message stringlengths 2 433k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 113 108k | cluster float64 12 12 | __index_level_0__ int64 226 217k |
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Provide a correct Python 3 solution for this coding contest problem.
There are N integers written on a blackboard. The i-th integer is A_i.
Takahashi and Aoki will arrange these integers in a row, as follows:
* First, Takahashi will arrange the integers as he wishes.
* Then, Aoki will repeatedly swap two adjacent integers that are coprime, as many times as he wishes.
We will assume that Takahashi acts optimally so that the eventual sequence will be lexicographically as small as possible, and we will also assume that Aoki acts optimally so that the eventual sequence will be lexicographically as large as possible. Find the eventual sequence that will be produced.
Constraints
* 1 ≦ N ≦ 2000
* 1 ≦ A_i ≦ 10^8
Input
The input is given from Standard Input in the following format:
N
A_1 A_2 … A_N
Output
Print the eventual sequence that will be produced, in a line.
Examples
Input
5
1 2 3 4 5
Output
5 3 2 4 1
Input
4
2 3 4 6
Output
2 4 6 3 | instruction | 0 | 85,615 | 12 | 171,230 |
"Correct Solution:
```
import sys
sys.setrecursionlimit(1000000000)
def gcd(a: int, b:int):
while b:
a,b=b,a%b
return a
def merge(a,us,vs):
i,j,res=0,0,[]
while i<len(us) and j<len(vs):
if a[us[i]]>=a[vs[j]]:
res.append(us[i])
i+=1
else:
res.append(vs[j])
j+=1
return res+us[i:]+vs[j:]
def dfs(g,a,u,vis):
vis[u]=True
res=[]
for v in g[u]:
if not vis[v]:
res=merge(a,res,dfs(g,a,v,vis))
return [u]+res
while 1:
try:
n=int(input())
a=sorted(map(int,input().split()))
except: break
g=[[] for _ in range(n)]
for i in range(n):
for j in range(i+1,n):
if gcd(a[i],a[j])!=1:
g[i].append(j)
g[j].append(i)
vis=[False]*n
res=[]
for u in range(n):
if not vis[u]:
res=merge(a,res,dfs(g,a,u,vis))
print(' '.join(str(a[u]) for u in res))
``` | output | 1 | 85,615 | 12 | 171,231 |
Provide tags and a correct Python 2 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,775 | 12 | 171,550 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
mod=10**9+7
def ni():
return int(raw_input())
def li():
return map(int,raw_input().split())
def pn(n):
stdout.write(str(n)+'\n')
def pa(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
n=ni()
BITTree=[0]*(n+1)
def gs(i):
s = 0
i = i+1
while i > 0:
s += BITTree[i]
i -= i & (-i)
return s
def ub(i ,v):
i += 1
while i <= n:
BITTree[i] += v
i += i & (-i)
def fun(x):
ret=0
sm=0
for i in range(21,-1,-1):
pw=1<<i
if ret+pw<=n and sm+BITTree[ret+pw]<=x:
ret+=pw
sm+=BITTree[ret]
return ret
l=li()
for i in range(n):
ub(i,i)
ans=[0]*n
for i in range(n-1,-1,-1):
ans[i]=fun(l[i])
ub(ans[i],-ans[i])
pa(ans)
``` | output | 1 | 85,775 | 12 | 171,551 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,776 | 12 | 171,552 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
# 1208D
class segTree():
def __init__(self, n):
self.t = [0] * (n << 2)
def update(self, node, l, r, index, value):
if l == r:
self.t[node] = value
return
mid = (l + r) >> 1
if index <= mid:
self.update(node*2, l, mid, index, value)
else:
self.update(node*2 + 1, mid + 1, r, index, value)
self.t[node] = self.t[node*2] + self.t[node*2 + 1]
def query(self, node, l, r, value):
if l == r:
return self.t[node]
mid = (l + r) >> 1
if self.t[node*2] >= value:
return self.query(node*2, l, mid, value)
return self.query(node*2 + 1, mid + 1, r, value - self.t[node*2])
def do():
n = int(input())
nums = [int(i) for i in input().split(" ")]
res = [0]*n
weightTree = segTree(n)
for i in range(1, n+1):
weightTree.update(1, 1, n, i, i)
# print(weightTree.t)
for i in range(n-1, -1, -1):
res[i] = weightTree.query(1, 1, n, nums[i] + 1)
weightTree.update(1, 1, n, res[i], 0)
return " ".join([str(c) for c in res])
print(do())
``` | output | 1 | 85,776 | 12 | 171,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,777 | 12 | 171,554 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
# https://codeforces.com/contest/1208/problem/D
from sys import stdin, stdout
input = stdin.readline
# print = stdout.write
# si is the sum of elements before the i-th element that are smaller than the i-th element.
# For every i from N to 1, let's say the value of the si is x.
# So it means there are k smallest unused numbers whose sum is x.
# We simply put the k+1st number in the output permutation at this i, and continue to move left.
# BIT and binary lifting
# https://codeforces.com/contest/1208/submission/59526098
class BIT:
def __init__(self, nums):
