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.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1 | instruction | 0 | 40,414 | 5 | 80,828 |
"Correct Solution:
```
import sys
readline = sys.stdin.readline
write = sys.stdout.write
N, Q = map(int, input().split())
INF = 2**31-1
LV = (N-1).bit_length()
N0 = 2**LV
data = [0]*(2*N0)
lazy = [0]*(2*N0)
def gindex(l, r):
L = l + N0; R = r + N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
def propagates(*ids):
for i in reversed(ids):
v = lazy[i-1]
if not v:
continue
lazy[2*i-1] += v; lazy[2*i] += v
data[2*i-1] += v; data[2*i] += v
lazy[i-1] = 0
def update(l, r, x):
L = N0 + l; R = N0 + r
while L < R:
if R & 1:
R -= 1
lazy[R-1] += x; data[R-1] += x
if L & 1:
lazy[L-1] += x; data[L-1] += x
L += 1
L >>= 1; R >>= 1
for i in gindex(l, r):
data[i-1] = min(data[2*i-1], data[2*i]) + lazy[i-1]
def query(l, r):
propagates(*gindex(l, r))
L = N0 + l; R = N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, data[R-1])
if L & 1:
s = min(s, data[L-1])
L += 1
L >>= 1; R >>= 1
return s
ans = []
for q in range(Q):
t, *cmd = map(int, readline().split())
if t:
s, t = cmd
ans.append(str(query(s, t+1)))
else:
s, t, x = cmd
update(s, t+1, x)
write("\n".join(ans))
write("\n")
``` | output | 1 | 40,414 | 5 | 80,829 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1 | instruction | 0 | 40,415 | 5 | 80,830 |
"Correct Solution:
```
import sys
input = sys.stdin.readline
INF = 2**31 - 1
class StarrySkyTree:
def __init__(self, n):
LV = (n-1).bit_length()
self.N0 = 2 ** LV
self.data = [0] * (2*self.N0)
self.lazy = [0] * (2*self.N0)
def gindex(self, l, r):
L = l + self.N0; R = r + self.N0
lm = (L // (L & -L)) >> 1
rm = (R // (R & -R)) >> 1
while L < R:
if R <= rm:
yield R
if L <= lm:
yield L
L >>= 1; R >>= 1
while L:
yield L
L >>= 1
def propagates(self, *ids):
for i in reversed(ids):
v = self.lazy[i-1]
if not v:
continue
self.lazy[2*i-1] += v; self.lazy[2*i] += v
self.data[2*i-1] += v; self.data[2*i] += v
self.lazy[i-1] = 0
def update(self, l, r, x):
L = self.N0 + l; R = self.N0 + r
while L < R:
if R & 1:
R -= 1
self.lazy[R-1] += x; self.data[R-1] += x
if L & 1:
self.lazy[L-1] += x; self.data[L-1] += x
L += 1
L >>= 1; R >>= 1
for i in self.gindex(l, r):
self.data[i-1] = min(self.data[2*i-1], self.data[2*i]) + self.lazy[i-1]
def query(self, l, r):
self.propagates(*self.gindex(l, r))
L = self.N0 + l; R = self.N0 + r
s = INF
while L < R:
if R & 1:
R -= 1
s = min(s, self.data[R-1])
if L & 1:
s = min(s, self.data[L-1])
L += 1
L >>= 1; R >>= 1
return s
n, q = map(int, input().split())
sst = StarrySkyTree(n)
for _ in range(q):
que = list(map(int, input().split()))
if que[0] == 0:
s, t, x = que[1], que[2], que[3]
sst.update(s, t+1, x)
elif que[0] == 1:
s, t = que[1], que[2]
print(sst.query(s, t+1))
``` | output | 1 | 40,415 | 5 | 80,831 |
Provide a correct Python 3 solution for this coding contest problem.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1 | instruction | 0 | 40,416 | 5 | 80,832 |
"Correct Solution:
```
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians, log2
from itertools import accumulate, permutations, combinations, product
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
mod = 10 ** 9 + 7
INF = float('inf')
class Lazysegtree:
#RAQ
def __init__(self, A, intv, initialize=True, segf=min):
#区間は 1-indexed で管理
self.N = len(A)
self.N0 = 2**(self.N-1).bit_length()
self.intv = intv
self.segf = segf
self.lazy = [0]*(2*self.N0)
if initialize:
self.data = [intv]*self.N0 + A + [intv]*(self.N0 - self.N)
for i in range(self.N0-1, 0, -1):
self.data[i] = self.segf(self.data[2*i], self.data[2*i+1])
else:
self.data = [intv]*(2*self.N0)
def _ascend(self, k):
k = k >> 1
c = k.bit_length()
for j in range(c):
idx = k >> j
self.data[idx] = self.segf(self.data[2*idx], self.data[2*idx+1]) \
+ self.lazy[idx]
def _descend(self, k):
k = k >> 1
idx = 1
c = k.bit_length()
for j in range(1, c+1):
idx = k >> (c - j)
ax = self.lazy[idx]
if not ax:
continue
self.lazy[idx] = 0
self.data[2*idx] += ax
self.data[2*idx+1] += ax
self.lazy[2*idx] += ax
self.lazy[2*idx+1] += ax
def query(self, l, r):
L = l+self.N0
R = r+self.N0
Li = L//(L & -L)
Ri = R//(R & -R)
self._descend(Li)
self._descend(Ri - 1)
s = self.intv
while L < R:
if R & 1:
R -= 1
s = self.segf(s, self.data[R])
if L & 1:
s = self.segf(s, self.data[L])
L += 1
L >>= 1
R >>= 1
return s
def update(self, l, r, x):
L = l+self.N0
R = r+self.N0
Li = L//(L & -L)
Ri = R//(R & -R)
while L < R :
if R & 1:
R -= 1
self.data[R] += x
self.lazy[R] += x
if L & 1:
self.data[L] += x
self.lazy[L] += x
L += 1
L >>= 1
R >>= 1
self._ascend(Li)
self._ascend(Ri-1)
n, q = MAP()
lazyseg = Lazysegtree([0]*n, INF)
for _ in range(q):
query = input().split()
if query[0] == "0": # add
_, s, t, x = query
s, t, x = map(int, [s, t, x])
lazyseg.update(s, t+1, x)
else: # find
_, s, t = query
s, t = map(int, [s, t])
print(lazyseg.query(s, t+1))
``` | output | 1 | 40,416 | 5 | 80,833 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
import time,random
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
class RangeAddMin():
def __init__(self, n):
i = 1
while 2**i <= n:
i += 1
self.N = 2**i
self.A = [0] * (self.N*2)
self.B = [0] * (self.N*2)
def add(self, a, b, x, k, l, r):
def ina(k, l, r):
if b <= l or r <= a:
pass
elif a <= l and r <= b:
self.A[k] += x
else:
m = (l+r) // 2
ina(k*2+1, l, m)
ina(k*2+2, m, r)
self.B[k] = min(self.B[k*2+1] + self.A[k*2+1], self.B[k*2+2] + self.A[k*2+2])
ina(k, l, r)
def query(self, a, b, k, l, r):
def inq(k, l, r):
if b <= l or r <= a:
return inf
if a <= l and r <= b:
return self.A[k] + self.B[k]
m = (l+r) // 2
rl = inq(k*2+1, l, m)
rr = inq(k*2+2, m, r)
return min(rl, rr) + self.A[k]
return inq(k, l, r)
def main():
n,q = LI()
qa = [LI() for _ in range(q)]
ram = RangeAddMin(n)
rr = []
for qi in qa:
s = qi[1]
t = qi[2] + 1
if qi[0] == 0:
x = qi[3]
ram.add(s, t, x, 0, 0, n)
else:
rr.append(ram.query(s, t, 0, 0, n))
# print(qi,ras.A,ras.B)
return JA(rr, "\n")
# start = time.time()
print(main())
# pe(time.time() - start)
``` | instruction | 0 | 40,417 | 5 | 80,834 |
Yes | output | 1 | 40,417 | 5 | 80,835 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
from array import array
class SegmentTree(object):
MAXV = 1000 * 10**5 + 1
def __init__(self, n: int) -> None:
size = 1
while (size < n):
size *= 2
self.size = 2 * size - 1
self.data = array('i', [0] * self.size)
