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 |
|---|---|---|---|---|---|
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 = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
class segmentTreeSum():
initValue = 0
dat = []
lenData = -1
def __init__(self):
pass
def load(self, l):
# len(l)個よりも大きい2の二乗を得る
self.lenData = 2 ** (len(l) - 1).bit_length() # len 5 なら 2^3 = 8
self.dat = [self.initValue] * (self.lenData * 2)
# 値のロード
for i in range(len(l)):
self.dat[self.lenData - 1 + i] = l[i]
self.build()
def build(self):
for i in range(self.lenData - 2, -1, -1):
self.dat[i] = self.dat[2 * i + 1] + self.dat[2 * i + 2]
# 若干無駄なインストラクションあるのでコードレベルでマージしたほうがいいかも
def addValue(self, i, a):
nodeId = (self.lenData - 1) + i
self.dat[nodeId] += a
self.setValue(i, self.dat[nodeId])
def setValue(self, i, a):
"""
set a to list[i]
"""
#print("setValue: {0}, {1}".format(i, a))
nodeId = (self.lenData - 1) + i
#print(" first nodeId: {0}".format(nodeId))
self.dat[nodeId] = a
while nodeId != 0:
nodeId = (nodeId - 1) // 2
#print(" next nodeId: {0}".format(nodeId))
self.dat[nodeId] = self.dat[nodeId * 2 + 1] + self.dat[nodeId * 2 + 2]
def querySub(self, a, b, nodeId, l, r):
"""
[a,b) 区間の親クエリに対するノードnodeへ[l, r)の探索をデリゲート
区間については、dataの添え字は0,1,2,3,4としたときに、
[0,3)なら0,1,2の結果を返す
"""
# print("querySub: a={0}, b={1}, nodeId={2}, l={3}, r={4}".format(a, b, nodeId, l, r))
if (r <= a or b <= l):
return self.initValue
if a <= l and r <= b:
return self.dat[nodeId]
resLeft = self.querySub(a, b, 2 * nodeId + 1, l, (l + r) // 2)
resRight = self.querySub(a, b, 2 * nodeId + 2, (l + r) // 2, r)
return resLeft + resRight
def query(self, a, b):
return self.querySub(a, b, 0, 0, self.lenData)
initVal = 0
n, q = map(int, input().split())
dat = [initVal] * n
st = segmentTreeSum()
st.load(dat)
st.build()
#print(st.dat)
for _ in range(q):
# indexの入力が1 originなので注意 (以下で-1しているのはそのため)
cmd, a, b = map(int, input().split())
if cmd == 0: # update
st.addValue(a - 1, b)
else:
print(st.query(a - 1 , b + 1 -1))
``` | instruction | 0 | 1,646 | 5 | 3,292 |
Yes | output | 1 | 1,646 | 5 | 3,293 |
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 = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
class RSQ:
def __init__(self, length, ini_num = float("inf")):
self.length = 1
self.ini_num = ini_num
while self.length < length:
self.length <<= 1
self.segtree = [ini_num] * (2 * self.length - 1)
def update(self, index, num):
leaf_index = index + self.length - 1
self.segtree[leaf_index] = num
while leaf_index > 0:
leaf_index = (leaf_index - 1) // 2
self.segtree[leaf_index] = self.segtree[2 * leaf_index + 1] + self.segtree[2 * leaf_index + 2]
def call_search(self, a, b):
return self.search(a, b, index = 0, l = 0, r = self.length - 1)
def search(self, a, b, index, l, r):
if a <= l <= r <= b:
return self.segtree[index]
elif r < a or b < l:
return self.ini_num
else:
return self.search(a, b, index * 2 + 1, l, (l + r) // 2) + self.search(a, b, index * 2 + 2, (l + r) // 2 + 1, r)
v_num, query = map(int, input().split(" "))
rsq = RSQ(v_num, 0)
for _ in range(query):
comm, index, num = map(int, input().split(" "))
if comm == 0:
rsq.update(index, num)
else:
print(rsq.call_search(index, num))
``` | instruction | 0 | 1,647 | 5 | 3,294 |
No | output | 1 | 1,647 | 5 | 3,295 |
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 = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
sys.setrecursionlimit(10 ** 9)
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()))
INF=2**31-1
class BIT:
def __init__(self, n):
# 0-indexed
self.size = n
self.tree = [0] * (n+1)
# [0, i]を合計する
def sum(self, i):
s = 0
i += 1
while i > 0:
s += self.tree[i-1]
i -= i & -i
return s
# 値の追加：添字, 値
def add(self, i, x):
i += 1
while i <= self.size:
self.tree[i-1] += x
i += i & -i
# 区間和の取得 [l, r)
def get(self, l, r=None):
# 引数が1つなら一点の値を取得
if r is None: r = l+1
res = 0
if r: res += self.sum(r-1)
if l: res -= self.sum(l-1)
return res
N,Q=MAP()
bit=BIT(N)
for _ in range(Q):
com,x,y=MAP()
if com==0:
bit.add(x, y)
else:
print(bit.get(x, y+1))
``` | instruction | 0 | 1,648 | 5 | 3,296 |
No | output | 1 | 1,648 | 5 | 3,297 |
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 = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
a = list(map(int,input().split()))
b = [0]*a[0]
for i in range(a[1]):
c = list(map(int,input().split()))
if c[0] == 0:
for j in range(a[0]-c[1]):
b[c[1]+j] += c[2]
else:
if c[1] == 1:
print(b[c[2]-1])
else:
print(b[c[2]]-b[c[1]-1])
``` | instruction | 0 | 1,649 | 5 | 3,298 |
No | output | 1 | 1,649 | 5 | 3,299 |
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 = {a1, a2, . . . , an} with the following operations:
* add(i, x): add x to ai.
* getSum(s, t): print the sum of as, as+1,...,at.
Note that the initial values of ai (i = 1, 2, . . . , n) are 0.
Constraints
* 1 ≤ n ≤ 100000
* 1 ≤ q ≤ 100000
* If comi is 0, then 1 ≤ xi ≤ n, 0 ≤ yi ≤ 1000.
* If comi is 1, then 1 ≤ xi ≤ n, 1 ≤ yi ≤ n.
Input
n q
com1 x1 y1
com2 x2 y2
...
comq xq yq
In the first line, n (the number of elements in A) and q (the number of queries) are given. Then, q queries are given where com represents the type of queries. '0' denotes add(xi, yi) and '1' denotes getSum(xi, yi).
Output
For each getSum operation, print the sum in a line.
Example
Input
3 5
0 1 1
0 2 2
0 3 3
1 1 2
1 2 2
Output
3
2
Submitted Solution:
```
n, q = map(int, input().split())
answer = []
# create segtree
size = 1
while size < n :
size *= 2
size = size * 2 - 1
segtree = [0 for x in range(size)]
# update query
def update(i, x) :
ind = size // 2 + i
segtree[ind] = x
while ind :
ind = (ind - 1) // 2 # parent
ch1 = segtree[ind * 2 + 1]
ch2 = segtree[ind * 2 + 2]
segtree[ind] = ch1 + ch2
def query(s, t, l, r, p) :
if s > r or t < l :
return 0
if s <= l and t >= r :
return segtree[p]
else :
vl = query(s, t, l, (l + r) // 2, p * 2 + 1)
vr = query(s, t, (l + r) // 2 + 1, r, p * 2 + 2)
return vl + vr
for i in range(q) :
# print("DBG : ", segtree)
com, x, y = map(int, input().split())
if com == 0 :
update(x, y)
else :
answer.append(query(x, y, 0, size // 2, 0))
for a in answer :
print(a)
``` | instruction | 0 | 1,650 | 5 | 3,300 |
No | output | 1 | 1,650 | 5 | 3,301 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a very important day for Katya. She has a test in a programming class. As always, she was given an interesting problem that she solved very fast. Can you solve that problem?
You are given n ordered segments sets. Each segment can be represented as a pair of two integers [l, r] where l≤ r. Each set can contain an arbitrary number of segments (even 0). It is possible that some segments are equal.
You are also given m queries, each of them can be represented as four numbers: a, b, x, y. For each segment, find out whether it is true that each set p (a≤ p≤ b) contains at least one segment [l, r] that lies entirely on the segment [x, y], that is x≤ l≤ r≤ y.
Find out the answer to each query.
Note that you need to solve this problem online. That is, you will get a new query only after you print the answer for the previous query.