# we store the sum information in bit 1.
# so the indices should be 1 based.
# here we assume nums[0] = 0
self.nums = [0] * (len(nums))
for i, x in enumerate(nums):
if i == 0:
continue
self.update(i, x)
def low_bit(self, x):
return x & (-x)
def update(self, i, diff):
while i < len(self.nums):
self.nums[i] += diff
i += self.low_bit(i)
def prefix_sum(self, i):
ret = 0
while i != 0:
ret += self.nums[i]
i -= self.low_bit(i)
return ret
def search(self, x):
# find the index i such that prefix_sum(i) == x
cur_index, cur_sum = 0, 0
delta = len(self.nums) - 1
while delta - self.low_bit(delta):
delta -= self.low_bit(delta)
while delta:
m = cur_index + delta
if m < len(self.nums):
sm = cur_sum + self.nums[m]
if sm <= x:
cur_index, cur_sum = m, sm
delta //= 2
return cur_index + 1
n = int(input())
bit = BIT(range(n+1))
ans = [0 for _ in range(n)]
nums = list(map(int, input().split()))
for i in range(n - 1, -1, -1):
index = bit.search(nums[i])
bit.update(index, -index)
ans[i] = index
print(*ans)
``` | output | 1 | 85,777 | 12 | 171,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,778 | 12 | 171,556 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
class FTree:
def __init__(self, f):
self.n = len(f)
self.ft = [0] * (self.n + 1)
for i in range(1, self.n + 1):
self.ft[i] += f[i - 1]
if i + self.lsone(i) <= self.n:
self.ft[i + self.lsone(i)] += self.ft[i]
def lsone(self, s):
return s & (-s)
def query(self, i, j):
if i > 1:
return self.query(1, j) - self.query(1, i - 1)
s = 0
while j > 0:
s += self.ft[j]
j -= self.lsone(j)
return s
def update(self, i, v):
while i <= self.n:
self.ft[i] += v
i += self.lsone(i)
def select(self, k):
lo = 1
hi = self.n
for i in range(19): ######## 30
mid = (lo + hi) // 2
if self.query(1, mid) < k:
lo = mid
else:
hi = mid
return hi
n = int(input())
data = [int(i) for i in input().split()]
ft = FTree(list(range(1, n+1)))
ans = [""]*n
for i in range(n-1, -1, -1):
val = data[i]
ind = ft.select(val+1)
ans[i] = str(ind)
ft.update(ind, -ind)
print(" ".join(ans))
``` | output | 1 | 85,778 | 12 | 171,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,779 | 12 | 171,558 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
# https://codeforces.com/contest/1208/problem/D
from sys import stdin, stdout
input = stdin.readline
print = stdout.write
# For every i from N to 1, let's say the value of the si is x.
# So it means there are k smallest unused numbers whose sum is x.
# We simply put the (k+1)st number in the output permutation at this i, and continue to move left.
# segment tree and binary search
_ = input()
x = [int(i) for i in input().split()]
res = []
from math import log
class SegmentTree(object):
def __init__(self, nums):
self.arr = nums
self.l = len(nums)
self.tree = [0] * self.l + nums
for i in range(self.l - 1, 0, -1):
self.tree[i] = self.tree[i << 1] + self.tree[i << 1 | 1]
def update(self, i, val):
n = self.l + i
self.tree[n] = val
while n > 1:
self.tree[n >> 1] = self.tree[n] + self.tree[n ^ 1]
n >>= 1
def query(self, i, j):
m = self.l + i
n = self.l + j
res = 0
while m <= n:
if m & 1:
res += self.tree[m]
m += 1
m >>= 1
if n & 1 == 0:
res += self.tree[n]
n -= 1
n >>= 1
return res
tree = SegmentTree(list(range(1, len(x) + 1)))
org = len(x)
while x:
# from back to forth
q = x.pop()
lo = 0
hi = org - 1
while lo < hi:
mid = (lo + hi) // 2
# print(lo, hi, mid)
sm = tree.query(0, mid)
# print(sm, mid)
if sm > q:
hi = mid
else:
lo = mid + 1
# print(tree.arr, lo, hi)
idx = tree.arr[lo]
# print(idx)
tree.update(lo, 0)
# also from back to forth
res.append(idx)
print(' '.join(str(i) for i in res[::-1]))
``` | output | 1 | 85,779 | 12 | 171,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,780 | 12 | 171,560 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=list(map(int,input().split()))
BIT=[0]*(n+1)
def update(v,w):
while v<=n:
BIT[v]+=w
v+=(v&(-v))
def getvalue(v):
ANS=0
while v!=0:
ANS+=BIT[v]
v-=(v&(-v))
return ANS
for i in range(1,n+1):
update(i,i)
ANS=[-1]*n
for i in range(n-1,-1,-1):
MIN=0
MAX=n
k=A[i]
while True:
x=(MIN+MAX+1)//2
if getvalue(x)>k:
if getvalue(x-1)==k:
ANS[i]=x
break
else:
MAX=x
else:
MIN=x
update(x,-x)
print(*ANS)
``` | output | 1 | 85,780 | 12 | 171,561 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,781 | 12 | 171,562 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
from sys import stdin, stdout
from collections import defaultdict