self.lazy = array('i', [0] * self.size)
def add(self, l: int, h: int, v: int) -> None:
def _add(r, i: int, j: int, lz: int) -> int:
left, right = r * 2 + 1, r * 2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if (l <= i and j <= h): # noqa: E741
lz += v
if lz:
data[r] += lz
if (i < j):
lazy[left] += lz
lazy[right] += lz
else:
mid = (i + j) // 2
if (mid >= l):
lv = _add(left, i, mid, lz)
else:
lazy[left] += lz
lv = data[left] + lazy[left]
if (mid < h):
rv = _add(right, mid + 1, j, lz)
else:
lazy[right] += lz
rv = data[right] + lazy[right]
if (lv < rv):
data[r] = lv
else:
data[r] = rv
return data[r]
lazy = self.lazy
data = self.data
_add(0, 0, self.size // 2, 0)
def min(self, l: int, h: int) -> int:
def _min(r, i: int, j: int, lz: int) -> int:
left, right = r * 2 + 1, r * 2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if lz:
data[r] += lz
if (l <= i and j <= h): # noqa: E741
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return data[r]
else:
mid = (i + j) // 2
if (mid >= l):
lv = _min(left, i, mid, lz)
else:
lazy[left] += lz
lv = self.MAXV
if (mid < h):
rv = _min(right, mid + 1, j, lz)
else:
lazy[right] += lz
rv = self.MAXV
if (lv < rv):
return lv
else:
return rv
lazy = self.lazy
data = self.data
return _min(0, 0, self.size // 2, 0)
if __name__ == "__main__":
n, q = map(lambda x: int(x), input().split())
segtree = SegmentTree(n)
ans = []
for _ in range(q):
com, *v = map(lambda x: int(x), input().split())
if (0 == com):
segtree.add(v[0], v[1], v[2])
else:
ans.append(segtree.min(v[0], v[1]))
print("\n".join(map(str, ans)))
``` | instruction | 0 | 40,418 | 5 | 80,836 |
Yes | output | 1 | 40,418 | 5 | 80,837 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
import sys,queue,math,copy,itertools,bisect
sys.setrecursionlimit(10**7)
INF = 10**10
LI = lambda : [int(x) for x in sys.stdin.readline().split()]
_LI = lambda : [int(x)-1 for x in sys.stdin.readline().split()]
NI = lambda : int(sys.stdin.readline())
N,Q = LI()
N0 = 2 ** (N.bit_length())
node = [0 for _ in range(N0*2)]
lazy = [0 for _ in range(N0*2)]
def gindex(l,r):
L = l + N0; R = r + N0
lm = (L // (L & -L)) // 2
rm = (R // (R & -R)) // 2
while L < R:
if R <= rm:
yield R-1
if L <= lm:
yield L-1
L //= 2; R //= 2
while L > 0:
yield L-1
L //= 2
def eval(*ids):
for i in reversed(ids):
v = lazy[i]
if v:
lazy[i*2+1] += v
lazy[i*2+2] += v
node[i*2+1] += v
node[i*2+2] += v
lazy[i] = 0
def data_add(l,r,v):
*ids, = gindex(l,r)
eval(*ids)
L = l + N0; R = r + N0
while L < R:
if R % 2:
R -= 1
lazy[R-1] += v
node[R-1] += v
if L % 2:
lazy[L-1] += v
node[L-1] += v
L += 1
L //= 2; R //= 2
for i in ids:
node[i] = min(node[i*2+1],node[i*2+2])
def data_get(l,r):
eval(*gindex(l,r))
ret = INF
L = l + N0; R = r + N0
while L < R:
if R % 2:
R -= 1
ret = min(ret,node[R-1])
if L % 2:
ret = min(ret,node[L-1])
L += 1
L //= 2; R //= 2
return ret
for _ in range(Q):
s = LI()
if s[0] == 0:
data_add(s[1],s[2]+1,s[3])
else:
print (data_get(s[1],s[2]+1))
``` | instruction | 0 | 40,419 | 5 | 80,838 |
Yes | output | 1 | 40,419 | 5 | 80,839 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
#!/usr/bin/env python3
import sys
sys.setrecursionlimit(10**6)
input = sys.stdin.buffer.readline
INF = 10 ** 9 + 1 # sys.maxsize # float("inf")
MOD = 10 ** 9 + 7
def debug(*x):
print(*x, file=sys.stderr)
def set_depth(depth):
global DEPTH, SEGTREE_SIZE, NONLEAF_SIZE
DEPTH = depth
SEGTREE_SIZE = 1 << DEPTH
NONLEAF_SIZE = 1 << (DEPTH - 1)
def set_width(width):
set_depth((width - 1).bit_length() + 1)
def get_size(pos):
ret = pos.bit_length()
return (1 << (DEPTH - ret))
def up(pos):
pos += SEGTREE_SIZE // 2
return pos // (pos & -pos)
def up_propagate(table, pos, binop):
while pos > 1:
pos >>= 1
table[pos] = binop(
table[pos * 2],
table[pos * 2 + 1]
)
def force_down_propagate(
action_table, value_table, pos,
action_composite, action_force, action_unity
):
max_level = pos.bit_length() - 1
size = NONLEAF_SIZE
for level in range(max_level):
size //= 2
i = pos >> (max_level - level)
action = action_table[i]
if action != action_unity:
# action_table[i * 2] = action_composite(
# action, action_table[i * 2])
# action_table[i * 2 + 1] = action_composite(
# action, action_table[i * 2 + 1])
action_table[i * 2] += action
action_table[i * 2 + 1] += action
action_table[i] = action_unity
# value_table[i * 2] = action_force(
# action, value_table[i * 2], size)
# value_table[i * 2 + 1] = action_force(
# action, value_table[i * 2 + 1], size)
value_table[i * 2] += action
value_table[i * 2 + 1] += action
def force_range_update(
value_table, action_table, left, right,
action, action_force, action_composite, action_unity
):
"""
action_force: action, value, cell_size => new_value
action_composite: new_action, old_action => composite_action
"""
left += NONLEAF_SIZE
right += NONLEAF_SIZE
while left < right:
if left & 1:
value_table[left] = action_force(
action, value_table[left], get_size(left))
action_table[left] = action_composite(action, action_table[left])
left += 1
if right & 1:
right -= 1
value_table[right] = action_force(
action, value_table[right], get_size(right))
action_table[right] = action_composite(action, action_table[right])
left //= 2
right //= 2
def range_reduce(table, left, right, binop, unity):
ret_left = unity
ret_right = unity
left += NONLEAF_SIZE
right += NONLEAF_SIZE
while left < right:
if left & 1:
ret_left = binop(ret_left, table[left])
left += 1
if right & 1:
right -= 1
ret_right = binop(table[right], ret_right)
left //= 2
right //= 2
return binop(ret_left, ret_right)
def lazy_range_update(
action_table, value_table, start, end,
action, action_composite, action_force, action_unity, value_binop):
"update [start, end)"
L = up(start)
R = up(end)
force_down_propagate(
action_table, value_table, L,
action_composite, action_force, action_unity)
force_down_propagate(
action_table, value_table, R,
action_composite, action_force, action_unity)
# print("action", file=sys.stderr)
# debugprint(action_table)
# print("value", file=sys.stderr)
# debugprint(value_table)
# print(file=sys.stderr)
force_range_update(
value_table, action_table, start, end,
action, action_force, action_composite, action_unity)
up_propagate(value_table, L, value_binop)
up_propagate(value_table, R, value_binop)
def lazy_range_reduce(
action_table, value_table, start, end,
action_composite, action_force, action_unity,
