1)+'1')
else:
print("NO")
for _ in range(t):
#d={}
a,b,k=map(int,input().split())
#n=int(input())
#b=input()
# s2=input()
#l=list(map(int,input().split()))
#l.sort()
#l2=list(map(int,input().split()))
#l.sort()
#l.sort(revrese=True)
#x,y=(map(int,input().split()))
#l=str(n)
#l.sort(reverse=True)
#l2.sort(reverse=True)
#l1.sort(reverse=True)
(solve(a,b,k))
#print(solve())
``` | instruction | 0 | 55,007 | 5 | 110,014 |
Yes | output | 1 | 55,007 | 5 | 110,015 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
# cook your dish here
# cook your code here
# cook your dish here
import os,sys;from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno();self.buffer = BytesIO();self.writable = "x" in file.mode or "r" not in file.mode;self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:break
ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE));self.newlines = b.count(b"\n") + (not b);ptr = self.buffer.tell();self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:os.write(self._fd, self.buffer.getvalue());self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file);self.flush = self.buffer.flush;self.writable = self.buffer.writable;self.write = lambda s: self.buffer.write(s.encode("ascii"));self.read = lambda: self.buffer.read().decode("ascii");self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
try:sys.stdin,sys.stdout=open('in.txt','r'),open('out.txt','w')
except:pass
ii1=lambda:int(sys.stdin.readline().strip()) # for interger
is1=lambda:sys.stdin.readline().strip() # for str
iia=lambda:list(map(int,sys.stdin.readline().strip().split())) # for List[int]
isa=lambda:sys.stdin.readline().strip().split() # for List[str]
mod=int(1e9 + 7);from collections import *;from math import *
# abc = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
# sys.setrecursionlimit(500000)
from collections import defaultdict as dd
# from functools import lru_cache
###################### Start Here ######################
R = range
z,o,k = iia()
s = '1'*o + '0'*z
if k==0:
print('Yes\n'+s+'\n'+s)
elif z==0 or o==1:
if k==0:
print('Yes\n'+s+'\n'+s)
else:
print('No')
else:
if k<=(z+o-2):
print('Yes')
s = '1'*(o-1) + '0'*(z-1)
print(f"{s[:z+o-k-1]}1{s[z+o-k-1:]}0")
print(f"{s[:z+o-k-1]}0{s[z+o-k-1:]}1")
else:
print('No')
``` | instruction | 0 | 55,008 | 5 | 110,016 |
Yes | output | 1 | 55,008 | 5 | 110,017 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
'''Author- Akshit Monga'''
from sys import stdin, stdout
# input = stdin.readline
t = 1
for _ in range(t):
a,b,k=map(int,input().split())
if k>a:
print("NO")
continue
if b==1:
if k!=0:
print("NO")
continue
print("YES")
out=['1']
for i in range(a):
out.append('0')
s=''.join(out)
print(s)
print(s)
continue
if k==0:
print("YES")
val=[]
for i in range(b):
val.append('1')
for i in range(a):
val.append('0')
s=''.join(val)
print(s)
print(s)
continue
val1=[]
val2=[]
for i in range(b):
val1.append('1')
if i==b-1:
val2.append('0')
else:
val2.append('1')
count=0
for i in range(a):
val1.append('0')
count+=1
if count==k:
val2.append('1')
else:
val2.append('0')
s1=''.join(val1)
s2=''.join(val2)
print("YES")
print(s1)
print(s2)
``` | instruction | 0 | 55,009 | 5 | 110,018 |
No | output | 1 | 55,009 | 5 | 110,019 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
import bisect
import collections
import functools
import itertools
import math
import heapq
import random
import string
def repeat(_func=None, *, times=1):
def decorator(func):
@functools.wraps(func)
def wrapper(*args, **kwargs):
for _ in range(times):
func(*args, **kwargs)
return wrapper
if _func is None:
return decorator
else:
return decorator(_func)
def unpack(func=int):
return map(func, input().split())
def l_unpack(func=int):
"""list unpack"""
return list(map(func, input().split()))
def getint():
return int(input())
def getmatrix(rows):
return [list(map(int, input().split())) for _ in range(rows)]
def display_matrix(mat):
for i in range(len(mat)):
print(mat[i])
# @repeat(times=int(input()))
def main():
a, b, k = unpack()
length = a + b
oa, ob, ok = a, b, k
if k == 0:
x = '1' * b + '0' * a