from collections import deque
import math
import copy
#T = int(input())
N = int(input())
#s1 = input()
#s2 = input()
#N,Q = [int(x) for x in stdin.readline().split()]
arr = [int(x) for x in stdin.readline().split()]
bit = [0]*(N+1)
series = [0] + [x for x in range(N)]
def lowbit(x):
return x&(-x)
def update(idx,delta):
while idx<=N:
bit[idx] += delta
idx += lowbit(idx)
def query(x):
s = 0
while x>0:
s += bit[x]
x -= lowbit(x)
return s
# init
for i in range(1,N+1):
bit[i] += series[i]
y = i + lowbit(i)
if y<=N:
series[y] += series[i]
visited = [0]*(N+1)
ans = [0]*N
for i in range(N-1,-1,-1):
# find
left = 1
right = N
target = arr[i]
while left<=right:
mid = (left+right)//2
q = query(mid)
#print(mid,q)
if q<target:
left = mid + 1
elif q>target:
right = mid - 1
else:
if visited[mid]==1:
left = mid + 1
else:
visited[mid] = 1
ans[i] = mid
break
# update
update(mid+1,-mid)
print(*ans)
``` | output | 1 | 85,781 | 12 | 171,563 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,782 | 12 | 171,564 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
from operator import add
class Stree:
def __init__(self, f, n, default, init_data):
self.ln = 2**(n-1).bit_length()
self.data = [default] * (self.ln * 2)
self.f = f
for i, d in init_data.items():
self.data[self.ln + i] = d
for j in range(self.ln - 1, 0, -1):
self.data[j] = f(self.data[j*2], self.data[j*2+1])
def update(self, i, a):
p = self.ln + i
self.data[p] = a
while p > 1:
p = p // 2
self.data[p] = self.f(self.data[p*2], self.data[p*2+1])
def get(self, i, j):
def _get(l, r, p):
if i <= l and j >= r:
return self.data[p]
else:
m = (l+r)//2
if j <= m:
return _get(l, m, p*2)
elif i >= m:
return _get(m, r, p*2+1)
else:
return self.f(_get(l, m, p*2), _get(m, r, p*2+1))
return _get(0, self.ln, 1)
def find_value(self, v):
def _find_value(l, r, p, v):
if r == l+1:
return l
elif self.data[p*2] <= v:
return _find_value((l+r)//2, r, p*2+1, v - self.data[p*2])
else:
return _find_value(l, (l+r)//2, p*2, v)
return _find_value(0, self.ln, 1, v)
def main():
n = int(input())
sums = {i:i for i in range(n+1)}
stree = Stree(add, n+1, 0, sums)
ss = list(map(int, input().split()))
ss.reverse()
pp = []
for s in ss:
sval = stree.find_value(s)
pp.append(sval)
stree.update(sval,0)
print(*(reversed(pp)))
if __name__ == "__main__":
main()
``` | output | 1 | 85,782 | 12 | 171,565 |
Provide tags and a correct Python 3 solution for this coding contest problem.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10. | instruction | 0 | 85,783 | 12 | 171,566 |
Tags: binary search, data structures, greedy, implementation
Correct Solution:
```
def update(x,val):
while x<=n:
BIT[x]+=val
x+=(x&-x)
def query(x):
s=0
while x>0:
s=(s+BIT[x])
x-=(x&-x)
return s
n=int(input())
BIT=[0]*(n+1)
for i in range(1,n+1):
update(i,i)
arr=list(map(int,input().split()))
answers=[0]*(n)
#print(BIT)
for i in range(n-1,-1,-1):
lol=arr[i]
low=0
fjf=0
high=n
# print(lol)
while True:
mid=(high+low+1)//2
j=query(mid)
# print(mid,j)
# print(answers)
# break
if j>lol:
if query(mid-1)==lol:
answers[i]=mid
update(mid,-mid)
break
else:
high=mid
else:
low=mid
print(*answers)
``` | output | 1 | 85,783 | 12 | 171,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
def sumsegtree(l,seg,st,en,x):
if st==en:
seg[x]=l[st]
else:
mid=(st+en)>>1
sumsegtree(l,seg,st,mid,2*x)
sumsegtree(l,seg,mid+1,en,2*x+1)
seg[x]=seg[2*x]+seg[2*x+1]
def query(seg,st,en,val,x):
if st==en:
return seg[x]
mid=(st+en)>>1
if seg[2*x]>=val:
return query(seg,st,mid,val,2*x)
return query(seg,mid+1,en,val-seg[2*x],2*x+1)
def upd(seg,st,en,ind,val,x):
if st==en:
seg[x]=val
return
mid=(st+en)>>1
if mid>=ind:
upd(seg,st,mid,ind,val,2*x)
else:
upd(seg,mid+1,en,ind,val,2*x+1)
seg[x]=seg[2*x]+seg[2*x+1]
n=int(input())
l=list(map(int,range(1,n+1)))
s=[0]*n
p=list(map(int,input().split()))
seg=["#"]*(n<<2)
sumsegtree(l,seg,0,len(l)-1,1)
for i in range(n-1,-1,-1):
s[i]=query(seg,1,n,p[i]+1,1)
upd(seg,1,n,s[i],0,1)
print (*s)
``` | instruction | 0 | 85,784 | 12 | 171,568 |
Yes | output | 1 | 85,784 | 12 | 171,569 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
from sys import setrecursionlimit as SRL, stdin
SRL(10 ** 7)
rd = stdin.readline
rrd = lambda: map(int, rd().strip().split())
n = int(rd())
bit = [0] * 200005
def add(x, val):
while x <= n:
bit[x] += val
x += (x & -x)
def query(x):
num = 0
for i in range(30, -1, -1):