value_binop, value_unity
):
"reduce [start, end)"
force_down_propagate(
action_table, value_table, up(start),
action_composite, action_force, action_unity)
force_down_propagate(
action_table, value_table, up(end),
action_composite, action_force, action_unity)
return range_reduce(value_table, start, end, value_binop, value_unity)
def debugprint(xs, minsize=0, maxsize=None):
global DEPTH
strs = [str(x) for x in xs]
if maxsize != None:
for i in range(NONLEAF_SIZE, SEGTREE_SIZE):
strs[i] = strs[i][:maxsize]
s = max(len(s) for s in strs[NONLEAF_SIZE:])
if s > minsize:
minsize = s
result = ["|"] * DEPTH
level = 0
next_level = 2
for i in range(1, SEGTREE_SIZE):
if i == next_level:
level += 1
next_level *= 2
width = ((minsize + 1) << (DEPTH - 1 - level)) - 1
result[level] += strs[i].center(width) + "|"
print(*result, sep="\n", file=sys.stderr)
def full_up(table, binop):
for i in range(NONLEAF_SIZE - 1, 0, -1):
table[i] = binop(
table[2 * i],
table[2 * i + 1])
def set_items(table, xs):
for i, x in enumerate(xs, NONLEAF_SIZE):
table[i] = x
def main():
# parse input
N, Q = map(int, input().split())
set_width(N)
value_unity = 10 ** 9
value_table = [0] * SEGTREE_SIZE
set_items(value_table, [0] * N)
full_up(value_table, min)
value_binop = min
action_unity = 0
action_table = [action_unity] * SEGTREE_SIZE
def force(action, value, size):
return action + value
def composite(new_action, old_action):
return new_action + old_action
for time in range(Q):
q, *args = map(int, input().split())
if q == 0:
# add
s, t, value = args
lazy_range_update(
action_table, value_table, s, t + 1,
value, composite, force, action_unity, value_binop)
else:
# getSum
s, t = args
print(lazy_range_reduce(
action_table, value_table, s, t + 1,
composite, force, action_unity, value_binop, value_unity))
# debugprint(action_table)
# debugprint(value_table)
# tests
T1 = """
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
"""
TEST_T1 = """
>>> as_input(T1)
>>> main()
-2
0
1
-1
"""
def _test():
import doctest
doctest.testmod()
g = globals()
for k in sorted(g):
if k.startswith("TEST_"):
doctest.run_docstring_examples(g[k], g)
def as_input(s):
"use in test, use given string as input file"
import io
global read, input
f = io.StringIO(s.strip())
def input():
return bytes(f.readline(), "ascii")
def read():
return bytes(f.read(), "ascii")
input = sys.stdin.buffer.readline
read = sys.stdin.buffer.read
if sys.argv[-1] == "-t":
print("testing")
_test()
sys.exit()
main()
``` | instruction | 0 | 40,420 | 5 | 80,840 |
Yes | output | 1 | 40,420 | 5 | 80,841 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
#!/usr/bin/env python3
# DSL_2_H: RMQ and RAQ
# Range Minimum Query and Range Add Query
# Lazy propagate segment tree
import sys
class SegmentTree:
MAXV = 1000 * 10**5 + 1
def __init__(self, n):
size = 1
while size < n:
size *= 2
self.size = 2*size - 1
self.data = [0] * self.size
self.lazy = [0] * self.size
def add(self, lo, hi, v):
def _add(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if j < lo or i > hi:
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
elif lo <= i and j <= hi:
lz += v
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
else:
mid = (i + j) // 2
lv = _add(left, i, mid, lz)
rv = _add(right, mid+1, j, lz)
if lv < rv:
data[r] = lv
else:
data[r] = rv
return data[r]
lazy = self.lazy
data = self.data
_add(0, 0, self.size//2, 0)
def min(self, lo, hi):
def _min(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if lz:
data[r] += lz
if j < lo or i > hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return self.MAXV
elif lo <= i and j <= hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return data[r]
else:
mid = (i + j) // 2
lv = _min(left, i, mid, lz)
rv = _min(right, mid+1, j, lz)
if lv < rv:
return lv
else:
return rv
lazy = self.lazy
data = self.data
return _min(0, 0, self.size//2, 0)
def run():
n, q = [int(i) for i in input().split()]
tree = SegmentTree(n)
ret = []
for line in sys.stdin:
com, *args = line.split()
if com == '0':
s, t, x = map(int, args)
tree.add(s, t, x)
elif com == '1':
s, t = map(int, args)
ret.append(tree.min(s, t))
sys.stdout.write("\n".join([str(i) for i in ret]))
sys.stdout.write("\n")
if __name__ == '__main__':
run()
``` | instruction | 0 | 40,421 | 5 | 80,842 |
No | output | 1 | 40,421 | 5 | 80,843 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
#!/usr/bin/env python3
# DSL_2_H: RMQ and RAQ
# Range Minimum Query and Range Add Query
# Lazy propagate segment tree
from array import array
import sys
class SegmentTree:
MAXV = 1000 * 10**5 + 1
def __init__(self, n):
size = 1
while size < n:
size *= 2
self.size = 2*size - 1
# self.data = [0] * self.size
# self.lazy = [0] * self.size
self.data = array('i', [0] * self.size)
self.lazy = array('i', [0] * self.size)
def add(self, lo, hi, v):
def _add(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if j < lo or i > hi:
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
elif lo <= i and j <= hi:
lz += v
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
else:
mid = (i + j) // 2
lv = _add(left, i, mid, lz)
rv = _add(right, mid+1, j, lz)
if lv < rv:
data[r] = lv
else:
data[r] = rv
return data[r]
lazy = self.lazy
data = self.data
_add(0, 0, self.size//2, 0)
def min(self, lo, hi):
def _min(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if lz:
data[r] += lz
if j < lo or i > hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return self.MAXV
elif lo <= i and j <= hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return data[r]
else:
mid = (i + j) // 2
lv = _min(left, i, mid, lz)
rv = _min(right, mid+1, j, lz)
if lv < rv:
return lv
else:
return rv
lazy = self.lazy
data = self.data
return _min(0, 0, self.size//2, 0)
def run():
n, q = [int(i) for i in input().split()]
tree = SegmentTree(n)
ret = []
for line in sys.stdin:
com, *args = line.split()
if com == '0':
s, t, x = map(int, args)
tree.add(s, t, x)
elif com == '1':
s, t = map(int, args)
ret.append(tree.min(s, t))
sys.stdout.write("\n".join([str(i) for i in ret]))
sys.stdout.write("\n")
if __name__ == '__main__':
run()
``` | instruction | 0 | 40,422 | 5 | 80,844 |
No | output | 1 | 40,422 | 5 | 80,845 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Write a program which manipulates a sequence $A$ = {$a_0, a_1, ..., a_{n-1}$} with the following operations:
* $add(s, t, x)$ : add $x$ to $a_s, a_{s+1}, ..., a_t$.