Input
The first line contains three integers n, m, and k (1≤ n,m≤ 10^5, 1≤ k≤ 3⋅10^5) — the number of sets, queries, and segments respectively.
Each of the next k lines contains three integers l, r, and p (1≤ l≤ r≤ 10^9, 1≤ p≤ n) — the limits of the segment and the index of a set, to which this segment belongs.
Each of the next m lines contains four integers a, b, x, y (1≤ a≤ b≤ n, 1≤ x≤ y≤ 10^9) — the description of the query.
Output
For each query, print "yes" or "no" in a new line.
Interaction
After printing a query, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
5 5 9
3 6 3
1 3 1
2 4 2
1 2 3
4 6 5
2 5 3
7 9 4
2 3 1
4 10 4
1 2 2 3
1 2 2 4
1 3 1 5
2 3 3 6
2 4 2 9
Output
no
yes
yes
no
yes
Note
For the first query, the answer is negative since the second set does not contain a segment that lies on the segment [2, 3].
In the second query, the first set contains [2, 3], and the second set contains [2, 4].
In the third query, the first set contains [2, 3], the second set contains [2, 4], and the third set contains [2, 5].
In the fourth query, the second set does not contain a segment that lies on the segment [3, 6].
In the fifth query, the second set contains [2, 4], the third set contains [2, 5], and the fourth contains [7, 9].
Submitted Solution:
```
def function(a,b,x,y,ai):
answer=[]
for i in range(a,b+1):
for j in (ai[i]):
if(x<=j[0]<= j[1]<=y):
answer.append("yes")
break
print(answer)
if(b+1-a==answer.count("yes")):
return("yes")
else:
return("no")
sets,queries,segments=map(int,input().split())
ai={}
for i in range(1,sets+1):
ai[i]=[]
for i in range(segments):
a,b,c=map(int,input().split())
ai[c].append([a,b])
for i in range(1,sets+1):
ai[i].sort()
answers=[]
for i in range(queries):
a,b,x,y=map(int,input().split())
t=function(a,b,x,y,ai)
answers.append(t)
for i in answers:
print(i,end="\n")
``` | instruction | 0 | 1,688 | 5 | 3,376 |
No | output | 1 | 1,688 | 5 | 3,377 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a very important day for Katya. She has a test in a programming class. As always, she was given an interesting problem that she solved very fast. Can you solve that problem?
You are given n ordered segments sets. Each segment can be represented as a pair of two integers [l, r] where l≤ r. Each set can contain an arbitrary number of segments (even 0). It is possible that some segments are equal.
You are also given m queries, each of them can be represented as four numbers: a, b, x, y. For each segment, find out whether it is true that each set p (a≤ p≤ b) contains at least one segment [l, r] that lies entirely on the segment [x, y], that is x≤ l≤ r≤ y.
Find out the answer to each query.
Note that you need to solve this problem online. That is, you will get a new query only after you print the answer for the previous query.
Input
The first line contains three integers n, m, and k (1≤ n,m≤ 10^5, 1≤ k≤ 3⋅10^5) — the number of sets, queries, and segments respectively.
Each of the next k lines contains three integers l, r, and p (1≤ l≤ r≤ 10^9, 1≤ p≤ n) — the limits of the segment and the index of a set, to which this segment belongs.
Each of the next m lines contains four integers a, b, x, y (1≤ a≤ b≤ n, 1≤ x≤ y≤ 10^9) — the description of the query.
Output
For each query, print "yes" or "no" in a new line.
Interaction
After printing a query, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
5 5 9
3 6 3
1 3 1
2 4 2
1 2 3
4 6 5
2 5 3
7 9 4
2 3 1
4 10 4
1 2 2 3
1 2 2 4
1 3 1 5
2 3 3 6
2 4 2 9
Output
no
yes
yes
no
yes
Note
For the first query, the answer is negative since the second set does not contain a segment that lies on the segment [2, 3].
In the second query, the first set contains [2, 3], and the second set contains [2, 4].
In the third query, the first set contains [2, 3], the second set contains [2, 4], and the third set contains [2, 5].
In the fourth query, the second set does not contain a segment that lies on the segment [3, 6].
In the fifth query, the second set contains [2, 4], the third set contains [2, 5], and the fourth contains [7, 9].
Submitted Solution:
```
class Set:
def __init__(self,l,r,p):
self.l = l
self.r = r
self.p = p
def __lt__(self,other):
return self.r - other.r < 0
def print(self):
print(f"l = {self.l}, r = {self.r} , in set {self.p}")
class NumList:
def __init__(self):
self.numList = [-1,int(1e9+7)]
self._numdic = {-1:0}
def append(self,x):
if x not in self._numdic.keys():
self.numList.append(x)
self._numdic[x] = 0
def sort(self):
self.numList.sort()
for index,value in enumerate(self.numList):
self._numdic[self.numList[index]] = index
def print(self):
print(self.numList)
print(self._numdic)
def id(self,x):
return self._numdic[x]
#返回左右两个值
def binary_search(self,x):
low = -1
high = len(self.numList)
while high - low > 1:
mid = (low + high) >> 1
if self.numList[mid] == x:
return mid,mid
elif self.numList[mid] > x:
high = mid
else:
low = mid
return low,high
class Node:
# 左子树序号，右子树序号
def __init__(self,l,r,val):
self.l = l
self.r = r
self.val = val
def print(self):
print(f"node: l = {self.l}, r = {self.r}, val = {self.val}")
MAX_VAL = int(1e9+7)
class SegmentTreeGroup:
MAXN = int(3e5+7)
def __init__(self):
self.head = []
self.nodeGroup = []
self.nodeNumber = 0
self.root = -1
self.rootGroup = []
def newNode(self,node):
self.nodeGroup.append(node)
self.nodeNumber += 1
return self.nodeNumber - 1
def _build(self,l,r):
if l == r:
node = Node(-1,-1,-1)
return self.newNode(node)
else:
mid = (l+r) >> 1
lchild = self._build(l,mid)
rchild = self._build(mid+1,r)
node = Node(lchild,rchild,-1)
return self.newNode(node)
def build(self):
self.root = self._build(0,SegmentTreeGroup.MAXN)
return self.root
# 单点更新，将第index的值更新为value
def _update(self,pos,val,index,l,r):
if l == r:
node = Node(-1,-1,val)
# node.print()
return self.newNode(node)
else:
mid = (l+r) >> 1
if index <= mid:
lchild = self._update(self.nodeGroup[pos].l,val,index,l,mid)
rchild = self.nodeGroup[pos].r
val = min(self.nodeGroup[lchild].val,self.nodeGroup[rchild].val)
# Node(lchild,rchild,val).print()
return self.newNode(Node(lchild,rchild,val))
else:
lchild = self.nodeGroup[pos].l
rchild = self._update(self.nodeGroup[pos].r,val,index,mid+1,r)
val = min(self.nodeGroup[lchild].val,self.nodeGroup[rchild].val)
return self.newNode(Node(lchild,rchild,val))
def update(self,val,index):
self.root = self._update(self.root,val,index,0,SegmentTreeGroup.MAXN)
# self.nodeGroup[self.root].print()
return self.root
def _query(self,nodeId,ql,qr,l,r):
# print(f"val of id = {nodeId} is {self.nodeGroup[nodeId].val} ; l = {self.nodeGroup[nodeId].l} ,r = {self.nodeGroup[nodeId].r}, ql = {l}, qr = {r}")
if ql <= l and r <= qr:
return self.nodeGroup[nodeId].val
else:
mid = (l+r) >> 1
valL = MAX_VAL
valR = MAX_VAL
if ql <= mid:
valL = self._query(self.nodeGroup[nodeId].l,ql,qr,l,mid)
if qr > mid:
valR = self._query(self.nodeGroup[nodeId].r,ql,qr,mid+1,r)
return min(valL,valR)
def query(self,treeId,ql,qr):
return self._query(self.rootGroup[treeId],ql,qr,0,SegmentTreeGroup.MAXN)
def record(self):
self.rootGroup.append(self.root)
return len(self.rootGroup) - 1
setList = [] # Set集合
numList = NumList()
def debug():
segmentTree = SegmentTreeGroup()
num = segmentTree.build()
print(f"node id in seg = {num}")
segmentTree.update(107,7700)
print(f"root = {segmentTree.root}")
segmentTree.nodeGroup[segmentTree.root].print()
segmentTree.update(109,7901)
tid = segmentTree.record()
segmentTree.update(7,7700)
tid2 = segmentTree.record()
print("tid = {}, q1 = {}".format(tid,segmentTree.query(tid,6800,8000)))
print("tid2 = {}, q2 = {}".format(tid2,segmentTree.query(tid2,6800,8000)))
def main():
#debug()
n,m,k = list(map(int,input().strip().split()))
# print(f"n = {n}, m = {m} ,k = {k}")
for i in range(0,k):
l,r,p = list(map(int,input().strip().split()))
setList.append(Set(l,r,p))
numList.append(l)
numList.append(r)
numList.sort()
setList.sort()
for i in range(0,len(setList)):
setList[i].l = numList.id(setList[i].l)
setList[i].r = numList.id(setList[i].r)
# setList[i].print()
segmentTree = SegmentTreeGroup()
segmentTree.build()
setMaxRecord = [ -1 for x in range(0,n+7) ]
p = 0
for i in numList.numList:
while p < len(setList) and setList[p].r <= i:
segL = setList[p].l
segR = setList[p].r
setNo = setList[p].p
p += 1
if segL > setMaxRecord[setNo]:
setMaxRecord[setNo] = segL
segmentTree.update(segL,setNo)
segmentTree.record()
# print("{}".format(numList._numdic))
for i in range(0,m):
a,b,x,y = list(map(int,input().strip().split()))
# print(f"a = {a},b = {b} , x = {x}, y = {y}")
_,x = numList.binary_search(x)
y,_ = numList.binary_search(y)
# print(f" -- a = {a},b = {b} , x = {x}, y = {y}")
Lmin = segmentTree.query(y,a,b)
# print(f"-- Lmin = {Lmin} ,in set({a},{b})")
if Lmin >= x:
print("yes",flush=True)
else:
print("no",flush=True)
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,689 | 5 | 3,378 |
No | output | 1 | 1,689 | 5 | 3,379 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a very important day for Katya. She has a test in a programming class. As always, she was given an interesting problem that she solved very fast. Can you solve that problem?