if num+(1 << i) <= n and bit[num + (1 << i)] <= x:
x -= bit[num + (1 << i)]
num += (1 << i)
return num + 1
for i in range(1, n + 1):
add(i, i)
s = list(rrd())
ans = []
for i in range(len(s) - 1, -1, -1):
q = query(s[i])
ans.append(q)
add(q, -q)
ans = ans[::-1]
print(*ans)
``` | instruction | 0 | 85,785 | 12 | 171,570 |
Yes | output | 1 | 85,785 | 12 | 171,571 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
import sys
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
#sys.setrecursionlimit(300000)
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")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default=0, func=lambda a, b: max(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] < key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def binary(x, length):
y = bin(x)[2:]
return y if len(y) >= length else "0" * (length - len(y)) + y
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
n=int(input())
l=list(map(int,input().split()))
f=[i for i in range(1,n+1)]
s=SegmentTree(f)
ans=[0]*n
for i in range(n-1,-1,-1):
st=1
end=n
while(st<=end):
mid=(st+end)//2
su=s.query(0,mid-2)
if su==l[i]:
an=mid
st=mid+1
elif su<l[i]:
st=mid+1
else:
end=mid-1
ans[i]=an
s.__setitem__(an-1,0)
print(*ans,sep=" ")
``` | instruction | 0 | 85,786 | 12 | 171,572 |
Yes | output | 1 | 85,786 | 12 | 171,573 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
from sys import stdin,stdout
class Tree(object):
def __init__(self,n):
self.tree=[0]*(4*n+10)
def update(self,L,C,l,r,rt):
if l==r:
self.tree[rt]+=C
return
mid=(l+r)//2
if L<=mid:
self.update(L,C,l,mid,rt<<1)
else:
self.update(L,C,mid+1,r,rt<<1|1)
self.tree[rt]=self.tree[rt<<1]+self.tree[rt<<1|1]
def query(self,s,l,r,rt):
if l==r:
return l
mid=(l+r)//2
if self.tree[rt<<1]>s:
return self.query(s,l,mid,rt<<1)
else:
return self.query(s-self.tree[rt<<1],mid+1,r,rt<<1|1)
if __name__ == '__main__':
n=int(stdin.readline())
a=list(map(int,stdin.readline().split()))
b=[0]*(n+10)
seg=Tree(n)
for i in range(n):
seg.update(i+1,i+1,1,n,1)
for i in range(n,0,-1):
b[i]=seg.query(a[i-1],1,n,1)
seg.update(b[i],-b[i],1,n,1)
for i in range(n):
stdout.write('%d '%(b[i+1]))
``` | instruction | 0 | 85,787 | 12 | 171,574 |
Yes | output | 1 | 85,787 | 12 | 171,575 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
n=int(input())
a=[]
b={}
for i in input().split():
a.append(int(i))
b[int(i)]=True
qq=list(b.keys())
qq=sorted(qq)
ii=1
b=[0 for i in range(n)]
for i in qq:
for i2 in range(n-1,-1,-1):
if a[i2]==i:
b[i2]=ii
ii+=1
for i in b:
print(i,end=' ')
``` | instruction | 0 | 85,788 | 12 | 171,576 |
No | output | 1 | 85,788 | 12 | 171,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
n = int(input())
arr = list(map(int, input().split()))
res = [0]*n
for i in range(n-1,-1,-1):
le = int((2*arr[i])**0.5)
res[i] = le+1
index = n-1
while index >= 0:
temp = res[index]
l = 0
while res[index] == temp and index >= 0:
res[index] = str(temp+l)
l += 1
index -= 1
print(' '.join(res))
``` | instruction | 0 | 85,789 | 12 | 171,578 |
No | output | 1 | 85,789 | 12 | 171,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
t=int(input())
l1=[int(num) for num in input().split()]
k=0
c=1
l3=[]
l=0
for i in range(1,t):
if(l1[i]==l1[i-1]) or (l1[i]==0):
c+=1
continue
else:
if(l1[i-1]==0):
if(i!=1):
l=c
c=c-1
p=int(((l1[i]-k)-(((c-1)*c)/2))/c)
for j in range(1,l+1):
l3.append(p+(c-j))
if(i==t-1):
l3.append(p+c)
else:
l=c
p=int(((l1[i]-k)-(((c-1)*c)/2))/c)
for j in range(1,l+1):
l3.append(p+(l-j))
if(i==t-1):
l3.append(p+c)
else:
l=c
p=int(((l1[i]-k)-(((c-1)*c)/2))/c)
for j in range(1,l+1):
l3.append(p+(l-j))
if(i==t-1):
l3.append(p+c)
c=1
k=l1[i]
if(c==t):
for i in range(1,t+1):
l3.append(t-i+1)
for i in range(0,t):
print(l3[i],end=" ")
``` | instruction | 0 | 85,790 | 12 | 171,580 |
No | output | 1 | 85,790 | 12 | 171,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
An array of integers p_{1},p_{2}, …,p_{n} is called a permutation if it contains each number from 1 to n exactly once. For example, the following arrays are permutations: [3,1,2], [1], [1,2,3,4,5] and [4,3,1,2]. The following arrays are not permutations: [2], [1,1], [2,3,4].
There is a hidden permutation of length n.
For each index i, you are given s_{i}, which equals to the sum of all p_{j} such that j < i and p_{j} < p_{i}. In other words, s_i is the sum of elements before the i-th element that are smaller than the i-th element.