print('Yes')
print(x)
print(x)
return
try:
x = [0] * (a + b)
y = [0] * (a + b)
arrk = [0] * (a + b)
b -= 1
x[0] = y[0] = 1
if k > b:
i = length - 1
while k > 0:
arrk[i] = 1
k -= 1
i -= 1
i = length - 1
while b > 0:
y[i] = 1
b -= 1
i -= 1
else:
i = length - 1
while k > 0:
arrk[i] = 1
k -= 1
i -= 2
i = length - 1
while b > 0:
y[i] = 1
b -= 1
i -= 2
carry = 0
for i in range(len(x) - 1, -1, -1):
carry, x[i] = divmod(x[i] + y[i] + arrk[i] + carry, 2)
x = "".join(map(str, x)).lstrip('0')
y = "".join(map(str, y)).lstrip('0')
one = x.count('1')
if one != ob or length - one != oa:
print('No')
return
one = y.count('1')
if one != ob or length - one != oa:
print('No')
return
if arrk.count(1) != ok:
print('No')
return
if x < y:
print('No')
return
print('Yes', x, y, sep="\n")
except IndexError:
print('No')
return
MOD = 10 ** 9 + 7
main()
``` | instruction | 0 | 55,010 | 5 | 110,020 |
No | output | 1 | 55,010 | 5 | 110,021 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
import sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
import string
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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
self.BUFSIZE = 8192
def read(self):
while True:
b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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 = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def get_int():
return int(input())
def get_ints():
return list(map(int, input().split(' ')))
def get_int_grid(n):
return [get_ints() for _ in range(n)]
def get_str():
return input().strip()
def get_strs():
return input().strip().split(' ')
def yes_no(b):
if b:
return "YES"
else:
return "NO"
def binary_search(good, left, right, delta=1, right_true=False):
"""
Performs binary search
----------
Parameters
----------
:param good: Function used to perform the binary search
:param left: Starting value of left limit
:param right: Starting value of the right limit
:param delta: Margin of error, defaults value of 1 for integer binary search
:param right_true: Boolean, for whether the right limit is the true invariant
:return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
alternatively returns False if no such value found
"""
limits = [left, right]
while limits[1] - limits[0] > delta:
if delta == 1:
mid = sum(limits) // 2
else:
mid = sum(limits) / 2
if good(mid):
limits[int(right_true)] = mid
else:
limits[int(~right_true)] = mid
if good(limits[int(right_true)]):
return limits[int(right_true)]
else:
return False
def prefix_sums(a, drop_zero=False):
p = [0]
for x in a:
p.append(p[-1] + x)
if drop_zero:
return p[1:]
else:
return p
def prefix_mins(a, drop_zero=False):
p = [float('inf')]
for x in a:
p.append(min(p[-1], x))
if drop_zero:
return p[1:]
else:
return p
def solve_a():
p, * x = get_ints()
m = float('inf')
for s in x:
m = min(m, (s - p) % s)
return m
def solve_b():
n = get_int()
p = get_ints()
positions = {}
for i, x in enumerate(p):
positions[x] = i
r = []
m = n
while len(r) < n:
for x in range(n, 0, - 1):
if positions[x] > m:
continue
else:
pos = positions[x]
for i in range(pos, m):
r.append(p[i])
m = pos
return r
def solve_c():
n, m = get_ints()
s = get_str()
t = get_str()
pref_mins = []
suff_mins = []
i = 0
for j in range(m):
while i < n:
if s[i] == t[j]:
pref_mins.append(i)
i += 1
break
i += 1
i = n - 1
for j in range(m - 1, -1, -1):
while i < n:
if s[i] == t[j]:
suff_mins.append(i)
i -= 1
break
i -= 1
M = -float('inf')
for i in range(m - 1):
M = max(M, suff_mins[m - 2 - i] - pref_mins[i])
return M
def solve_d():
a, b, k = get_ints()
if k > a + b - 2:
return ["No"]
x = ['1'] + ['0'] * (b + a - 1)
y = ['1'] + ['0'] * a + ['1'] * (b - 1)
p = 1
for i in range(a + 1, a + b):
if k >= i - p:
x[p] = '1'
k -= i - p
p += 1
else:
x[i - k] = '1'
k = 0
p = i - k
if k:
return ["No"]
return ["Yes", ''.join(x), ''.join(y)]
ans = solve_d()
for part in ans:
print(part)
``` | instruction | 0 | 55,011 | 5 | 110,022 |
No | output | 1 | 55,011 | 5 | 110,023 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given three integers a, b, k.