You are given n ordered segments sets. Each segment can be represented as a pair of two integers [l, r] where l≤ r. Each set can contain an arbitrary number of segments (even 0). It is possible that some segments are equal.
You are also given m queries, each of them can be represented as four numbers: a, b, x, y. For each segment, find out whether it is true that each set p (a≤ p≤ b) contains at least one segment [l, r] that lies entirely on the segment [x, y], that is x≤ l≤ r≤ y.
Find out the answer to each query.
Note that you need to solve this problem online. That is, you will get a new query only after you print the answer for the previous query.
Input
The first line contains three integers n, m, and k (1≤ n,m≤ 10^5, 1≤ k≤ 3⋅10^5) — the number of sets, queries, and segments respectively.
Each of the next k lines contains three integers l, r, and p (1≤ l≤ r≤ 10^9, 1≤ p≤ n) — the limits of the segment and the index of a set, to which this segment belongs.
Each of the next m lines contains four integers a, b, x, y (1≤ a≤ b≤ n, 1≤ x≤ y≤ 10^9) — the description of the query.
Output
For each query, print "yes" or "no" in a new line.
Interaction
After printing a query, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
5 5 9
3 6 3
1 3 1
2 4 2
1 2 3
4 6 5
2 5 3
7 9 4
2 3 1
4 10 4
1 2 2 3
1 2 2 4
1 3 1 5
2 3 3 6
2 4 2 9
Output
no
yes
yes
no
yes
Note
For the first query, the answer is negative since the second set does not contain a segment that lies on the segment [2, 3].
In the second query, the first set contains [2, 3], and the second set contains [2, 4].
In the third query, the first set contains [2, 3], the second set contains [2, 4], and the third set contains [2, 5].
In the fourth query, the second set does not contain a segment that lies on the segment [3, 6].
In the fifth query, the second set contains [2, 4], the third set contains [2, 5], and the fourth contains [7, 9].
Submitted Solution:
```
print("yes\nyes\nyes\nyes\nyes\n")
``` | instruction | 0 | 1,690 | 5 | 3,380 |
No | output | 1 | 1,690 | 5 | 3,381 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
It is a very important day for Katya. She has a test in a programming class. As always, she was given an interesting problem that she solved very fast. Can you solve that problem?
You are given n ordered segments sets. Each segment can be represented as a pair of two integers [l, r] where l≤ r. Each set can contain an arbitrary number of segments (even 0). It is possible that some segments are equal.
You are also given m queries, each of them can be represented as four numbers: a, b, x, y. For each segment, find out whether it is true that each set p (a≤ p≤ b) contains at least one segment [l, r] that lies entirely on the segment [x, y], that is x≤ l≤ r≤ y.
Find out the answer to each query.
Note that you need to solve this problem online. That is, you will get a new query only after you print the answer for the previous query.
Input
The first line contains three integers n, m, and k (1≤ n,m≤ 10^5, 1≤ k≤ 3⋅10^5) — the number of sets, queries, and segments respectively.
Each of the next k lines contains three integers l, r, and p (1≤ l≤ r≤ 10^9, 1≤ p≤ n) — the limits of the segment and the index of a set, to which this segment belongs.
Each of the next m lines contains four integers a, b, x, y (1≤ a≤ b≤ n, 1≤ x≤ y≤ 10^9) — the description of the query.
Output
For each query, print "yes" or "no" in a new line.
Interaction
After printing a query, do not forget to output end of line and flush the output. Otherwise, you will get Idleness limit exceeded. To do this, use:
* fflush(stdout) or cout.flush() in C++;
* System.out.flush() in Java;
* flush(output) in Pascal;
* stdout.flush() in Python;
* see documentation for other languages.
Example
Input
5 5 9
3 6 3
1 3 1
2 4 2
1 2 3
4 6 5
2 5 3
7 9 4
2 3 1
4 10 4
1 2 2 3
1 2 2 4
1 3 1 5
2 3 3 6
2 4 2 9
Output
no
yes
yes
no
yes
Note
For the first query, the answer is negative since the second set does not contain a segment that lies on the segment [2, 3].
In the second query, the first set contains [2, 3], and the second set contains [2, 4].
In the third query, the first set contains [2, 3], the second set contains [2, 4], and the third set contains [2, 5].
In the fourth query, the second set does not contain a segment that lies on the segment [3, 6].
In the fifth query, the second set contains [2, 4], the third set contains [2, 5], and the fourth contains [7, 9].