* $find(s, t)$ : report the minimum value in $a_s, a_{s+1}, ..., a_t$.
Note that the initial values of $a_i ( i = 0, 1, ..., n-1 )$ are 0.
Constraints
* $1 ≤ n ≤ 100000$
* $1 ≤ q ≤ 100000$
* $0 ≤ s ≤ t < n$
* $-1000 ≤ x ≤ 1000$
Input
$n$ $q$
$query_1$
$query_2$
:
$query_q$
In the first line, $n$ (the number of elements in $A$) and $q$ (the number of queries) are given. Then, $i$th query $query_i$ is given in the following format:
0 $s$ $t$ $x$
or
1 $s$ $t$
The first digit represents the type of the query. '0' denotes $add(s, t, x)$ and '1' denotes $find(s, t)$.
Output
For each $find$ query, print the minimum value.
Example
Input
6 7
0 1 3 1
0 2 4 -2
1 0 5
1 0 1
0 3 5 3
1 3 4
1 0 5
Output
-2
0
1
-1
Submitted Solution:
```
#!/usr/bin/env python3
# DSL_2_H: RMQ and RAQ
# Range Minimum Query and Range Add Query
# Lazy propagate segment tree
import sys
class SegmentTree:
MAXV = 1000 * 10**5 + 1
def __init__(self, n):
size = 1
while size < n:
size *= 2
self.size = 2*size - 1
self.data = [0] * self.size
self.lazy = [0] * self.size
def add(self, lo, hi, v):
def _add(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if j < lo or i > hi:
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
elif lo <= i and j <= hi:
lz += v
if lz:
data[r] += lz
if i < j:
lazy[left] += lz
lazy[right] += lz
else:
mid = (i + j) // 2
lv = _add(left, i, mid, lz)
rv = _add(right, mid+1, j, lz)
if lv < rv:
data[r] = lv
else:
data[r] = rv
return data[r]
lazy = self.lazy
data = self.data
_add(0, 0, self.size//2, 0)
def min(self, lo, hi):
def _min(r, i, j, lz):
left, right = r*2 + 1, r*2 + 2
if lazy[r]:
lz += lazy[r]
lazy[r] = 0
if lz:
data[r] += lz
if j < lo or i > hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return self.MAXV
elif lo <= i and j <= hi:
if lz and i < j:
lazy[left] += lz
lazy[right] += lz
return data[r]
else:
mid = (i + j) // 2
lv = _min(left, i, mid, lz)
rv = _min(right, mid+1, j, lz)
if lv < rv:
return lv
else:
return rv
lazy = self.lazy
data = self.data
return _min(0, 0, self.size//2, 0)
def run():
n, q = [int(i) for i in input().split()]
tree = SegmentTree(n)
ret = []
for line in sys.stdin:
com, *args = line.split()
if com == '0':
s, t, x = [int(i) for i in args]
tree.add(s, t, x)
elif com == '1':
s, t = [int(i) for i in args]
ret.append(tree.min(s, t))
sys.stdout.write("\n".join([str(i) for i in ret]))
sys.stdout.write("\n")
if __name__ == '__main__':
run()
``` | instruction | 0 | 40,423 | 5 | 80,846 |
No | output | 1 | 40,423 | 5 | 80,847 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
#import sys
# import math
# import bisect
# import collections
# import itertools
# #from sys import stdin,stdout
# from math import gcd,floor,sqrt,log
# from collections import defaultdict as dd, Counter as ctr
# from bisect import bisect_left as bl, bisect_right as br
# from itertools import permutations as pr, combinations as cb
#sys.setrecursionlimit(100000000)
#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$#$
inp = lambda: int(input())
# strng = lambda: input().strip()
# jn = lambda x,l: x.join(map(str,l))
seq = lambda: list(map(int,input().strip().split()))
def results(rr):
a, b = rr[0], rr[1]
j, x =0, 0
while a or b:
if ((a & 1) and (a & 1)):
x += (1 << j)
a >>= 1
b >>= 1
j += 1
return(x)
def main():
t = inp()
for _ in range(t):
cd = seq()
c, d = cd[0], cd[1]
x = results(cd)
print((c ^ x) + (d ^ x))
if __name__ == '__main__':
main()
``` | instruction | 0 | 40,631 | 5 | 81,262 |
Yes | output | 1 | 40,631 | 5 | 81,263 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
T = int(input())
ans_ls = [0] * T
for t in range(T):
a,b = map(int,input().split())
base = 0
ans = 0
while a > 0 or b > 0:
a,r1 = divmod(a,2)
b,r2 = divmod(b,2)
if r1 != r2:
ans += 2**base
base += 1
ans_ls[t] = ans
for ans in ans_ls:
print(ans)
``` | instruction | 0 | 40,632 | 5 | 81,264 |
Yes | output | 1 | 40,632 | 5 | 81,265 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
def binary(n):
return bin(n).replace('0b', "")
def decimal(x):
value = 0
for i in range(len(x)):
temp = x.pop()
if temp == '1':
value += pow(2, i)
return value
t = int(input())
for i in range(t):
a, b = map(int, input().split())
a = list(binary(a))
b = list(binary(b))
for i in range(1, 1+min(len(a), len(b))):
if a[-i] == b[-i] and a[-i] == '1':
a[-i] = '0'
b[-i] = '0'
print(decimal(a) + decimal(b))
``` | instruction | 0 | 40,633 | 5 | 81,266 |
Yes | output | 1 | 40,633 | 5 | 81,267 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
import math
t = int(input())
for _ in range(t):
a,b = list(map(int,input().split(" ")))
m = min(a,b)
v = 0
while m > 0:
l = math.log2(m)
s = 1<< int(l)
v = v + s
m = m - s
print((a^v) + (b^v))
``` | instruction | 0 | 40,634 | 5 | 81,268 |
Yes | output | 1 | 40,634 | 5 | 81,269 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
import os
import 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)
input = lambda: sys.stdin.readline().rstrip("\r\n")
from collections import defaultdict
from math import ceil,floor
for _ in range(int(input())):
m,n=map(int,input().split())
a=str(bin(m))
b=str(bin(n))
a=a[2:]
b=b[2:]
for i in range(40-len(a)):
a='0'+a
for i in range(40-len(b)):
b='0'+b
ans=""