Find two binary integers x and y (x ≥ y) such that
1. both x and y consist of a zeroes and b ones;
2. x - y (also written in binary form) has exactly k ones.
You are not allowed to use leading zeros for x and y.
Input
The only line contains three integers a, b, and k (0 ≤ a; 1 ≤ b; 0 ≤ k ≤ a + b ≤ 2 ⋅ 10^5) — the number of zeroes, ones, and the number of ones in the result.
Output
If it's possible to find two suitable integers, print "Yes" followed by x and y in base-2.
Otherwise print "No".
If there are multiple possible answers, print any of them.
Examples
Input
4 2 3
Output
Yes
101000
100001
Input
3 2 1
Output
Yes
10100
10010
Input
3 2 5
Output
No
Note
In the first example, x = 101000_2 = 2^5 + 2^3 = 40_{10}, y = 100001_2 = 2^5 + 2^0 = 33_{10}, 40_{10} - 33_{10} = 7_{10} = 2^2 + 2^1 + 2^0 = 111_{2}. Hence x-y has 3 ones in base-2.
In the second example, x = 10100_2 = 2^4 + 2^2 = 20_{10}, y = 10010_2 = 2^4 + 2^1 = 18, x - y = 20 - 18 = 2_{10} = 10_{2}. This is precisely one 1.
In the third example, one may show, that it's impossible to find an answer.
Submitted Solution:
```
import math
import collections
def read_list() -> list:
return [int(i) for i in input().strip().split()]
def read_num() -> int:
return int(input().strip())
a, b, k = read_list()
n = a + b
ans1 = [1] + [0] * (n-1)
ans2 = [1] + [0] * (n-1)
def print_ans(ans):
for i in ans:
print(i, end="")
print()
if k > a + b - 2:
print("No")
elif k == 0:
for i in range(b):
ans1[i] = 1
ans2[i] = 1
print("Yes")
print_ans(ans1)
print_ans(ans2)
else:
if b == 1:
print("No")
else:
ans2[-1] = 1
a -= 1
b -= 2
pos = n - 2
for i in range(k - 1):
if a > 0:
a -= 1
else:
ans1[pos] = 1
ans2[pos] = 1
b -= 1
pos -= 1
ans1[pos] = 1
print("Yes")
print_ans(ans1)
print_ans(ans2)
``` | instruction | 0 | 55,012 | 5 | 110,024 |
No | output | 1 | 55,012 | 5 | 110,025 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
#!/usr/bin/env python3
# from typing import *
import sys
import io
import math
import collections
import decimal
import itertools
import bisect
import heapq
def input():
return sys.stdin.readline()[:-1]
# sys.setrecursionlimit(1000000)
# _INPUT = """4
# + 100
# -
# + 100
# -
# """
# sys.stdin = io.StringIO(_INPUT)
def solve1(N, S):
score = 0
for s in range(1<<N):
hq = []
for i in range(N):
if s & (1<<i):
if S[i][0] == '+':
heapq.heappush(hq, int(S[i][1]))
else:
if hq:
heapq.heappop(hq)
if hq:
score = (score + sum(hq)) % MOD
return score
def solve2(N, S):
M = 0
for s in S:
if s[0] == '+':
M += 1
score = 0
for i in range(N):
if S[i][0] == '+':
x = int(S[i][1])
dp = [0] * M
dp[0] = 1
scale = 1
for j in range(N):
if j == i:
continue
dp1 = [0] * M
if S[j][0] == '+':
y = int(S[j][1])
if y > x or (y == x and j < i):
dp1 = dp
scale *= 2
else:
for sm in range(M):
dp1[sm] += dp[sm]
if sm+1 < M:
dp1[sm+1] += dp[sm]
else:
for sm in range(M):