Submitted Solution:
```
class Set:
def __init__(self,l,r,p):
self.l = l
self.r = r
self.p = p
def __lt__(self,other):
return self.r - other.r < 0
def print(self):
print(f"l = {self.l}, r = {self.r} , in set {self.p}")
class NumList:
def __init__(self):
self.numList = [-1,int(1e9+7)]
self._numdic = {-1:0}
def append(self,x):
if x not in self._numdic.keys():
self.numList.append(x)
self._numdic[x] = 0
def sort(self):
self.numList.sort()
for index,value in enumerate(self.numList):
self._numdic[self.numList[index]] = index
def print(self):
print(self.numList)
print(self._numdic)
def id(self,x):
return self._numdic[x]
#返回左右两个值
def binary_search(self,x):
low = -1
high = len(self.numList)
while high - low > 1:
mid = (low + high) >> 1
if self.numList[mid] == x:
return mid,mid
elif self.numList[mid] > x:
high = mid
else:
low = mid
return low,high
class Node:
# 左子树序号，右子树序号
def __init__(self,l,r,val):
self.l = l
self.r = r
self.val = val
def print(self):
print(f"node: l = {self.l}, r = {self.r}, val = {self.val}")
MAX_VAL = int(1e9+7)
class SegmentTreeGroup:
MAXN = int(1e5+7)
def __init__(self):
self.head = []
self.nodeGroup = []
self.nodeNumber = 0
self.root = -1
self.rootGroup = []
def newNode(self,node):
self.nodeGroup.append(node)
self.nodeNumber += 1
return self.nodeNumber - 1
def _build(self,l,r):
if l == r:
node = Node(-1,-1,-1)
return self.newNode(node)
else:
mid = (l+r) >> 1
lchild = self._build(l,mid)
rchild = self._build(mid+1,r)
node = Node(lchild,rchild,-1)
return self.newNode(node)
def build(self):
self.root = self._build(0,SegmentTreeGroup.MAXN)
return self.root
# 单点更新，将第index的值更新为value
def _update(self,pos,val,index,l,r):
if l == r:
node = Node(-1,-1,val)
# node.print()
return self.newNode(node)
else:
mid = (l+r) >> 1
if index <= mid:
lchild = self._update(self.nodeGroup[pos].l,val,index,l,mid)
rchild = self.nodeGroup[pos].r
val = min(self.nodeGroup[lchild].val,self.nodeGroup[rchild].val)
# Node(lchild,rchild,val).print()
return self.newNode(Node(lchild,rchild,val))
else:
lchild = self.nodeGroup[pos].l
rchild = self._update(self.nodeGroup[pos].r,val,index,mid+1,r)
val = min(self.nodeGroup[lchild].val,self.nodeGroup[rchild].val)
return self.newNode(Node(lchild,rchild,val))
def update(self,val,index):
self.root = self._update(self.root,val,index,0,SegmentTreeGroup.MAXN)
# self.nodeGroup[self.root].print()
return self.root
def _query(self,nodeId,ql,qr,l,r):
# print(f"val of id = {nodeId} is {self.nodeGroup[nodeId].val} ; l = {self.nodeGroup[nodeId].l} ,r = {self.nodeGroup[nodeId].r}, ql = {l}, qr = {r}")
if ql <= l and r <= qr:
return self.nodeGroup[nodeId].val
else:
mid = (l+r) >> 1
valL = MAX_VAL
valR = MAX_VAL
if ql <= mid:
valL = self._query(self.nodeGroup[nodeId].l,ql,qr,l,mid)
if qr > mid:
valR = self._query(self.nodeGroup[nodeId].r,ql,qr,mid+1,r)
return min(valL,valR)
def query(self,treeId,ql,qr):
return self._query(self.rootGroup[treeId],ql,qr,0,SegmentTreeGroup.MAXN)
def record(self):
self.rootGroup.append(self.root)
return len(self.rootGroup) - 1
setList = [] # Set集合
numList = NumList()
def debug():
segmentTree = SegmentTreeGroup()
num = segmentTree.build()
print(f"node id in seg = {num}")
segmentTree.update(107,7700)
print(f"root = {segmentTree.root}")
segmentTree.nodeGroup[segmentTree.root].print()
segmentTree.update(109,7901)
tid = segmentTree.record()
segmentTree.update(7,7700)
tid2 = segmentTree.record()
print("tid = {}, q1 = {}".format(tid,segmentTree.query(tid,6800,8000)))
print("tid2 = {}, q2 = {}".format(tid2,segmentTree.query(tid2,6800,8000)))
def main():
#debug()
n,m,k = list(map(int,input().strip().split()))
# print(f"n = {n}, m = {m} ,k = {k}")
for i in range(0,k):
l,r,p = list(map(int,input().strip().split()))
setList.append(Set(l,r,p))
numList.append(l)
numList.append(r)
numList.sort()
setList.sort()
for i in range(0,len(setList)):
setList[i].l = numList.id(setList[i].l)
setList[i].r = numList.id(setList[i].r)
# setList[i].print()
segmentTree = SegmentTreeGroup()
segmentTree.build()
setMaxRecord = [ -1 for x in range(0,n+7) ]
p = 0
for i in numList.numList:
while p < len(setList) and setList[p].r <= i:
segL = setList[p].l
segR = setList[p].r
setNo = setList[p].p
p += 1
if segL > setMaxRecord[setNo]:
setMaxRecord[setNo] = segL
segmentTree.update(segL,setNo)
segmentTree.record()
# print("{}".format(numList._numdic))
for i in range(0,m):
a,b,x,y = list(map(int,input().strip().split()))
# print(f"a = {a},b = {b} , x = {x}, y = {y}")
_,x = numList.binary_search(x)
y,_ = numList.binary_search(y)
# print(f" -- a = {a},b = {b} , x = {x}, y = {y}")
Lmin = segmentTree.query(y,a,b)
# print(f"-- Lmin = {Lmin} ,in set({a},{b})")
if Lmin >= x:
print("yes",flush=True)
else:
print("no",flush=True)
if __name__ == "__main__":
main()
``` | instruction | 0 | 1,691 | 5 | 3,382 |
No | output | 1 | 1,691 | 5 | 3,383 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).
Output
Print a single integer — the number of suitable pairs.
Example
Input
6 3
1 3 9 8 24 1
Output
5
Note
In the sample case, the suitable pairs are:
* a_1 ⋅ a_4 = 8 = 2^3;
* a_1 ⋅ a_6 = 1 = 1^3;
* a_2 ⋅ a_3 = 27 = 3^3;
* a_3 ⋅ a_5 = 216 = 6^3;
* a_4 ⋅ a_6 = 8 = 2^3.
Submitted Solution:
```
from collections import Counter as C
NN = 100000
X = [-1] * (NN+1)
L = [[] for _ in range(NN+1)]
k = 2
while k <= NN:
X[k] = 1
L[k].append(k)
for i in range(k*2, NN+1, k):
X[i] = 0
L[i].append(k)
d = 2
while k**d <= NN:
for i in range(k**d, NN+1, k**d):
L[i].append(k)
d += 1
while k <= NN and X[k] >= 0:
k += 1
P = [i for i in range(NN+1) if X[i] == 1]
K = 4
def free(ct):
a = 1
for c in ct:
a *= c ** (ct[c] % K)
return a
def inv(ct):
a = 1
for c in ct:
a *= c ** (-ct[c] % K)
return a
N, K = map(int, input().split())
A = [int(a) for a in input().split()]
D = {}
ans = 0
for a in A:
c = C(L[a])
invc = inv(c)
if invc in D:
ans += D[invc]
freec = free(c)
if freec in D:
D[freec] += 1
else:
D[freec] = 1
print(ans)
``` | instruction | 0 | 1,766 | 5 | 3,532 |
Yes | output | 1 | 1,766 | 5 | 3,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).
Output
Print a single integer — the number of suitable pairs.
Example
Input
6 3
1 3 9 8 24 1
Output
5
Note
In the sample case, the suitable pairs are:
* a_1 ⋅ a_4 = 8 = 2^3;
* a_1 ⋅ a_6 = 1 = 1^3;
* a_2 ⋅ a_3 = 27 = 3^3;
* a_3 ⋅ a_5 = 216 = 6^3;
* a_4 ⋅ a_6 = 8 = 2^3.
Submitted Solution:
```
import math
from collections import Counter
n, k = map(int, input().split())
a = list(map(int, input().split()))
ans = 0
prev = Counter()
for x in a:
sig = []
p = 2
while p <= math.sqrt(x):
cnt = 0
while x % p == 0:
cnt += 1
x = x // p
cnt = cnt % k
if cnt > 0:
sig.append((p, cnt))
p += 1
if x > 1:
sig.append((x, 1))
com_sig = []
for p, val in sig:
com_sig.append((p, (k - val) % k))
ans += prev[tuple(sig)]
prev[tuple(com_sig)] += 1
print(ans)
``` | instruction | 0 | 1,767 | 5 | 3,534 |
Yes | output | 1 | 1,767 | 5 | 3,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).
Output
Print a single integer — the number of suitable pairs.
Example
Input
6 3
1 3 9 8 24 1
Output
5
Note
In the sample case, the suitable pairs are:
* a_1 ⋅ a_4 = 8 = 2^3;
* a_1 ⋅ a_6 = 1 = 1^3;
* a_2 ⋅ a_3 = 27 = 3^3;
* a_3 ⋅ a_5 = 216 = 6^3;
* a_4 ⋅ a_6 = 8 = 2^3.