for i in range(40):
if a[i]==b[i]:
ans+=a[i]
else:
ans+='1'
c=ans
ai=0
bi=0
ci=0
for i in range(40):
ai+=int(a[i])*(2**i)
bi+=int(b[i])*(2**i)
ci+=int(c[i])*(2**i)
a_=m^ci+n^ci
print(a_)
``` | instruction | 0 | 40,635 | 5 | 81,270 |
No | output | 1 | 40,635 | 5 | 81,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
reps = input()
def xor(a,b):
return "1" if a != b else "0"
for i in range(int(reps)):
b1, b2 = input().split(" ")
b1 = list(bin(int(b1)).replace("0b",""))
if len(b1) < 8:
dif = 8 - len(b1)
b1 = list(str(0)*dif) + b1
b2 = list(bin(int(b2)).replace("0b",""))
if len(b2) < 8:
dif2 = 8 - len(b2)
b2 = list(str(0)*dif2) + b2
binRep = "".join(map(xor,b1,b2))
print(int(binRep,2))
``` | instruction | 0 | 40,636 | 5 | 81,272 |
No | output | 1 | 40,636 | 5 | 81,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
def findX(A,B):
j = 0
x = 0
while (A or B):
if ((A & 1) and (B & 1)):
x += (1 << j)
A >>= 1
B >>= 1
j += 1
return x
if __name__ == '__main__':
A = 2
B = 3
X = findX(A, B)
print("X =",X,", Sum =",(A ^ X) + (B ^ X))
``` | instruction | 0 | 40,637 | 5 | 81,274 |
No | output | 1 | 40,637 | 5 | 81,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In order to celebrate Twice's 5th anniversary, Tzuyu and Sana decided to play a game.
Tzuyu gave Sana two integers a and b and a really important quest.
In order to complete the quest, Sana has to output the smallest possible value of (a ⊕ x) + (b ⊕ x) for any given x, where ⊕ denotes the [bitwise XOR operation](http://en.wikipedia.org/wiki/Bitwise_operation#XOR).
Input
Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^{4}). Description of the test cases follows.
The only line of each test case contains two integers a and b (1 ≤ a, b ≤ 10^{9}).
Output
For each testcase, output the smallest possible value of the given expression.
Example
Input
6
6 12
4 9
59 832
28 14
4925 2912
1 1
Output
10
13
891
18
6237
0
Note
For the first test case Sana can choose x=4 and the value will be (6 ⊕ 4) + (12 ⊕ 4) = 2 + 8 = 10. It can be shown that this is the smallest possible value.
Submitted Solution:
```
def s(n):
if (n == 0):
return 0;
msb = 0;
n = int(n / 2);
while (n > 0):
n = int(n / 2);
msb += 1
return (1 << msb)
for _ in range(int(input())):
a, b = map(int,input().split())
if(a==1 and b == 1):
print(0)
else:
k = max(a,b)
p = s(k)
print(min(p^a+p^b,a^p+b^p,p^a+b^p,a^p+p^b))
``` | instruction | 0 | 40,638 | 5 | 81,276 |
No | output | 1 | 40,638 | 5 | 81,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:
Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i \& x_j) ⋅ (x_j | x_k)$$$
Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.
The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.
The sum of n over all test cases will not exceed 5 ⋅ 10^5.
Output
Print t lines. The i-th line should contain the answer to the i-th text case.
Example
Input
8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973
Output
128
91
1600
505
0
1
502811676
264880351
Submitted Solution:
```
pows = [pow(2, i, 10 ** 9 + 7) for i in range(60)]
def main():
t = int(input())
for _ in range(t):
n = int(input())
x = [int(a) for a in input().split()]
counts = [0] * 60
for xi in x:
for i in range(60):
if xi & 1:
counts[i] += 1
xi >>= 1
answer = 0
for xi in x:
at = 0
ot = 0
for j in range(60):
if xi & 1:
at += counts[j] * pows[j]
at %= (10 ** 9) + 7
ot += n * pows[j]
ot %= (10 ** 9) + 7
else:
ot += counts[j] * pows[j]
ot %= (10 ** 9) + 7
xi >>= 1
answer += (at * ot) % (10 ** 9 + 7)
answer %= 10 ** 9 + 7
print(answer)
print(answer)
main()
``` | instruction | 0 | 40,639 | 5 | 81,278 |
No | output | 1 | 40,639 | 5 | 81,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:
Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i \& x_j) ⋅ (x_j | x_k)$$$
Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.
The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.
The sum of n over all test cases will not exceed 5 ⋅ 10^5.
Output
Print t lines. The i-th line should contain the answer to the i-th text case.
Example
Input
8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973
Output
128
91
1600
505
0
1
502811676
264880351
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def power(x, y, p):
res = 1 # Initialize result
# Update x if it is more
# than or equal to p
x = x % p
if (x == 0):
return 0
while (y > 0):
# If y is odd, multiply
# x with result
if ((y & 1) == 1):
res = (res * x) % p
# y must be even now
y = y >> 1 # y = y/2
x = (x * x) % p
return res
t = int(input())
p=10**9+7
for _ in range(t):
n = int(input())
b = list(map(int, input().split()))
val=[0]*(61)
for i in range(n):
for j in range(61):
if b[i]&(1<<j):
val[j]+=1
ans=0
res=[]
for j in range(61):
res.append(power(2,j,p)%p)
for i in range(n):
fi=0
se=0
k=bin(b[i])[2:][::-1]
for j in range(len(k)):
if k[j]=="1":
fi=(fi+(res[j]*val[j])%p)%p
se=(se+(res[j]*n))%p
else:
se = (se + res[j]* val[j]) % p
ans=(ans+(fi*se)%p)
print(ans%p)
``` | instruction | 0 | 40,640 | 5 | 81,280 |
No | output | 1 | 40,640 | 5 | 81,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:
Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i \& x_j) ⋅ (x_j | x_k)$$$
Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.