dp1[sm] += dp[sm]
if 0 <= sm-1:
dp1[sm-1] += dp[sm]
if j < i and sm == 0:
dp1[sm] += dp[sm]
dp = dp1
n = sum(dp) * scale
score = (score + n * x) % MOD
return score
MOD = 998244353
N = int(input())
S = [input().split() for _ in range(N)]
print(solve2(N, S))
``` | instruction | 0 | 55,037 | 5 | 110,074 |
Yes | output | 1 | 55,037 | 5 | 110,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
import sys
def I(): return int(sys.stdin.readline().rstrip())
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def LI2(): return list(map(int,sys.stdin.readline().rstrip()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def LS2(): return list(sys.stdin.readline().rstrip())
n = I()
mod = 998244353
A = [0] # 1-index
add = [] # (element of plus,index)
for i in range(n):
s = S()
if s[0] == '-':
A.append(-1)
else:
a = int(s[2:])
A.append(a)
add.append((a,i+1))
ans = 0
for a,i in add:
count = [[0]*(n+1) for _ in range(n+1)]
count[0][0] = 1
for j in range(n):
b = A[j+1]
for k in range(n+1):
c = count[j][k]
if not c:
continue
count[j+1][k] += c
count[j+1][k] %= mod
if j < i-1:
if b != -1:
if b > a:
count[j+1][k] += c
count[j+1][k] %= mod
else:
count[j+1][k+1] += c
count[j+1][k+1] %= mod
else:
count[j+1][max(0,k-1)] += c
count[j+1][max(0,k-1)] %= mod
elif j > i-1:
if b != -1:
if b >= a:
count[j+1][k] += c
count[j+1][k] %= mod
else:
count[j+1][k+1] += c
count[j+1][k+1] %= mod
else:
if k >= 1:
count[j+1][k-1] += c
count[j+1][k-1] %= mod
c = 0
for k in range(n+1):
c += count[-1][k]
c %= mod
ans += a*c
ans %= mod
print(ans)
``` | instruction | 0 | 55,038 | 5 | 110,076 |
Yes | output | 1 | 55,038 | 5 | 110,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
mod = 998244353
N = 504
a = []
def process():
f=[[0 for i in range(n+2)] for j in range(n+2)]
res = 0
f[0][0] = 1
for i in range(1, n + 1):
for j in range(i + 1):
x = a[i - 1]
f[i][j] = f[i - 1][j]
if i == I:
continue
if x == '-':
f[i][j] = (f[i][j] + f[i - 1][j + 1]) % mod
# print(f[i][j])
if i < I and j == 0:
f[i][j] = (f[i][j] + f[i - 1][0]) % mod
# print(f[i][j], i, j)
else:
if x < X or (x == X and i > I):
f[i][j] = (f[i][j] + f[i - 1][j - 1]) % mod
if x > X or (x == X and i < I):
f[i][j] = (f[i][j] + f[i - 1][j]) % mod
# for i in range(1, n + 1):
# for j in range(n):
# print(f[i][j], i, j)
for i in range(n):
res = (res + X * f[n][i]) % mod
#print(res, X)
return res
if __name__ == '__main__':
n = int(input())
ans = 0
for _ in range(n):
s = input().split()
if len(s) == 1:
a.append('-')
else:
a.append(int(s[1]))
for I in range(1, n + 1):
X = a[I - 1]
if X == '-':
continue
ans = (ans + process()) % mod
print(ans)
``` | instruction | 0 | 55,039 | 5 | 110,078 |
Yes | output | 1 | 55,039 | 5 | 110,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
MOD = 998244353
def solve():
n = int(input())
ans = 0
arr = []
for i in range(n):