Submitted Solution:
```
import math
def F(m):
L=[]
k=2
while True:
if m%k==0:
L.append(k)
m=m//k
else:
k+=1
if k>math.sqrt(m):
break
L.append(m)
k=0
while True:
if k>len(L)-KK:
break
else:
q=False
jt=k
while jt<k+KK-1:
if L[jt]!=L[jt+1]:
q=True
break
jt+=1
if q==False:
L=L[KK:]
k+=1
if not L:
L=[1]
return L
def FFF(a,b):
if len(a)!=len(b):
return False
if a[0][0]==1 and b[0][0]==1:
return True
k=0
while k<len(a):
if a[k][0]!=b[k][0]:
return False
if a[k][1]+b[k][1]!=KK:
return False
k+=1
return True
n,KK=map(int,input().split())
A=map(int,input().split())
B=[]
for i in A:
B.append(F(i))
B.sort()
C=[]
i=0
while i<len(B):
tmp=[]
e=0
while e < len(B[i]):
if len(tmp)==0:
tmp.append([B[i][e],1])
else:
if tmp[-1][0]==B[i][e]:
tmp[-1][1]+=1
else:
tmp.append([B[i][e],1])
e+=1
i+=1
C.append(tmp)
D=[]
last_i=0
i=1
while i<len(C):
if C[i]!=C[i-1]:
D.append([C[i-1],i-last_i])
last_i=i
i+=1
D.append([C[i-1],i-last_i])
ans=0
i=0
while True:
if i>=len(D):
break
k=i
while True:
if k==len(D):
i+=1
break
if FFF(D[i][0],D[k][0]):
if i==k:
ans+=D[i][1]*(D[k][1]-1)//2
i+=1
else:
ans+=D[i][1]*D[k][1]
D.pop(k)
i+=1
break
k+=1
print (ans)
``` | instruction | 0 | 1,771 | 5 | 3,542 |
No | output | 1 | 1,771 | 5 | 3,543 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).
Output
Print a single integer — the number of suitable pairs.
Example
Input
6 3
1 3 9 8 24 1
Output
5
Note
In the sample case, the suitable pairs are:
* a_1 ⋅ a_4 = 8 = 2^3;
* a_1 ⋅ a_6 = 1 = 1^3;
* a_2 ⋅ a_3 = 27 = 3^3;
* a_3 ⋅ a_5 = 216 = 6^3;
* a_4 ⋅ a_6 = 8 = 2^3.
Submitted Solution:
```
n,k=input().split()
n=int(n)
k=int(k)
lst=input().split()
s=0
for i in range(n):
for j in range(n):
if j>i:
if pow(int(lst[i])*int(lst[j]),1.0/k).is_integer() or int(pow(int(lst[i])*int(lst[j]),1.0/k)*10)%10==9:
s+=1
#print(int(lst[i])*int(lst[j]))
print(s)
``` | instruction | 0 | 1,772 | 5 | 3,544 |
No | output | 1 | 1,772 | 5 | 3,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given n positive integers a_1, …, a_n, and an integer k ≥ 2. Count the number of pairs i, j such that 1 ≤ i < j ≤ n, and there exists an integer x such that a_i ⋅ a_j = x^k.
Input
The first line contains two integers n and k (2 ≤ n ≤ 10^5, 2 ≤ k ≤ 100).
The second line contains n integers a_1, …, a_n (1 ≤ a_i ≤ 10^5).
Output
Print a single integer — the number of suitable pairs.
Example
Input
6 3
1 3 9 8 24 1
Output
5
Note
In the sample case, the suitable pairs are:
* a_1 ⋅ a_4 = 8 = 2^3;
* a_1 ⋅ a_6 = 1 = 1^3;
* a_2 ⋅ a_3 = 27 = 3^3;
* a_3 ⋅ a_5 = 216 = 6^3;
* a_4 ⋅ a_6 = 8 = 2^3.
Submitted Solution:
```
import math
n,k = [int(i) for i in input().split()]
ch = sorted([int(i) for i in input().split()])
c = 0
sch = 0
for i in ch[:-1]:
c += 1
for s in ch[c:]:
if s >= i:
r = math.exp(math.log(s*i)/k)
if (r - int(r)) <= 0.000000000000001:
sch += 1
print(sch)
``` | instruction | 0 | 1,773 | 5 | 3,546 |
No | output | 1 | 1,773 | 5 | 3,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's define the function f of multiset a as the multiset of number of occurences of every number, that is present in a.
E.g., f(\{5, 5, 1, 2, 5, 2, 3, 3, 9, 5\}) = \{1, 1, 2, 2, 4\}.
Let's define f^k(a), as applying f to array a k times: f^k(a) = f(f^{k-1}(a)), f^0(a) = a.
E.g., f^2(\{5, 5, 1, 2, 5, 2, 3, 3, 9, 5\}) = \{1, 2, 2\}.
You are given integers n, k and you are asked how many different values the function f^k(a) can have, where a is arbitrary non-empty array with numbers of size no more than n. Print the answer modulo 998 244 353.
Input
The first and only line of input consists of two integers n, k (1 ≤ n, k ≤ 2020).
Output
Print one number — the number of different values of function f^k(a) on all possible non-empty arrays with no more than n elements modulo 998 244 353.
Examples
Input
3 1
Output
6
Input
5 6
Output
1
Input
10 1
Output
138
Input
10 2
Output
33
Submitted Solution:
```
n,k = map(int, input().split())
arr = [5,5,1,2,5,2,3,3,9,5]
``` | instruction | 0 | 1,787 | 5 | 3,574 |
No | output | 1 | 1,787 | 5 | 3,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27. | instruction | 0 | 1,938 | 5 | 3,876 |
Tags: implementation, math, number theory
Correct Solution:
```
n = int(input())
nums = list(map(int, input().split()))
s = input()
ans = 0
for i in range(n):
if s[i] == '1':
ans += nums[i]
cur = 0
pa = ans
for i in range(n):
if s[i] == '1':
ans = max(ans, pa+cur-nums[i])
else:
cur += nums[i]
print(ans)
``` | output | 1 | 1,938 | 5 | 3,877 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27. | instruction | 0 | 1,939 | 5 | 3,878 |
Tags: implementation, math, number theory
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
k = input()
pref = [a[0]]
for i in range(1, n):
pref.append(pref[-1] + a[i])
cur = 0
ans = 0
for i in range(n - 1, -1, -1):
if k[i] == '0':
continue
if cur + (pref[i - 1] if i > 0 else 0) > ans:
ans = cur + (pref[i - 1] if i > 0 else 0)
cur += a[i]
cur = 0
for i in range(n):
if k[i] == '1':
cur += a[i]
ans = max(ans, cur)
print(ans)
``` | output | 1 | 1,939 | 5 | 3,879 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27. | instruction | 0 | 1,942 | 5 | 3,884 |
Tags: implementation, math, number theory
Correct Solution:
```
#!/usr/bin/python3
def readln(): return tuple(map(int, input().split()))
n, = readln()
a = readln()
s = [0] * (n + 1)
sm = [0] * (n + 1)
m = list(input())
for i in range(n):
s[i + 1] = s[i] + a[i]
sm[i + 1] = sm[i] + (a[i] if m[i] == '1' else 0)
ans = sm[n]
for i in range(n - 1, -1, -1):
if m[i] == '1':
ans = max(ans, s[i] + sm[n] - sm[i + 1])
print(ans)
``` | output | 1 | 1,942 | 5 | 3,885 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27. | instruction | 0 | 1,943 | 5 | 3,886 |
Tags: implementation, math, number theory
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = input()
ans = 0
sm = 0
for i in range(n):
if s[i] == '1':
ans = max(ans + a[i], sm)
sm += a[i]
print(ans)
``` | output | 1 | 1,943 | 5 | 3,887 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27. | instruction | 0 | 1,944 | 5 | 3,888 |
Tags: implementation, math, number theory
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
s = input()
m = len(s)
prevSum = [0]*n
prevSum[0] = a[0]
for i in range(1,n):
prevSum[i] = prevSum[i-1]+a[i]
dp = [0]*m
lastOne = -1
if (s[0]!="0"):
lastOne = 0
dp[0] = a[0]
for i in range(1,m):
if (s[i]=="0"):
dp[i] = dp[i-1]
else:
y = 0
if (lastOne!=-1):
y = dp[lastOne]
dp[i] = max(prevSum[i-1],y+a[i])
lastOne = i
print(dp[m-1])
``` | output | 1 | 1,944 | 5 | 3,889 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
b = [0] + a
for i in range(1, n + 1):
b[i] += b[i - 1]
t = input()
s, m, i = 0, 0, t.rfind('1')
while i >= 0:
d = b[i] + s
if d > m: m = d
if d + a[i] < m: break
s += a[i]
i = t[: i].rfind('1')
print(max(m, s))
``` | instruction | 0 | 1,946 | 5 | 3,892 |
Yes | output | 1 | 1,946 | 5 | 3,893 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
import sys
N=int(sys.stdin.readline())
A=list(map(int,sys.stdin.readline().split()))
Sums=[0]*(N)
m=sys.stdin.readline()
maxx=0
s=0
for i in range(N):
s+=A[i]
Sums[i]=s
if(m[i]=="1"):
maxx+=A[i]
ind=N-1
for i in range(N):
if(m[i]!="1"):
ind=i
break
temp_ans=0
conf=""
for i in range(N-1,0,-1):
if(i==ind):
break
if(m[i]=="1"):
temp_ans+=A[i]
if(A[i]<Sums[i-1]):
ans=temp_ans-A[i]+Sums[i-1]
if(ans>maxx):
maxx=ans
print(maxx)
``` | instruction | 0 | 1,947 | 5 | 3,894 |
Yes | output | 1 | 1,947 | 5 | 3,895 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
def makePref(a):
pref = a[:]
for i in range(1,len(pref)):
pref[i] += pref[i-1]
return pref
def makeSuff(a, m):
suff = a[:]
for i in range(len(suff)):
suff[i] *= m[i]
for i in range(len(suff)-2, -1, -1):
suff[i] += suff[i+1]
return suff
def solve(pref, suff, m):
ans = suff[0]
for i in range(1,len(m)):
if m[i] == 1:
if i == len(m)-1:
ans = max(ans, pref[i-1])
else:
ans = max(ans, pref[i-1] + suff[i+1])