Your task is to restore the permutation.
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^{5}) — the size of the permutation.
The second line contains n integers s_{1}, s_{2}, …, s_{n} (0 ≤ s_{i} ≤ (n(n-1))/(2)).
It is guaranteed that the array s corresponds to a valid permutation of length n.
Output
Print n integers p_{1}, p_{2}, …, p_{n} — the elements of the restored permutation. We can show that the answer is always unique.
Examples
Input
3
0 0 0
Output
3 2 1
Input
2
0 1
Output
1 2
Input
5
0 1 1 1 10
Output
1 4 3 2 5
Note
In the first example for each i there is no index j satisfying both conditions, hence s_i are always 0.
In the second example for i = 2 it happens that j = 1 satisfies the conditions, so s_2 = p_1.
In the third example for i = 2, 3, 4 only j = 1 satisfies the conditions, so s_2 = s_3 = s_4 = 1. For i = 5 all j = 1, 2, 3, 4 are possible, so s_5 = p_1 + p_2 + p_3 + p_4 = 10.
Submitted Solution:
```
n=int(input())
a=[]
b={}
for i in input().split():
a.append(int(i))
b[int(i)]=True
qq=list(b.keys())
ii=1
b=[0 for i in range(n)]
for i in qq:
for i2 in range(n-1,-1,-1):
if a[i2]==i:
b[i2]=ii
ii+=1
for i in b:
print(i,end=' ')
``` | instruction | 0 | 85,791 | 12 | 171,582 |
No | output | 1 | 85,791 | 12 | 171,583 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,792 | 12 | 171,584 |
Tags: data structures, greedy
Correct Solution:
```
n=int(input())
s=sorted([[v,-i] for i,v in enumerate(map(int,input().split()))])
for _ in range(int(input())):
k,i=map(int,input().split())
ans=sorted(s[-k:],key=lambda x:-x[1])
print(ans[i-1][0])
``` | output | 1 | 85,792 | 12 | 171,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,793 | 12 | 171,586 |
Tags: data structures, greedy
Correct Solution:
```
def main():
n = int(input())
a = list(enumerate(map(int, (input().split()))))
a.sort(key = lambda item: (item[1], -item[0]))
#print(a)
m = int(input())
for i in range(m):
k, pos = map(int, input().split())
s = a[-k:]
s = sorted(s)
print(s[pos - 1][1])
main()
``` | output | 1 | 85,793 | 12 | 171,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,794 | 12 | 171,588 |
Tags: data structures, greedy
Correct Solution:
```
n = int(input())
a = [int(x) for x in input().split()]
a_sorted = sorted(a, reverse=True)
m = int(input())
for i in range(m):
k, pos = [int(x) for x in input().split()]
vals = a_sorted[0:k]
result = []
for val in a:
if val in vals:
result.append(val)
vals.remove(val)
if len(vals) == 0:
break
print(result[pos - 1])
``` | output | 1 | 85,794 | 12 | 171,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,795 | 12 | 171,590 |
Tags: data structures, greedy
Correct Solution:
```
n = int(input())
b = list(map(int,input().split()))
a = [[0] * 2 for i in range(n)]
for i in range(n):
a[i][0] = b[i]
a[i][1] = i
for i in range(n-1):
for j in range(n-i-1):
if a[j][0] > a[j+1][0]:
a[j], a[j+1] = a[j+1], a[j]
elif (a[j][0] == a[j + 1][0]) and (a[j][1] < a[j + 1][1]):
a[j], a[j+1] = a[j+1], a[j]
m = int(input())
for k in range(m):
k, ind = map(int,input().split())
ans = [[0] * 2 for i in range(k)]
for i in range(k):
ans[i][0] = a[n - i - 1][1]
ans[i][1] = a[n - i - 1][0]
ans.sort()
print(ans[ind - 1][1])
``` | output | 1 | 85,795 | 12 | 171,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,796 | 12 | 171,592 |
Tags: data structures, greedy
Correct Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
from itertools import permutations
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")
#-------------------game starts now-----------------------------------------------------
n=int(input())
a1=list(map(int,input().split()))
a=list()
for i in range (n):
a.append((a1[i],n-i))
a.sort(reverse=True)
m=int(input())
for i in range (m):
k,pos=map(int,input().split())
c=list()
for j in range (k):
c.append(n-a[j][1])
c.sort()
res=a1[c[pos-1]]
print(res)
``` | output | 1 | 85,796 | 12 | 171,593 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,797 | 12 | 171,594 |
Tags: data structures, greedy
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
b = sorted(a)
b.reverse()
m = int(input())
for _ in range(m):
k, pos = map(int, input().split())
x = 0
j = 0
d = dict()
ans = 0
for i in b[:k]:
d[i] = d.get(i, 0) + 1
while x != pos:
u = a[j]
if u in d:
d[u] -= 1
x += 1
if d[u] == 0:
del d[u]
j += 1
ans = u
print(ans)
``` | output | 1 | 85,797 | 12 | 171,595 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,798 | 12 | 171,596 |
Tags: data structures, greedy