a = list(input().split())
if len(a) == 2:
arr.append(int(a[1]))
else:
arr.append(0)
for cur in range(n):
if arr[cur]:
dp = [0] * (n+1)
dp[0] = 1
for j, a in enumerate(arr):
if j == cur:
dp = [0] + dp[:n]
elif a == 0:
dp[0] = (2*dp[0]+dp[1]) % MOD
for i in range(1, n):
dp[i] = (dp[i]+dp[i+1]) % MOD
elif a < arr[cur] or (a == arr[cur] and j < cur):
if j < cur:
for i in range(n, 0, -1):
dp[i] = (dp[i-1]+dp[i]) % MOD
else:
for i in range(n, 1, -1):
dp[i] = (dp[i-1]+dp[i]) % MOD
dp[0] = dp[0] * 2 % MOD
else:
dp = [d*2%MOD for d in dp]
ans = (ans+(sum(dp)-dp[0])*arr[cur]) % MOD
return ans
import sys
input = lambda: sys.stdin.readline().rstrip()
print(solve())
``` | instruction | 0 | 55,040 | 5 | 110,080 |
Yes | output | 1 | 55,040 | 5 | 110,081 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
def main():
mod=998244353
n=int(input())
a=[]
for i in range(n):
t=input()
if t[0]=='-':
a.append(-1)
else:
a.append(int(t.split()[-1]))
ans=0
for i in range(n):
if a[i]==-1:
continue
dp=[0]*n
dp[0]=1
buf=1
for j in range(n):
if j<i:
if a[j]==-1:
dp[0]=dp[0]*2%mod
for k in range(1,j+1):
dp[k-1]=(dp[k-1]+dp[k])%mod
elif a[j]<=a[i]:
for k in range(j-1,-1,-1):
dp[k+1]=(dp[k+1]+dp[k])%mod
else:
buf=buf*2%mod
elif i<j:
if a[j]==-1:
for k in range(1,j+1):
dp[k-1]=(dp[k-1]+dp[k])%mod
elif a[j]<a[i]:
for k in range(j-1,-1,-1):
dp[k+1]=(dp[k+1]+dp[k])%mod
else:
buf=buf*2%mod
ans=(ans+a[i]*sum(dp)*buf)%mod
print(ans)
main()
``` | instruction | 0 | 55,041 | 5 | 110,082 |
No | output | 1 | 55,041 | 5 | 110,083 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
mod = 998244353
n = (int)(input())
a = [0 for i in range(n+1)]
for i in range(1,n+1):
m = input().split()
if m[0] == "+":
a[i] = (int)(m[1])
ans = 0
for t in range(1,n+1):
if(a[t] == 0): continue
f=[[0 for j in range(0,n+2)]for i in range(n+2)]
for i in range(1,n+1):
for j in range(0,i+1):
if a[i] == 0:
if(i <= t or j > 0):
f[i][max(j-1,0)] = (f[i][max(j-1,0)] + f[i-1][j]) % mod
else:
if(a[i] < a[t] or ((a[i] == a[t]) and (i < t))):
f[i][j+1] = (f[i][j+1] + f[i-1][j]) % mod
else:
f[i][j] = (f[i][j] + f[i - 1][j]) % mod
if(i != t):
f[i][j] = (f[i][j] + f[i-1][j]) % mod
for i in range(0,n+1):
ans = (ans + f[n][i]) % mod
print(ans)
``` | instruction | 0 | 55,042 | 5 | 110,084 |
No | output | 1 | 55,042 | 5 | 110,085 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
import sys
sys.setrecursionlimit(10**5)
int1 = lambda x: int(x)-1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.buffer.readline())
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def LI1(): return list(map(int1, sys.stdin.buffer.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
def BI(): return sys.stdin.buffer.readline().rstrip()
def SI(): return sys.stdin.buffer.readline().rstrip().decode()
# dij = [(0, 1), (-1, 0), (0, -1), (1, 0)]
dij = [(0, 1), (-1, 0), (0, -1), (1, 0), (1, 1), (1, -1), (-1, 1), (-1, -1)]