return ans
n = int(input())
a = [int(x) for x in input().split()]
m = [int(x) for x in input()]
pref = makePref(a)
suff = makeSuff(a, m)
print(solve(pref, suff, m))
``` | instruction | 0 | 1,948 | 5 | 3,896 |
Yes | output | 1 | 1,948 | 5 | 3,897 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
a=list(map(int,input().split()))
s=list(input().rstrip())
for i in range(len(s)):
s[i]=int(s[i])
ans,x=0,0
for i in range(n):
if s[i]==1:
ans=max(ans+a[i],x)
x+=a[i]
print(ans)
``` | instruction | 0 | 1,949 | 5 | 3,898 |
Yes | output | 1 | 1,949 | 5 | 3,899 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
m = input()
target = 0
k = 0
for i in range(len(m)):
target += (2**i)*int(m[i])
if m[i] == '1':
k = i
total = sum(a[0:k+1])
ele = []
for i in range(k+1):
ele.append([a[i], i])
ele.sort()
num = 2**(k+1) - 1
for i in range(n):
if num > target:
num -= 2**ele[i][1]
total -= ele[i][0]
else:
break
print(total)
``` | instruction | 0 | 1,950 | 5 | 3,900 |
No | output | 1 | 1,950 | 5 | 3,901 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
m = int(input())
listValue = input().split()
x = input().rstrip('\n')
reverse_x = list(x)
reverse_int = [int(s) for s in reverse_x]
while reverse_int[-1] == 0 and len(reverse_int) > 1:
reverse_int.pop(-1)
reverse_len = len(reverse_int)
values = [int(ele) for ele in listValue]
sum_primary = 0
sum_total = 0
for i in range(reverse_len):
sum_total += values[i]
if reverse_int[i] == 1:
sum_primary += values[i]
index = reverse_len - 1
mini = values[index]
while reverse_int[index] == 1 and index >= 1:
mini = min(mini, values[index])
index -= 1
sum_total -= mini
print(max(sum_primary, sum_total))
``` | instruction | 0 | 1,951 | 5 | 3,902 |
No | output | 1 | 1,951 | 5 | 3,903 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
m = input()
i = n-1
while i >= 0:
if m[i] == '1':
break
else:
i -= 1
total_1 = 0
for j in range(i):
total_1 += a[j]
total_2 = 0
for j in range(n):
if m[j] == '1':
total_2 += a[j]
print(max(total_1, total_2))
``` | instruction | 0 | 1,952 | 5 | 3,904 |
No | output | 1 | 1,952 | 5 | 3,905 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Valera has array a, consisting of n integers a0, a1, ..., an - 1, and function f(x), taking an integer from 0 to 2n - 1 as its single argument. Value f(x) is calculated by formula <image>, where value bit(i) equals one if the binary representation of number x contains a 1 on the i-th position, and zero otherwise.
For example, if n = 4 and x = 11 (11 = 20 + 21 + 23), then f(x) = a0 + a1 + a3.
Help Valera find the maximum of function f(x) among all x, for which an inequality holds: 0 ≤ x ≤ m.
Input
The first line contains integer n (1 ≤ n ≤ 105) — the number of array elements. The next line contains n space-separated integers a0, a1, ..., an - 1 (0 ≤ ai ≤ 104) — elements of array a.
The third line contains a sequence of digits zero and one without spaces s0s1... sn - 1 — the binary representation of number m. Number m equals <image>.
Output
Print a single integer — the maximum value of function f(x) for all <image>.
Examples
Input
2
3 8
10
Output
3
Input
5
17 0 10 2 1
11010
Output
27
Note
In the first test case m = 20 = 1, f(0) = 0, f(1) = a0 = 3.
In the second sample m = 20 + 21 + 23 = 11, the maximum value of function equals f(5) = a0 + a2 = 17 + 10 = 27.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
s = input()
ans = 0
sm = 0
for i in range(n):
if s[i] == '1':
ans = max(ans + a[i], sm)
sm += i
print(ans)
``` | instruction | 0 | 1,953 | 5 | 3,906 |
No | output | 1 | 1,953 | 5 | 3,907 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
C*++ language is quite similar to C++. The similarity manifests itself in the fact that the programs written in C*++ sometimes behave unpredictably and lead to absolutely unexpected effects. For example, let's imagine an arithmetic expression in C*++ that looks like this (expression is the main term):
* expression ::= summand | expression + summand | expression - summand
* summand ::= increment | coefficient*increment
* increment ::= a++ | ++a
* coefficient ::= 0|1|2|...|1000
For example, "5*a++-3*++a+a++" is a valid expression in C*++.
Thus, we have a sum consisting of several summands divided by signs "+" or "-". Every summand is an expression "a++" or "++a" multiplied by some integer coefficient. If the coefficient is omitted, it is suggested being equal to 1.
The calculation of such sum in C*++ goes the following way. First all the summands are calculated one after another, then they are summed by the usual arithmetic rules. If the summand contains "a++", then during the calculation first the value of the "a" variable is multiplied by the coefficient, then value of "a" is increased by 1. If the summand contains "++a", then the actions on it are performed in the reverse order: first "a" is increased by 1, then — multiplied by the coefficient.
The summands may be calculated in any order, that's why sometimes the result of the calculation is completely unpredictable! Your task is to find its largest possible value.
Input
The first input line contains an integer a ( - 1000 ≤ a ≤ 1000) — the initial value of the variable "a". The next line contains an expression in C*++ language of the described type. The number of the summands in the expression does not exceed 1000. It is guaranteed that the line describing the expression contains no spaces and tabulation.
Output
Output a single number — the maximal possible value of the expression.
Examples
Input
1
5*a++-3*++a+a++
Output
11
Input
3
a+++++a
Output
8
Note
Consider the second example. Initially a = 3. Suppose that at first the first summand is calculated, and then the second one is. The first summand gets equal to 3, and the value of a is increased by 1. At the calculation of the second summand a is increased once more (gets equal to 5). The value of the second summand is 5, and together they give 8. If we calculate the second summand first and the first summand later, then the both summands equals to 4, and the result is 8, too.
Submitted Solution:
```
import re
n = int(input())
s = input()
f1, f2 = s.count('a++'), s.count('++a')
res = 0
if s.startswith('844*a++'):
res += 1
s = s.replace('a++', 'a')
s = s.replace('++a', 'b')
a, m = [], []
z = 1
t = ''
for i in s:
if i in ['-', '+']:
if z:
a.append(t)
else:
m.append(t)
t = ''
z = 0 if i == '-' else 1
else:
t += i
if t != '':
if z:
a.append(t)
else:
m.append(t)
for i in range(len(a)):
t = a[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if not j.isalpha():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
a[i] = (r, k1, k2)
for i in range(len(m)):
t = m[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if j.isnumeric():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
m[i] = (r, k1, k2)
a.sort(key=lambda x: x[0])
m.sort(key=lambda x: -x[0])
ans = 0
t = n
for i in m:
if i[1] + i[2] == 0:
ans += i[0]
continue
z = i[0]
for j in range(i[1]):
z *= t
t += 1
for j in range(i[2]):
t += 1
z *= t
ans += z
for i in a:
if i[1] + i[2] == 0:
res += i[0]
continue
z = i[0]
for j in range(i[1]):
z *= t
t += 1
for j in range(i[2]):
t += 1
z *= t
res += z
print(res - ans)
``` | instruction | 0 | 1,962 | 5 | 3,924 |
No | output | 1 | 1,962 | 5 | 3,925 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
C*++ language is quite similar to C++. The similarity manifests itself in the fact that the programs written in C*++ sometimes behave unpredictably and lead to absolutely unexpected effects. For example, let's imagine an arithmetic expression in C*++ that looks like this (expression is the main term):
* expression ::= summand | expression + summand | expression - summand
* summand ::= increment | coefficient*increment
* increment ::= a++ | ++a
* coefficient ::= 0|1|2|...|1000
For example, "5*a++-3*++a+a++" is a valid expression in C*++.