The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.
The sum of n over all test cases will not exceed 5 ⋅ 10^5.
Output
Print t lines. The i-th line should contain the answer to the i-th text case.
Example
Input
8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973
Output
128
91
1600
505
0
1
502811676
264880351
Submitted Solution:
```
def fast2():
import os, sys, atexit
from io import BytesIO
sys.stdout = BytesIO()
_write = sys.stdout.write
sys.stdout.write = lambda s: _write(s.encode())
atexit.register(lambda: os.write(1, sys.stdout.getvalue()))
return BytesIO(os.read(0, os.fstat(0).st_size)).readline
input = fast2()
add = lambda x: x if x < mod else x - mod
mult = lambda a, b: (a * b) % mod
get_bit = lambda x, i: (x >> i) & 1
out, mod, msk = [], 10 ** 9 + 7, 2 ** 30 - 1
for _ in range(int(input())):
n, a, ans, mem = int(input()), [int(x) for x in input().split()], 0, [0] * 60
mi, ma = [x & msk for x in a], [x >> 30 for x in a]
for x in mi:
for j in range(30):
if x & 1:
mem[j] += 1
x >>= 1
for x in ma:
for j in range(30, 60):
if x & 1:
mem[j] += 1
x >>= 1
mem = [pow(2, i, mod) * mem[i] % mod for i in range(60)]
mem2 = [pow(2, i, mod) * n % mod for i in range(60)]
for i in range(n):
And, Or = 0, 0
x1, x2 = mi[i], ma[i]
for j in range(30):
if x1 & 1:
And = add(mem[j] + And)
Or = add(mem2[j] + Or)
else:
Or = add(mem[j] + Or)
x1 >>= 1
for j in range(30, 60):
if x2 & 1:
And = add(mem[j] + And)
Or = add(mem2[j] + Or)
else:
Or = add(mem[j] + Or)
x2 >>= 1
ans = add(And * Or + ans)
out.append(ans)
print('\n'.join(map(str, out)))
``` | instruction | 0 | 40,641 | 5 | 81,282 |
No | output | 1 | 40,641 | 5 | 81,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Only a few know that Pan and Apollo weren't only battling for the title of the GOAT musician. A few millenniums later, they also challenged each other in math (or rather in fast calculations). The task they got to solve is the following:
Let x_1, x_2, …, x_n be the sequence of n non-negative integers. Find this value: $$$∑_{i=1}^n ∑_{j=1}^n ∑_{k=1}^n (x_i \& x_j) ⋅ (x_j | x_k)$$$
Here \& denotes the [bitwise and,](https://en.wikipedia.org/wiki/Bitwise_operation#AND) and | denotes the [bitwise or.](https://en.wikipedia.org/wiki/Bitwise_operation#OR)
Pan and Apollo could solve this in a few seconds. Can you do it too? For convenience, find the answer modulo 10^9 + 7.
Input
The first line of the input contains a single integer t (1 ≤ t ≤ 1 000) denoting the number of test cases, then t test cases follow.
The first line of each test case consists of a single integer n (1 ≤ n ≤ 5 ⋅ 10^5), the length of the sequence. The second one contains n non-negative integers x_1, x_2, …, x_n (0 ≤ x_i < 2^{60}), elements of the sequence.
The sum of n over all test cases will not exceed 5 ⋅ 10^5.
Output
Print t lines. The i-th line should contain the answer to the i-th text case.
Example
Input
8
2
1 7
3
1 2 4
4
5 5 5 5
5
6 2 2 1 0
1
0
1
1
6
1 12 123 1234 12345 123456
5
536870912 536870911 1152921504606846975 1152921504606846974 1152921504606846973
Output
128
91
1600
505
0
1
502811676
264880351
Submitted Solution:
```
''' Design by Dinh Viet Anh(JOKER)
//_____________________________________$$$$$__
//___________________________________$$$$$$$$$
//___________________________________$$$___$
//___________________________$$$____$$$$
//_________________________$$$$$$$__$$$$$$$$$$$
//_______________________$$$$$$$$$___$$$$$$$$$$$
//_______________________$$$___$______$$$$$$$$$$
//________________$$$$__$$$$_________________$$$
//_____________$__$$$$__$$$$$$$$$$$_____$____$$$
//__________$$$___$$$$___$$$$$$$$$$$__$$$$__$$$$
//_________$$$$___$$$$$___$$$$$$$$$$__$$$$$$$$$
//____$____$$$_____$$$$__________$$$___$$$$$$$
//__$$$$__$$$$_____$$$$_____$____$$$_____$
//__$$$$__$$$_______$$$$__$$$$$$$$$$
//___$$$$$$$$$______$$$$__$$$$$$$$$
//___$$$$$$$$$$_____$$$$___$$$$$$
//___$$$$$$$$$$$_____$$$
//____$$$$$$$$$$$____$$$$
//____$$$$$__$$$$$___$$$
//____$$$$$___$$$$$$
//____$$$$$____$$$
//_____$$$$
//_____$$$$
//_____$$$$
'''
from math import *
from cmath import *
from itertools import *
from decimal import * # su dung voi so thuc
from fractions import * # su dung voi phan so
from sys import *
#from numpy import *
'''getcontext().prec = x # lay x-1 chu so sau giay phay (thuoc decimal)
Decimal('12.3') la 12.3 nhung Decimal(12.3) la 12.30000000012
Fraction(a) # tra ra phan so bang a (Fraction('1.23') la 123/100 Fraction(1.23) la so khac (thuoc Fraction)
a = complex(c, d) a = c + d(i) (c = a.real, d = a.imag)
a.capitalize() bien ki tu dau cua a(string) thanh chu hoa, a.lower() bien a thanh chu thuong, tuong tu voi a.upper()
a.swapcase() doi nguoc hoa thuong, a.title() bien chu hoa sau dau cach, a.replace('a', 'b', slg)
chr(i) ki tu ma i ord(c) ma ki tu c
a.join['a', 'b', 'c'] = 'a'a'b'a'c, a.strip('a') bo dau va cuoi ki tu 'a'(rstrip, lstrip)
a.split('a', slg = -1) cat theo ki tu 'a' slg lan(rsplit(), lsplit()), a.count('aa', dau = 0, cuoi= len(a)) dem slg
a.startswith('a', dau = 0, cuoi = len(a)) co bat dau bang 'a' ko(tuong tu endswith())
a.index("aa") vi tri dau tien xuat hien (rfind())
input = open(".inp", mode='r') a = input.readline()
out = open(".out", mode='w') a.index(val) '''
#inn = open(".inp", "r")
def f(x, c):
return (x >> c) & 1
b = [0 for x in range(60)]