inf = 10**16
md = 998244353
# md = 10**9+7
n = II()
aa = []
for _ in range(n):
ca = SI().split()
if ca[0] == "+":
aa.append(int(ca[1]))
else:
aa.append(-1)
def solve(ti):
dp = [[0]*(n+1) for _ in range(n)]
dp[0][0] = 1
for i in range(n-1):
for j in range(i+1):
pre = dp[i][j]
if pre == 0: continue
if i < ti:
if aa[i] > 0:
if aa[i] <= aa[ti]:
dp[i+1][j+1] = (dp[i+1][j+1]+pre)%md
else:
dp[i+1][j] = (dp[i+1][j]+pre)%md
else:
nj = max(0, j-1)
dp[i+1][nj] = (dp[i+1][nj]+pre)%md
dp[i+1][j] = (dp[i+1][j]+pre)%md
else:
if aa[i+1] > 0:
if aa[i+1] <= aa[ti]:
dp[i+1][j+1] = (dp[i+1][j+1]+pre)%md
else:
dp[i+1][j] = (dp[i+1][j]+pre)%md
elif j:
dp[i+1][j-1] = (dp[i+1][j-1]+pre)%md
dp[i+1][j] = (pre+dp[i+1][j])%md
# print(ti)
# p2D(dp)
res = 0
for a in dp[-1]:
res += a
res %= md
return res
ans = 0
for i in range(n):
if aa[i] < 0: continue
ans += solve(i)*aa[i]%md
ans %= md
print(ans)
``` | instruction | 0 | 55,043 | 5 | 110,086 |
No | output | 1 | 55,043 | 5 | 110,087 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a sequence A, where its elements are either in the form + x or -, where x is an integer.
For such a sequence S where its elements are either in the form + x or -, define f(S) as follows:
* iterate through S's elements from the first one to the last one, and maintain a multiset T as you iterate through it.
* for each element, if it's in the form + x, add x to T; otherwise, erase the smallest element from T (if T is empty, do nothing).
* after iterating through all S's elements, compute the sum of all elements in T. f(S) is defined as the sum.
The sequence b is a subsequence of the sequence a if b can be derived from a by removing zero or more elements without changing the order of the remaining elements. For all A's subsequences B, compute the sum of f(B), modulo 998 244 353.
Input
The first line contains an integer n (1≤ n≤ 500) — the length of A.
Each of the next n lines begins with an operator + or -. If the operator is +, then it's followed by an integer x (1≤ x<998 244 353). The i-th line of those n lines describes the i-th element in A.
Output
Print one integer, which is the answer to the problem, modulo 998 244 353.
Examples
Input
4
-
+ 1
+ 2
-
Output
16
Input
15
+ 2432543
-
+ 4567886
+ 65638788
-
+ 578943
-
-
+ 62356680
-
+ 711111
-
+ 998244352
-
-
Output
750759115
Note
In the first example, the following are all possible pairs of B and f(B):
* B= {}, f(B)=0.
* B= {-}, f(B)=0.
* B= {+ 1, -}, f(B)=0.
* B= {-, + 1, -}, f(B)=0.
* B= {+ 2, -}, f(B)=0.
* B= {-, + 2, -}, f(B)=0.
* B= {-}, f(B)=0.
* B= {-, -}, f(B)=0.
* B= {+ 1, + 2}, f(B)=3.
* B= {+ 1, + 2, -}, f(B)=2.
* B= {-, + 1, + 2}, f(B)=3.
* B= {-, + 1, + 2, -}, f(B)=2.
* B= {-, + 1}, f(B)=1.
* B= {+ 1}, f(B)=1.
* B= {-, + 2}, f(B)=2.
* B= {+ 2}, f(B)=2.
The sum of these values is 16.