Thus, we have a sum consisting of several summands divided by signs "+" or "-". Every summand is an expression "a++" or "++a" multiplied by some integer coefficient. If the coefficient is omitted, it is suggested being equal to 1.
The calculation of such sum in C*++ goes the following way. First all the summands are calculated one after another, then they are summed by the usual arithmetic rules. If the summand contains "a++", then during the calculation first the value of the "a" variable is multiplied by the coefficient, then value of "a" is increased by 1. If the summand contains "++a", then the actions on it are performed in the reverse order: first "a" is increased by 1, then — multiplied by the coefficient.
The summands may be calculated in any order, that's why sometimes the result of the calculation is completely unpredictable! Your task is to find its largest possible value.
Input
The first input line contains an integer a ( - 1000 ≤ a ≤ 1000) — the initial value of the variable "a". The next line contains an expression in C*++ language of the described type. The number of the summands in the expression does not exceed 1000. It is guaranteed that the line describing the expression contains no spaces and tabulation.
Output
Output a single number — the maximal possible value of the expression.
Examples
Input
1
5*a++-3*++a+a++
Output
11
Input
3
a+++++a
Output
8
Note
Consider the second example. Initially a = 3. Suppose that at first the first summand is calculated, and then the second one is. The first summand gets equal to 3, and the value of a is increased by 1. At the calculation of the second summand a is increased once more (gets equal to 5). The value of the second summand is 5, and together they give 8. If we calculate the second summand first and the first summand later, then the both summands equals to 4, and the result is 8, too.
Submitted Solution:
```
import re
class Term:
def __init__(self, factor, type):
self.factor = factor
self.type = type
def __str__(self):
return '(%d * %s)' % (self.factor, self.type)
a = int(input())
s = input().strip()
s = re.sub('\+\+a', 'pre', s)
s = re.sub('a\+\+', 'post', s)
parts = re.split('([+-])', s)
terms = []
for i in range(0, len(parts), 2):
part = parts[i]
sign = 1
if i != 0 and parts[i - 1] == '-':
sign = -1
if '*' not in part:
terms.append(Term(sign * 1, part))
else:
factor, type = part.split('*')
terms.append(Term(sign * int(factor), type))
terms.sort(key = lambda term: term.factor)
#print(', '.join(map(str, terms)))
result = 0
for term in terms:
if term.type == 'pre':
a += 1
result += term.factor * a
if term.type == 'post':
a += 1
print(result)
``` | instruction | 0 | 1,963 | 5 | 3,926 |
No | output | 1 | 1,963 | 5 | 3,927 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
C*++ language is quite similar to C++. The similarity manifests itself in the fact that the programs written in C*++ sometimes behave unpredictably and lead to absolutely unexpected effects. For example, let's imagine an arithmetic expression in C*++ that looks like this (expression is the main term):
* expression ::= summand | expression + summand | expression - summand
* summand ::= increment | coefficient*increment
* increment ::= a++ | ++a
* coefficient ::= 0|1|2|...|1000
For example, "5*a++-3*++a+a++" is a valid expression in C*++.
Thus, we have a sum consisting of several summands divided by signs "+" or "-". Every summand is an expression "a++" or "++a" multiplied by some integer coefficient. If the coefficient is omitted, it is suggested being equal to 1.
The calculation of such sum in C*++ goes the following way. First all the summands are calculated one after another, then they are summed by the usual arithmetic rules. If the summand contains "a++", then during the calculation first the value of the "a" variable is multiplied by the coefficient, then value of "a" is increased by 1. If the summand contains "++a", then the actions on it are performed in the reverse order: first "a" is increased by 1, then — multiplied by the coefficient.
The summands may be calculated in any order, that's why sometimes the result of the calculation is completely unpredictable! Your task is to find its largest possible value.
Input
The first input line contains an integer a ( - 1000 ≤ a ≤ 1000) — the initial value of the variable "a". The next line contains an expression in C*++ language of the described type. The number of the summands in the expression does not exceed 1000. It is guaranteed that the line describing the expression contains no spaces and tabulation.
Output
Output a single number — the maximal possible value of the expression.
Examples
Input
1
5*a++-3*++a+a++
Output
11
Input
3
a+++++a
Output
8
Note
Consider the second example. Initially a = 3. Suppose that at first the first summand is calculated, and then the second one is. The first summand gets equal to 3, and the value of a is increased by 1. At the calculation of the second summand a is increased once more (gets equal to 5). The value of the second summand is 5, and together they give 8. If we calculate the second summand first and the first summand later, then the both summands equals to 4, and the result is 8, too.
Submitted Solution:
```
import re
n = int(input())
s = input()
f1, f2 = s.count('a++'), s.count('++a')
s = s.replace('a++', 'a')
s = s.replace('++a', 'b')
a, m = [], []
z = 1
t = ''
for i in s:
if i in ['-', '+']:
if z:
a.append(t)
else:
m.append(t)
t = ''
z = 0 if i == '-' else 1
else:
t += i
if t != '':
if z:
a.append(t)
else:
m.append(t)
for i in range(len(a)):
t = a[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if not j.isalpha():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
a[i] = (r, k1, k2)
for i in range(len(m)):
t = m[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if j.isnumeric():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
m[i] = (r, k1, k2)
a.sort(key=lambda x: x[0])
m.sort(key=lambda x: -x[0])
ans = 0
t = n
for i in m:
if i[1] + i[2] == 0:
ans += i[0]
continue
z = i[0]
for j in range(i[1]):
z *= t
t += 1
for j in range(i[2]):
t += 1
z *= t
ans += z
res = 0
for i in a:
if i[1] + i[2] == 0:
res += i[0]
continue
z = i[0]
for j in range(i[1]):
z *= t
t += 1
for j in range(i[2]):
t += 1
z *= t
res += z
print(res - ans)
``` | instruction | 0 | 1,964 | 5 | 3,928 |
No | output | 1 | 1,964 | 5 | 3,929 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
C*++ language is quite similar to C++. The similarity manifests itself in the fact that the programs written in C*++ sometimes behave unpredictably and lead to absolutely unexpected effects. For example, let's imagine an arithmetic expression in C*++ that looks like this (expression is the main term):
* expression ::= summand | expression + summand | expression - summand
* summand ::= increment | coefficient*increment
* increment ::= a++ | ++a
* coefficient ::= 0|1|2|...|1000
For example, "5*a++-3*++a+a++" is a valid expression in C*++.
Thus, we have a sum consisting of several summands divided by signs "+" or "-". Every summand is an expression "a++" or "++a" multiplied by some integer coefficient. If the coefficient is omitted, it is suggested being equal to 1.
The calculation of such sum in C*++ goes the following way. First all the summands are calculated one after another, then they are summed by the usual arithmetic rules. If the summand contains "a++", then during the calculation first the value of the "a" variable is multiplied by the coefficient, then value of "a" is increased by 1. If the summand contains "++a", then the actions on it are performed in the reverse order: first "a" is increased by 1, then — multiplied by the coefficient.
The summands may be calculated in any order, that's why sometimes the result of the calculation is completely unpredictable! Your task is to find its largest possible value.
Input
The first input line contains an integer a ( - 1000 ≤ a ≤ 1000) — the initial value of the variable "a". The next line contains an expression in C*++ language of the described type. The number of the summands in the expression does not exceed 1000. It is guaranteed that the line describing the expression contains no spaces and tabulation.
Output
Output a single number — the maximal possible value of the expression.
Examples
Input
1
5*a++-3*++a+a++
Output
11
Input
3
a+++++a
Output
8
Note
Consider the second example. Initially a = 3. Suppose that at first the first summand is calculated, and then the second one is. The first summand gets equal to 3, and the value of a is increased by 1. At the calculation of the second summand a is increased once more (gets equal to 5). The value of the second summand is 5, and together they give 8. If we calculate the second summand first and the first summand later, then the both summands equals to 4, and the result is 8, too.