Correct Solution:
```
n = int(input())
line = list(map(int, input().split()))
m = int(input())
for _ in range(m):
k, pos = map(int, input().split())
lf = line[:]
while len(lf) > k:
j = -1
x = min(lf)
while j > -len(lf)-1:
if lf[j] == x:
del lf[j]
if len(lf) == k:
break
else:
j -= 1
print(lf[pos-1])
``` | output | 1 | 85,798 | 12 | 171,597 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10]. | instruction | 0 | 85,799 | 12 | 171,598 |
Tags: data structures, greedy
Correct Solution:
```
def mergesort(l, r, arr, pos):
if r - l == 1:
return arr, pos
m = (l + r) // 2
arr, pos = mergesort(l, m, arr, pos)
arr, pos = mergesort(m, r, arr, pos)
c = [0 for i in range(r)]
d = [0 for i in range(r)]
poi_a = l
poi_b = m
for i in range(l, r):
if poi_a == m:
c[i] = arr[poi_b]
d[i] = pos[poi_b]
poi_b += 1
elif poi_b == r:
c[i] = arr[poi_a]
d[i] = pos[poi_a]
poi_a += 1
elif a[poi_a] > arr[poi_b]:
c[i] = arr[poi_a]
d[i] = pos[poi_a]
poi_a += 1
else:
c[i] = arr[poi_b]
d[i] = pos[poi_b]
poi_b += 1
for i in range(l, r):
arr[i] = c[i]
pos[i] = d[i]
return arr, pos
n = int(input())
a = list(map(int, input().split()))
p = [i for i in range(n)]
temp = a[:]
a, p = mergesort(0, n, a, p)
for m in range(int(input())):
k, pos = map(int, input().split())
j = k
while j < n and a[j - 1] == a[j]:
j += 1
i = k - 1
l = 1
while i > 0 and a[i - 1] == a[i]:
i -= 1
l += 1
m = sorted(p[i:j])
res = sorted(m[:l] + p[:i])
print(temp[res[pos - 1]])
``` | output | 1 | 85,799 | 12 | 171,599 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n = int(input())
a = list(map(int,input().split()))
for i in range(n):
a[i] = (a[i],-i)
a.sort()
m = int(input())
for j in range(m):
t = list(map(int, input().split()))
k = t[0]
p = t[1]
print(sorted(a[-k:], key = lambda x: -x[1])[p-1][0])
``` | instruction | 0 | 85,800 | 12 | 171,600 |
Yes | output | 1 | 85,800 | 12 | 171,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n = int(input())
s = list(map(int, input().split()))
for i in range(n):
s[i] = (s[i], i)
mx = sorted(s, key=lambda y: y[0], reverse=True)
m = int(input())
for i in range(m):
k, pos = map(int, input().split())
pos -= 1
a = 0
last = mx[0][0]
res = []
d = dict()
d2 = dict()
for x in range(k):
if mx[x][0] not in d:
d[mx[x][0]] = 1
else:
d[mx[x][0]] += 1
for x in range(n):
if mx[x][0] != last:
a += 1
last = mx[x][0]
if a == k:
break
if mx[x][0] not in d2:
d2[mx[x][0]] = [mx[x][1]]
else:
d2[mx[x][0]].append(mx[x][1])
for a in list(d.keys()):
d2[a].sort()
for j in range(d[a]):
res.append((a, d2[a][j]))
res = sorted(res, key=lambda y: y[1])
print(res[pos][0])
``` | instruction | 0 | 85,801 | 12 | 171,602 |
Yes | output | 1 | 85,801 | 12 | 171,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n = int(input())
data = list(map(int, input().split()))
m = int(input())
for _ in range(m):
k, pos = map(int, input().split())
s = data[:]
ans = []
for i in range(k):
x = s.index(max(s))
ans.append(x)
s[x] = -1
ans.sort()
print(data[ans[pos - 1]])
``` | instruction | 0 | 85,802 | 12 | 171,604 |
Yes | output | 1 | 85,802 | 12 | 171,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n=int(input())
a=[int(i) for i in input().split()]
b=sorted(a)
c=[]
c.append(a)
for i in range(1,n+1):
k=len(c[i-1])-1-c[i-1][::-1].index(b[i-1])
c.append(c[i-1][0:k]+c[i-1][k+1::])
m=int(input())
for i in range (m):
k,pos=map(int,input().split())
print (c[len(c)-k-1][pos-1])
``` | instruction | 0 | 85,803 | 12 | 171,606 |
Yes | output | 1 | 85,803 | 12 | 171,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n = int(input())
arr = [int(i) for i in input().split()]
sor = [[arr[i], n - i] for i in range(n)]
sor.sort()
m = int(input())
print (sor)
for i in range(m):
op = []
[q, index] = [int(i) for i in input().split()]
for j in range(q):
op.append(n - sor[-1-j][1])
print (op)
op.sort()
print(arr[op[index - 1]])
``` | instruction | 0 | 85,804 | 12 | 171,608 |
No | output | 1 | 85,804 | 12 | 171,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
from sys import stdin,stdout
from math import gcd,sqrt,factorial,pi
from collections import deque,defaultdict
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
lcm=lambda x,y:(x*y)//gcd(x,y)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
sm=lambda x:(x**2+x)//2
N=10**9+7
n=I()
*a,=R()
d={}
for i in a:
if i not in d:d[i]=0
d[i]+=1
for _ in range(I()):
k,p=R()
m=[]
cnt=0
for i in sorted(d,reverse=True):
cnt+=d[i]
m+=i,
if cnt>=k:break
m=[i for i in a if i in m]