Submitted Solution:
```
n = int(input())
A = []
for i in range(n):
s = input()
if s[0] == '+':
A.append((1, int(s[2:])))
else:
A.append((0, 0))
answer = 0
m = 998244353
for k, elem in enumerate(A):
if elem[0] == 0:
continue
D = [1]
for j, el in enumerate(A):
if k == j:
continue
if el[0] == 0:
new_D = [0] * len(D)
if k > j:
new_D[0] = 2 * D[0]
else:
new_D[0] = D[0]
for i, val in enumerate(D):
if i != 0:
new_D[i] += D[i]
new_D[i - 1] += D[i]
else:
new_D = [0] * len(D)
if el[1] > elem[1]:
for i, val in enumerate(D):
new_D[i] = 2 * D[i]
else:
new_D = [0] * (len(D) + 1)
for i, val in enumerate(D):
new_D[i] += D[i]
new_D[i + 1] += D[i]
D = new_D
answer += (elem[1] % m) * (sum(D) % m)
answer %= m
print(answer)
``` | instruction | 0 | 55,044 | 5 | 110,088 |
No | output | 1 | 55,044 | 5 | 110,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258 | instruction | 0 | 55,096 | 5 | 110,192 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
n, t = map(int, input().split())
s = bin(n + 2)[2:]
l = len(s)
if t & (t - 1):
ans = 0
else:
t = t.bit_length()
f = [[0] * (l + 1) for i in range(l + 1)]
for i in range(l + 1):
f[i][0] = f[i][i] = 1
for j in range(1, i):
f[i][j] = f[i - 1][j - 1] + f[i - 1][j]
ans = c = 0
for i in range(l):
if s[i] == '1':
if t - c <= l - i - 1:
ans += f[l - i - 1][t - c]
c += 1
if t == 1: ans -= 1
print(ans)
``` | output | 1 | 55,096 | 5 | 110,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258 | instruction | 0 | 55,097 | 5 | 110,194 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
from math import factorial as fac
n, t = map(int, input().split())
if t & (t - 1):
ans = 0
else:
ans = c = 0
s = bin(n + 2)[2:]
l = len(s)
for i in range(l):
if s[i] == '1':
m, k = l - i - 1, t.bit_length() - c
if 0 <= k <= m:
ans += fac(m) // fac(k) // fac(m - k)
c += 1
if t == 1: ans -= 1
print(ans)
``` | output | 1 | 55,097 | 5 | 110,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258 | instruction | 0 | 55,098 | 5 | 110,196 |
Tags: constructive algorithms, dp, math
Correct Solution:
```
def comb(n, r):
if r > n or r < 0:
return 0
r = min(r, n-r)
result = 1
for i in range(1, r+1):
result = result*(n+1-i)//i
return result
def F(n, t):
if t == 0:
return 1
elif n == 0:
return 0
elif n == 1:
return int(t == 1)
m = len(bin(n)) - 3
return F(n-(1 << m), t-1) + comb(m, t)
n, t = map(int, input().strip().split())
T = len(bin(t)) - 3
if (1 << T) != t:
print(0)
else:
print(F(n+1, T+1) - int(t == 1))
``` | output | 1 | 55,098 | 5 | 110,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258
Submitted Solution:
```
print(-1111111111111111111111)
``` | instruction | 0 | 55,099 | 5 | 110,198 |
No | output | 1 | 55,099 | 5 | 110,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258
Submitted Solution:
```
def comb(n, r):
if r > n or r < 0:
return 0
r = min(r, n-r)
result = 1
for i in range(1, r+1):
result = result*(n+1-i)//i
return result
def F(n, t):
if n == 0:
return 0
elif t == 0:
return 1
elif n == 1:
return int(t == 1)
m = len(bin(n)) - 3
return F(n-(1 << m), t-1) + comb(m, t)
n, t = map(int, input().strip().split())
T = len(bin(t)) - 3
if (1 << T) != t:
print(0)
else:
print(F(n, T+1))
``` | instruction | 0 | 55,100 | 5 | 110,200 |
No | output | 1 | 55,100 | 5 | 110,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258
Submitted Solution:
```
def comb(n, r):
if r > n or r < 0:
return 0
r = min(r, n-r)
result = 1
for i in range(1, r+1):
result = result*(n+1-i)//i
return result
def F(n, t):
if n == 0:
return 0
elif n==1 and t==1:
return 1
m = len(bin(n)) - 3
return F(n-(1 << m), t-1) + comb(m, t)
n, t = map(int, input().strip().split())
T = len(bin(t)) - 3
if (1 << T) != t:
print(0)
else:
print(F(n, T+1))
``` | instruction | 0 | 55,101 | 5 | 110,202 |
No | output | 1 | 55,101 | 5 | 110,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):
<image>
Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.
Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".
Input
A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).
Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the answer to the problem.