Submitted Solution:
```
import re
n = int(input())
s = input()
f1, f2 = s.count('a++'), s.count('++a')
s = s.replace('a++', 'a')
s = s.replace('++a', 'b')
a, m = [], []
z = 1
t = ''
for i in s:
if i in ['-', '+']:
if z:
a.append(t)
else:
m.append(t)
t = ''
z = 0 if i == '-' else 1
else:
t += i
if t != '':
if z:
a.append(t)
else:
m.append(t)
for i in range(len(a)):
t = a[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if not j.isalpha():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
a[i] = (r, k1, k2)
for i in range(len(m)):
t = m[i].split('*')
r = 1
k1 = k2 = 0
for j in t:
if j.isnumeric():
r *= int(j)
elif j == 'a':
k1 += 1
else:
k2 += 1
m[i] = (r, k1, k2)
a.sort(key=lambda x: x[0])
m.sort(key=lambda x: -x[0])
ans = 0
t = n
for i in m:
if i[1] + i[2] == 0:
ans += i[0]
continue
for j in range(i[1]):
ans += i[0] * t
t += 1
for j in range(i[2]):
t += 1
ans += i[0] * t
res = 0
for i in a:
if i[1] + i[2] == 0:
res += i[0]
continue
for j in range(i[1]):
res += i[0] * t
t += 1
for j in range(i[2]):
t += 1
res += i[0] * t
print(res - ans)
``` | instruction | 0 | 1,965 | 5 | 3,930 |
No | output | 1 | 1,965 | 5 | 3,931 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,222 | 5 | 4,444 |
"Correct Solution:
```
l,r,d = map(int,input().split())
a = (l-1) // d
b = r // d
s = b - a
print(s)
``` | output | 1 | 2,222 | 5 | 4,445 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,223 | 5 | 4,446 |
"Correct Solution:
```
a = list(map(int,input().split()))
x = (a[1] // a[2]) - ((a[0] - 1) // a[2])
print (x)
``` | output | 1 | 2,223 | 5 | 4,447 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,224 | 5 | 4,448 |
"Correct Solution:
```
L, R, d = map(int, input().split())
print(R // d - L // d + (L % d == 0))
``` | output | 1 | 2,224 | 5 | 4,449 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,225 | 5 | 4,450 |
"Correct Solution:
```
l, r, d = map(int, input().split(' '))
a = (r - l) // d
if r % d == 0:
a += 1
print(a)
``` | output | 1 | 2,225 | 5 | 4,451 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,226 | 5 | 4,452 |
"Correct Solution:
```
l,r,n = map(int, input().split())
L = l//n if l%n!=0 else (l//n)-1
R = r//n
print(R-L)
``` | output | 1 | 2,226 | 5 | 4,453 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,227 | 5 | 4,454 |
"Correct Solution:
```
L, R, d = map(int, input().split())
print(len([x for x in range(L, R+1) if x%d==0]))
``` | output | 1 | 2,227 | 5 | 4,455 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,228 | 5 | 4,456 |
"Correct Solution:
```
l,r,d = map(int, input().split())
sr= r //d
sl= (l-1) //d
print(sr-sl)
``` | output | 1 | 2,228 | 5 | 4,457 |
Provide a correct Python 3 solution for this coding contest problem.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100 | instruction | 0 | 2,229 | 5 | 4,458 |
"Correct Solution:
```
l, r, d = [int(x) for x in input().rstrip().split(" ")]
print(r//d-l//d + (l%d==0))
``` | output | 1 | 2,229 | 5 | 4,459 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
a,b,c=map(int,input().split())
print(b//c-(a-1)//c)
``` | instruction | 0 | 2,230 | 5 | 4,460 |
Yes | output | 1 | 2,230 | 5 | 4,461 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
L, R, d = list(map(int, input().split()))
ans = R // d - (L -1) // d
print(ans)
``` | instruction | 0 | 2,231 | 5 | 4,462 |
Yes | output | 1 | 2,231 | 5 | 4,463 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
l,r,n=map(int,input().split())
a=(r//n)-(l//n if l!=n else 0)
if l!=r:print(a)
else: print(1)
``` | instruction | 0 | 2,232 | 5 | 4,464 |
Yes | output | 1 | 2,232 | 5 | 4,465 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
N, M ,L = map(int, input().split())
print(M//L-(N-1)//L)
``` | instruction | 0 | 2,233 | 5 | 4,466 |
Yes | output | 1 | 2,233 | 5 | 4,467 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
a,b,c=(int(x) for x in input().split())
if a != b:
if b !=c:
if 1 <= a:
if b <= 100:
if a < b:
if 1 <= c <= 100:
print((b-a)//c)
``` | instruction | 0 | 2,234 | 5 | 4,468 |
No | output | 1 | 2,234 | 5 | 4,469 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
l,r,d = map(int,input.split())
print((l-r+1) // d)
``` | instruction | 0 | 2,235 | 5 | 4,470 |
No | output | 1 | 2,235 | 5 | 4,471 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
s = input().split(" ")
L = int(s[0])
R = int(s[1])
d = int(s[2])
cnt = 0
for i in range(L:R+1):
if i%d == 0:
cnt += 1
print(cnt)
``` | instruction | 0 | 2,236 | 5 | 4,472 |
No | output | 1 | 2,236 | 5 | 4,473 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
How many multiples of d are there among the integers between L and R (inclusive)?
Constraints
* All values in input are integers.
* 1 \leq L \leq R \leq 100
* 1 \leq d \leq 100
Input
Input is given from Standard Input in the following format:
L R d
Output
Print the number of multiples of d among the integers between L and R (inclusive).
Examples
Input
5 10 2
Output
3
Input
6 20 7
Output
2
Input
1 100 1
Output
100
Submitted Solution:
```
l,r,d=map(int,input().split())
ans=0
for i in range(l,r+1):
if i%2==0:
ans+=1
print(ans)
``` | instruction | 0 | 2,237 | 5 | 4,474 |
No | output | 1 | 2,237 | 5 | 4,475 |
Provide a correct Python 3 solution for this coding contest problem.
A sequence a=\\{a_1,a_2,a_3,......\\} is determined as follows:
* The first term s is given as input.
* Let f(n) be the following function: f(n) = n/2 if n is even, and f(n) = 3n+1 if n is odd.
* a_i = s when i = 1, and a_i = f(a_{i-1}) when i > 1.
Find the minimum integer m that satisfies the following condition:
* There exists an integer n such that a_m = a_n (m > n).
Constraints
* 1 \leq s \leq 100
* All values in input are integers.
* It is guaranteed that all elements in a and the minimum m that satisfies the condition are at most 1000000.
Input
Input is given from Standard Input in the following format:
s
Output
Print the minimum integer m that satisfies the condition.
Examples
Input
8
Output
5
Input
7
Output
18
Input
54
Output
114 | instruction | 0 | 2,286 | 5 | 4,572 |
"Correct Solution:
```
a = [int(input())]
s = a[0]
while s not in a[0:len(a)-1]:
if s%2 == 0: a.append(s//2)
else:a.append(3*s+1)
s = a[len(a)-1]
print(len(a))
``` | output | 1 | 2,286 | 5 | 4,573 |
Provide a correct Python 3 solution for this coding contest problem.
A sequence a=\\{a_1,a_2,a_3,......\\} is determined as follows:
* The first term s is given as input.
* Let f(n) be the following function: f(n) = n/2 if n is even, and f(n) = 3n+1 if n is odd.
* a_i = s when i = 1, and a_i = f(a_{i-1}) when i > 1.
Find the minimum integer m that satisfies the following condition:
* There exists an integer n such that a_m = a_n (m > n).
Constraints
* 1 \leq s \leq 100
* All values in input are integers.
* It is guaranteed that all elements in a and the minimum m that satisfies the condition are at most 1000000.
Input
Input is given from Standard Input in the following format:
s
Output
Print the minimum integer m that satisfies the condition.
Examples
Input
8
Output
5
Input
7
Output
18
Input
54
Output
114 | instruction | 0 | 2,287 | 5 | 4,574 |
"Correct Solution:
```
s = int(input())
a = []
n=1
while not s in a:
a.append(s)
n += 1
if s%2==0:
s/=2
else: s = 3*s+1
print(n)
``` | output | 1 | 2,287 | 5 | 4,575 |
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