ans=m[:k]
sm=sum(ans)
for i in range(k,len(m)):
v=sum(m[k-i+1:i+1])
if v>sm:
sm=v
ans=m[k-i+1:i+1]
elif v==sm and m[k-i+1:i+1]<ans:
ans=m[k-i+1:i+1]
print(ans[p-1])
``` | instruction | 0 | 85,805 | 12 | 171,610 |
No | output | 1 | 85,805 | 12 | 171,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
def mergesort(l, r, arr, pos):
if r - l == 1:
return arr, pos
m = (l + r) // 2
arr, pos = mergesort(l, m, arr, pos)
arr, pos = mergesort(m, r, arr, pos)
c = [0 for i in range(r)]
d = [0 for i in range(r)]
poi_a = l
poi_b = m
for i in range(l, r):
if poi_a == m:
c[i] = arr[poi_b]
d[i] = pos[poi_b]
poi_b += 1
elif poi_b == r:
c[i] = arr[poi_a]
d[i] = pos[poi_a]
poi_a += 1
elif a[poi_a] > arr[poi_b]:
c[i] = arr[poi_a]
d[i] = pos[poi_a]
poi_a += 1
else:
c[i] = arr[poi_b]
d[i] = pos[poi_b]
poi_b += 1
for i in range(l, r):
arr[i] = c[i]
pos[i] = d[i]
return arr, pos
n = int(input())
a = list(map(int, input().split()))
p = [i for i in range(n)]
temp = a[:]
a, p = mergesort(0, n, a, p)
pref = []
i = 0
while i < n:
j = i + 1
if j < n and a[i] == a[j]:
j += 1
pref.append([i, j])
i = j
for m in range(int(input())):
k, pos = map(int, input().split())
for t in range(len(pref)):
if pref[t][0] <= k - 1 < pref[t][1]:
i = pref[t][0]
j = pref[t][1]
l = k - 1 - i + 1
m = sorted(p[i:j])
res = sorted(m[:l] + p[:i])
print(temp[res[pos - 1]])
``` | instruction | 0 | 85,806 | 12 | 171,612 |
No | output | 1 | 85,806 | 12 | 171,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the easier version of the problem. In this version 1 ≤ n, m ≤ 100. You can hack this problem only if you solve and lock both problems.
You are given a sequence of integers a=[a_1,a_2,...,a_n] of length n. Its subsequence is obtained by removing zero or more elements from the sequence a (they do not necessarily go consecutively). For example, for the sequence a=[11,20,11,33,11,20,11]:
* [11,20,11,33,11,20,11], [11,20,11,33,11,20], [11,11,11,11], [20], [33,20] are subsequences (these are just some of the long list);
* [40], [33,33], [33,20,20], [20,20,11,11] are not subsequences.
Suppose that an additional non-negative integer k (1 ≤ k ≤ n) is given, then the subsequence is called optimal if:
* it has a length of k and the sum of its elements is the maximum possible among all subsequences of length k;
* and among all subsequences of length k that satisfy the previous item, it is lexicographically minimal.
Recall that the sequence b=[b_1, b_2, ..., b_k] is lexicographically smaller than the sequence c=[c_1, c_2, ..., c_k] if the first element (from the left) in which they differ less in the sequence b than in c. Formally: there exists t (1 ≤ t ≤ k) such that b_1=c_1, b_2=c_2, ..., b_{t-1}=c_{t-1} and at the same time b_t<c_t. For example:
* [10, 20, 20] lexicographically less than [10, 21, 1],
* [7, 99, 99] is lexicographically less than [10, 21, 1],
* [10, 21, 0] is lexicographically less than [10, 21, 1].
You are given a sequence of a=[a_1,a_2,...,a_n] and m requests, each consisting of two numbers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j). For each query, print the value that is in the index pos_j of the optimal subsequence of the given sequence a for k=k_j.
For example, if n=4, a=[10,20,30,20], k_j=2, then the optimal subsequence is [20,30] — it is the minimum lexicographically among all subsequences of length 2 with the maximum total sum of items. Thus, the answer to the request k_j=2, pos_j=1 is the number 20, and the answer to the request k_j=2, pos_j=2 is the number 30.
Input
The first line contains an integer n (1 ≤ n ≤ 100) — the length of the sequence a.
The second line contains elements of the sequence a: integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
The third line contains an integer m (1 ≤ m ≤ 100) — the number of requests.
The following m lines contain pairs of integers k_j and pos_j (1 ≤ k ≤ n, 1 ≤ pos_j ≤ k_j) — the requests.
Output
Print m integers r_1, r_2, ..., r_m (1 ≤ r_j ≤ 10^9) one per line: answers to the requests in the order they appear in the input. The value of r_j should be equal to the value contained in the position pos_j of the optimal subsequence for k=k_j.
Examples
Input
3
10 20 10
6
1 1
2 1
2 2
3 1
3 2
3 3
Output
20
10
20
10
20
10
Input
7
1 2 1 3 1 2 1
9
2 1
2 2
3 1
3 2
3 3
1 1
7 1
7 7
7 4
Output
2
3
2
3
2
3
1
1
3
Note
In the first example, for a=[10,20,10] the optimal subsequences are:
* for k=1: [20],
* for k=2: [10,20],
* for k=3: [10,20,10].