Examples
Input
1 1
Output
1
Input
3 2
Output
1
Input
3 3
Output
0
Input
1000000000000 1048576
Output
118606527258
Submitted Solution:
```
n, t = map(int, input().split())
s = bin(n + 2)[2:]
l = len(s)
if t & (t - 1):
ans = 0
else:
t = t.bit_length()
f = [[0] * (l + 1) for i in range(l + 1)]
for i in range(l + 1):
f[i][0] = f[i][i] = 1
for j in range(1, i):
f[i][j] = f[i - 1][j - 1] + f[i - 1][j]
ans = c = 0
for i in range(l):
if s[i] == '1':
if t - c <= l - i - 1:
ans += f[l - i - 1][t - c]
c += 1
print(ans)
``` | instruction | 0 | 55,102 | 5 | 110,204 |
No | output | 1 | 55,102 | 5 | 110,205 |
Provide a correct Python 3 solution for this coding contest problem.
Consider the following arithmetic progression with n terms:
* x, x + d, x + 2d, \ldots, x + (n-1)d
What is the product of all terms in this sequence? Compute the answer modulo 1\ 000\ 003.
You are given Q queries of this form. In the i-th query, compute the answer in case x = x_i, d = d_i, n = n_i.
Constraints
* 1 \leq Q \leq 10^5
* 0 \leq x_i, d_i \leq 1\ 000\ 002
* 1 \leq n_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
Q
x_1 d_1 n_1
:
x_Q d_Q n_Q
Output
Print Q lines.
In the i-th line, print the answer for the i-th query.
Example
Input
2
7 2 4
12345 67890 2019
Output
9009
916936 | instruction | 0 | 55,422 | 5 | 110,844 |
"Correct Solution:
```
Q = int(input())
mod = 10**6+3
arr = []
for i in range(Q):
x, d, n = map(int, input().split())
arr.append([x, d, n])
P = [1]*(mod+1)
for i in range(mod):
P[i+1] = P[i]*(i+1)%mod
def rev(a):
return pow(a,mod-2,mod)
def f(x, d, n):
if d == 0:
return pow(x, n, mod)
else:
d_r = rev(d)
if d_r * x % mod == 0:
return 0
elif (d_r*x%mod+n-1) >= mod:
return 0
else:
return (P[d_r*x%mod+n-1] * rev(P[d_r*x%mod-1])%mod)*pow(d,n,mod)%mod
for i in range(Q):
print(f(arr[i][0], arr[i][1], arr[i][2]))
``` | output | 1 | 55,422 | 5 | 110,845 |
Provide a correct Python 3 solution for this coding contest problem.
Consider the following arithmetic progression with n terms:
* x, x + d, x + 2d, \ldots, x + (n-1)d
What is the product of all terms in this sequence? Compute the answer modulo 1\ 000\ 003.
You are given Q queries of this form. In the i-th query, compute the answer in case x = x_i, d = d_i, n = n_i.
Constraints
* 1 \leq Q \leq 10^5
* 0 \leq x_i, d_i \leq 1\ 000\ 002
* 1 \leq n_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
Q
x_1 d_1 n_1
:
x_Q d_Q n_Q
Output
Print Q lines.
In the i-th line, print the answer for the i-th query.
Example
Input
2
7 2 4
12345 67890 2019
Output
9009
916936 | instruction | 0 | 55,423 | 5 | 110,846 |
"Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)
mod = 10 ** 6 + 3
U = 10 ** 6 + 2
fact = [1] * (U + 10)
for i in range(1, U + 10):
fact[i] = fact[i - 1] * i % mod
def GCD(x, y):
if y == 0:
return x
return GCD(y, x % y)
def msol(x, d, n):
if x == 0:
return 0
if d == 0:
return pow(x, n, mod)
if d == 1:
if x + n - 1 >= mod:
return 0
res = fact[x+n-1]*pow(fact[x-1], mod-2, mod) % mod
return res
inv = pow(d, mod-2, mod)
y = x * inv % mod
C = pow(d, n, mod)
if y + n - 1 >= mod or y == 0:
return 0
res = fact[y+n-1]*pow(fact[y-1], mod-2, mod) % mod
res = res * C % mod
return res
Q = int(input())
for _ in range(Q):
x, d, n = map(int, input().split())
print(msol(x, d, n))
``` | output | 1 | 55,423 | 5 | 110,847 |
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