c = [0 for x in range(60)]
a = []
for _ in range(int(input())):
n = int(input())
a = [int(x) for x in input().split()]
sum = 0
maxx = -1
for x in a:
for i in range(60):
b[i] += f(x, i)
c[i] += (1 - f(x, i))
if f(x, i) and maxx < i:
maxx = i
for x in a:
suma = 0
sumb = 0
for i in range(maxx+1):
suma = (suma+ (2**i)*b[i]*f(x, i))%1000000007
sumb = (sumb+ (2**i)*(n - c[i] + c[i] * f(x, i)))%1000000007
sum += ((suma % 1000000007) * (sumb%1000000007))%1000000007
print(sum % 1000000007)
for x in range(maxx+1):
b[x] = 0
c[x] = 0
``` | instruction | 0 | 40,642 | 5 | 81,284 |
No | output | 1 | 40,642 | 5 | 81,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
'__author__'=='deepak Singh Mehta(learning brute force)) '
if __name__=='__main__':
s, x = map(int, input().split(' '))
if (s-x)%2 or s < x:
print(0)
else:
c = bin((s-x)//2)[2:][::-1]
t = bin(x)[2:][::-1]
for i in range(len(t)):
if t[i] == '1' and i < len(c) and c[i] == '1':
print(0)
exit(0)
print(pow(2, bin(x)[2:].count('1'))-(2 if s==x else 0))
``` | instruction | 0 | 40,828 | 5 | 81,656 |
Yes | output | 1 | 40,828 | 5 | 81,657 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
def foo(s, x):
if x == 0 and s % 2 == 0:
return 1
if s == 0:
return 0
if s % 2 == 1 and x % 2 == 1:
return 2 * foo(s // 2, x // 2)
if s % 2 == 0 and x % 2 == 1:
return 0
if s % 2 == 1 and x % 2 == 0:
return 0
if s % 2 == 0 and x % 2 == 0:
return foo(s // 2 - 1, x // 2) + foo(s // 2, x // 2)
s, x = map(int, input().split())
cnt = foo(s, x)
if s ^ 0 == x:
cnt -= 2
print(cnt)
``` | instruction | 0 | 40,829 | 5 | 81,658 |
Yes | output | 1 | 40,829 | 5 | 81,659 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
def main():
s, x = map(int, input().split())
if s < x:
print(0)
return
s, x = ([c == '1' for c in bin(int(_))[2:]] for _ in (s, x))
x = [False] * (len(s) - len(x)) + x
a, b = [False] * len(s), [False] * len(s)
def deeper(idx, carry):
if idx:
idx -= 1
if carry:
if s[idx] == x[idx]:
return
if s[idx]:
deeper(idx, False)
a[idx] = True
b[idx] = True
deeper(idx, True)
a[idx] = False
b[idx] = False
else:
b[idx] = True
deeper(idx, True)
b[idx] = False
else:
if s[idx] != x[idx]:
return
if s[idx]:
b[idx] = True
deeper(idx, False)
b[idx] = False
else:
a[idx] = True
b[idx] = True
deeper(idx, True)
a[idx] = False
b[idx] = False
deeper(idx, False)
else:
raise TabError
try:
deeper(len(s), False)
except TabError:
print((1 << (sum(b) - sum(a))) - (0 if any(a)else 2))
else:
print(0)
if __name__ == '__main__':
main()
``` | instruction | 0 | 40,830 | 5 | 81,660 |
Yes | output | 1 | 40,830 | 5 | 81,661 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
import io
import sys
import time
import random
#~ start = time.clock()
#~ test = '''3 3''' # 2 solutions, (1,2) and (2,1)
#~ test = '''9 5''' # 4 solutions
#~ test = '''5 2''' # 0 solutions
#~ test = '''6 0''' # 1 solution
#~ sys.stdin = io.StringIO(test)
s,x = list(map(int, input().split()))
def bit_of(x,i):
return int((1<<i)&x!=0)
def solve(s,x):
if s<x or (s^x)%2!=0:
return 0
d = (s-x)//2
zero_is_solution = True
count = 1
for i in range(s.bit_length()):
db = bit_of(d,i)
xb = bit_of(x,i)
if db==1:
if xb==1:
return 0
else:
zero_is_solution = False
if db==0 and xb==1:
count *= 2
if zero_is_solution:
count -= 2
return count
print(solve(s,x))
``` | instruction | 0 | 40,831 | 5 | 81,662 |
Yes | output | 1 | 40,831 | 5 | 81,663 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
s,x=in_arr()
if (s-x)%2 or x>s or (x<<1)&(s-x):
pr_num(0)
exit()
ans=bin(x)[2:]
pr_num(pow(2,ans.count('1'))-(2*(s==x)))
``` | instruction | 0 | 40,832 | 5 | 81,664 |
Yes | output | 1 | 40,832 | 5 | 81,665 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_num():
return int(raw_input())
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return map(int,stdin.read().split())
range = xrange # not for python 3.0+
# main code
s,x=in_arr()
if (s-x)%2 or x>s:
pr_num(0)
exit()
ans=bin(x)[2:]
pr_num(pow(2,ans.count('1'))-(2*(s==x)))
``` | instruction | 0 | 40,833 | 5 | 81,666 |
No | output | 1 | 40,833 | 5 | 81,667 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
s, x = map(int, input().split())
f = x
if (s - x) % 2 == 1:
print(0)
exit(0)
x = bin(x)[2::]
k = 0
for i in range(len(x)):
if x[i] == "1":
k += 1
if s == f:
print(2 ** k - 2)
exit(0)
print(2 ** k)
``` | instruction | 0 | 40,834 | 5 | 81,668 |
No | output | 1 | 40,834 | 5 | 81,669 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
s, x = map(int, input().split())
print(((s-x)&1==0 and x <= s) * (2 ** bin(x).count('1') - 2 * (s == x)))
``` | instruction | 0 | 40,835 | 5 | 81,670 |
No | output | 1 | 40,835 | 5 | 81,671 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
s, x = map(int, input().split())
if s < x:
print(0)
elif s == x:
print(2)
else:
data = bin(x).count('1')
if (s-x)/2 == (s-x) // 2:
if 2**data - 1 == x and x != 1 and x != 0:
print(0)
else:
print(2**data)
else:
print(0)
``` | instruction | 0 | 40,836 | 5 | 81,672 |
No | output | 1 | 40,836 | 5 | 81,673 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?
Input
The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Output
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.
Examples
Input
9 5
Output
4
Input
3 3
Output
2
Input
5 2
Output
0
Note
In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and (2, 1).
Submitted Solution:
```
def sol(s,x):
if x>s or (s-x)%2:
print(0)
return
n = str(bin(x))[2:].count('1')
res = 2**n
if x==s:
res-=2
print(res)
s ,k = list(map(int,input().split(' ')))
sol(s,k)
``` | instruction | 0 | 40,837 | 5 | 81,674 |
No | output | 1 | 40,837 | 5 | 81,675 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,991 | 5 | 81,982 |
"Correct Solution:
```
a, b, c = map(int, input().split())
print("Yes" if ((c-a-b) > 0 and 4*a*b < (c-a-b)**2) else "No")
``` | output | 1 | 40,991 | 5 | 81,983 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,992 | 5 | 81,984 |
"Correct Solution:
```
a,b,c=map(int,input().split())
if 4*a*b<(c-a-b)**2 and c-a-b>=0:
print('Yes')
else:
print('No')
``` | output | 1 | 40,992 | 5 | 81,985 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,993 | 5 | 81,986 |
"Correct Solution:
```
a,b,c=map(int,input().split())
print(["No","Yes"][4*a*b<(c-a-b)**2 and c-a-b>0])
``` | output | 1 | 40,993 | 5 | 81,987 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,994 | 5 | 81,988 |
"Correct Solution:
```
a, b, c = map(int, input().split())
if 4*a*b < (c-a-b)**2 and (c-a-b) > 0:
print("Yes")
else:
print("No")
``` | output | 1 | 40,994 | 5 | 81,989 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,995 | 5 | 81,990 |
"Correct Solution:
```
a,b,c = [int(hoge) for hoge in input().split()]
print("NYoe s"[4*a*b < (c-a-b)**2 and a+b<c::2])
``` | output | 1 | 40,995 | 5 | 81,991 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,996 | 5 | 81,992 |
"Correct Solution:
```
a,b,c=map(int, input().split())
r=c-a-b
if r > 0 and r**2>4*a*b:
print("Yes")
else:
print("No")
``` | output | 1 | 40,996 | 5 | 81,993 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,997 | 5 | 81,994 |
"Correct Solution:
```
a,b,c = map(int,input().split())
print('Yes' if c-b-a>0 and 0<(c-a-b)**2-4*a*b else 'No')
``` | output | 1 | 40,997 | 5 | 81,995 |
Provide a correct Python 3 solution for this coding contest problem.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes | instruction | 0 | 40,998 | 5 | 81,996 |
"Correct Solution:
```
a,b,c=map(int,input().split())
if c>=a+b and 4*a*b<(c-a-b)**2:
print('Yes')
else:
print('No')
``` | output | 1 | 40,998 | 5 | 81,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a,b,c = map(int, input().split())
print('Yes' if c-a-b > 0 and (c-a-b)**2 > 4*a*b else 'No')
``` | instruction | 0 | 40,999 | 5 | 81,998 |
Yes | output | 1 | 40,999 | 5 | 81,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a,b,c=map(int,input().split())
if (c-a-b>0)and(4*a*b<(c-a-b)**2):
print('Yes')
else:
print('No')
``` | instruction | 0 | 41,000 | 5 | 82,000 |
Yes | output | 1 | 41,000 | 5 | 82,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a,b,c = (int(x) for x in input().split())
if 4*a*b < (c-a-b)**2 and c-a-b > 0:
print('Yes')
else:
print('No')
``` | instruction | 0 | 41,001 | 5 | 82,002 |
Yes | output | 1 | 41,001 | 5 | 82,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
import math
a,b,c=map(int,input().split())
if c-a-b>=0 and 4*a*b<(c-a-b)**2:
print("Yes")
else:
print("No")
``` | instruction | 0 | 41,002 | 5 | 82,004 |
Yes | output | 1 | 41,002 | 5 | 82,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a,b,c = map(int,input().split())
import math
import decimal
decimal.getcontext().prec = 15
if decimal(math.sqrt(a))+decimal(math.sqrt(b)) < decimal(math.sqrt(c)):
print('Yes')
else:
print('No')
``` | instruction | 0 | 41,003 | 5 | 82,006 |
No | output | 1 | 41,003 | 5 | 82,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a,b,c = map(int,input().split())
import math
if math.sqrt(a) * math.sqrt(b) < (c - a - b) // 2:
print("Yes")
else:
print("No")
``` | instruction | 0 | 41,004 | 5 | 82,008 |
No | output | 1 | 41,004 | 5 | 82,009 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
A,B,C = (int(x) for x in input().split())
import math
a = math.sqrt(A)
b = math.sqrt(B)
c = math.sqrt(C)
if 0 < c - a - b:
print('Yes')
else:
print('No')
``` | instruction | 0 | 41,005 | 5 | 82,010 |
No | output | 1 | 41,005 | 5 | 82,011 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Does \sqrt{a} + \sqrt{b} < \sqrt{c} hold?
Constraints
* 1 \leq a, b, c \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
a \ b \ c
Output
If \sqrt{a} + \sqrt{b} < \sqrt{c}, print `Yes`; otherwise, print `No`.
Examples
Input
2 3 9
Output
No
Input
2 3 10
Output
Yes
Submitted Solution:
```
a, b, c = map(int, input().split())
if (a*b) ** (0.5) < (c-a-b) / 2:
print("Yes")
else:
print("No")
``` | instruction | 0 | 41,006 | 5 | 82,012 |
No | output | 1 | 41,006 | 5 | 82,013 |
Provide a correct Python 3 solution for this coding contest problem.
We have N weights indexed 1 to N. The mass of the weight indexed i is W_i.
We will divide these weights into two groups: the weights with indices not greater than T, and those with indices greater than T, for some integer 1 \leq T < N. Let S_1 be the sum of the masses of the weights in the former group, and S_2 be the sum of the masses of the weights in the latter group.
Consider all possible such divisions and find the minimum possible absolute difference of S_1 and S_2.
Constraints
* 2 \leq N \leq 100
* 1 \leq W_i \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
W_1 W_2 ... W_{N-1} W_N
Output
Print the minimum possible absolute difference of S_1 and S_2.
Examples
Input
3
1 2 3
Output
0
Input
4
1 3 1 1
Output
2
Input
8
27 23 76 2 3 5 62 52
Output
2 | instruction | 0 | 41,015 | 5 | 82,030 |
"Correct Solution:
```
n = int(input())
l = list(map(int, input().split()))
print(min([abs(sum(l[0:i]) - sum(l[i:])) for i in range(1, len(l))]))
``` | output | 1 | 41,015 | 5 | 82,031 |
Provide a correct Python 3 solution for this coding contest problem.
We have N weights indexed 1 to N. The mass of the weight indexed i is W_i.
We will divide these weights into two groups: the weights with indices not greater than T, and those with indices greater than T, for some integer 1 \leq T < N. Let S_1 be the sum of the masses of the weights in the former group, and S_2 be the sum of the masses of the weights in the latter group.
Consider all possible such divisions and find the minimum possible absolute difference of S_1 and S_2.
Constraints
* 2 \leq N \leq 100
* 1 \leq W_i \leq 100
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
W_1 W_2 ... W_{N-1} W_N
Output
Print the minimum possible absolute difference of S_1 and S_2.
Examples
Input
3
1 2 3
Output
0
Input
4
1 3 1 1
Output
2
Input
8
27 23 76 2 3 5 62 52
Output
2 | instruction | 0 | 41,016 | 5 | 82,032 |
"Correct Solution:
```
input()
l=list(map(int,input().split()))
s=sum(l)
a,t=s,0
for i in l: t+=i; a=min(a,abs(s-t*2))
print(a)
``` | output | 1 | 41,016 | 5 | 82,033 |
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