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
a1 = sorted(a, key=lambda x: -x)
m = int(input())
for i in range(m):
k, p = list(map(int, input().split()))
c = {}
s = []
for j in range(n+1):
s.append({})
pos = {}
for j in range(n):
c[a1[j]] = 0
s[0][a[j]] = 0
pos[a1[j]] = []
for j in range(n):
s[0][a[j]] += 1
for j in range(n):
for t in range(n):
s[j+1][a[t]] = s[j][a[t]]
s[j+1][a[j]] -= 1
b = []
for j in range(k):
c[a1[j]]+=1
ns = []
for j in range(n):
if c[a[j]] ^ 0:
pos[a[j]].append(j)
if len(pos[a[j]]) == 1:
ns.append(a[j])
ns = sorted(ns)
counter = 0
ans = []
minpos = 0
while counter ^ k:
flag = 0
for j in range(len(ns)):
c[ns[j]] -= 1
flag1 = 0
for f in range(len(pos[ns[j]])):
if minpos <= pos[ns[j]][f]:
for u in range(len(ns)):
if s[pos[ns[j]][f]+1][ns[u]] < c[ns[u]]:
flag1 = 1
break
if flag1 == 1:
break
else:
minpos = pos[ns[j]][f]+1
flag = 1
break
if flag == 1:
ans.append(ns[j])
counter += 1
break
else:
c[ns[j]] += 1
print(ans[p-1])
``` | instruction | 0 | 85,807 | 12 | 171,614 |
No | output | 1 | 85,807 | 12 | 171,615 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
Let f(k) be the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation.
You need to find f(1), f(2), …, f(n).
Input
The first line of input contains one integer n (1 ≤ n ≤ 200 000): the number of elements in the permutation.
The next line of input contains n integers p_1, p_2, …, p_n: given permutation (1 ≤ p_i ≤ n).
Output
Print n integers, the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation, for k=1, 2, …, n.
Examples
Input
5
5 4 3 2 1
Output
0 1 3 6 10
Input
3
1 2 3
Output
0 0 0 | instruction | 0 | 85,808 | 12 | 171,616 |
Tags: binary search, data structures
Correct Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
pos, pb, ps = [[0] * (n + 1) for x in range(3)]
def add(bit, i, val):
while i <= n:
bit[i] += val
i += i & -i
def sum(bit, i):
res = 0
while i > 0:
res += bit[i]
i -= i & -i
return res
def find(bit, sum):
i, t = 0, 0
if sum == 0:
return 0
for k in range(17, -1, -1):
i += 1 << k
if i <= n and t + bit[i] < sum:
t += bit[i]
else:
i -= 1 << k
return i + 1
for i in range(1, n + 1):
pos[a[i]] = i
invSum = 0
totalSum = 0
for i in range(1, n + 1):
totalSum += pos[i]
invSum += i - sum(pb, pos[i]) - 1
add(pb, pos[i], 1)
add(ps, pos[i], pos[i])
mid = find(pb, i // 2)
if i % 2 == 1:
mid2 = find(pb, i // 2 + 1)
seqSum = (i + 1) * (i // 2) // 2
else:
mid2 = mid
seqSum = i * (i // 2) // 2
leftSum = sum(ps, mid)
rightSum = totalSum - sum(ps, mid2)
print(rightSum - leftSum - seqSum + invSum, end=" ")
``` | output | 1 | 85,808 | 12 | 171,617 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
Let f(k) be the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation.
You need to find f(1), f(2), …, f(n).
Input
The first line of input contains one integer n (1 ≤ n ≤ 200 000): the number of elements in the permutation.
The next line of input contains n integers p_1, p_2, …, p_n: given permutation (1 ≤ p_i ≤ n).
Output
Print n integers, the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation, for k=1, 2, …, n.
Examples
Input
5
5 4 3 2 1
Output
0 1 3 6 10
Input
3
1 2 3
Output
0 0 0 | instruction | 0 | 85,809 | 12 | 171,618 |
Tags: binary search, data structures
Correct Solution:
```
from bisect import bisect_right, bisect_left
# instead of AVLTree
class BITbisect():
def __init__(self, InputProbNumbers):
# 座圧
self.ind_to_co = [-10**18]
self.co_to_ind = {}
for ind, num in enumerate(sorted(list(set(InputProbNumbers)))):
self.ind_to_co.append(num)
self.co_to_ind[num] = ind+1
self.max = len(self.co_to_ind)
self.data = [0]*(self.max+1)
def __str__(self):
retList = []
for i in range(1, self.max+1):
x = self.ind_to_co[i]
if self.count(x):
c = self.count(x)
for _ in range(c):
retList.append(x)
return "[" + ", ".join([str(a) for a in retList]) + "]"
def __getitem__(self, key):
key += 1
s = 0
ind = 0
l = self.max.bit_length()
for i in reversed(range(l)):
if ind + (1<<i) <= self.max:
if s + self.data[ind+(1<<i)] < key:
s += self.data[ind+(1<<i)]
ind += (1<<i)
if ind == self.max or key < 0:
raise IndexError("BIT index out of range")
return self.ind_to_co[ind+1]
def __len__(self):
return self._query_sum(self.max)
def __contains__(self, num):
if not num in self.co_to_ind:
return False
return self.count(num) > 0
# 0からiまでの区間和
# 左に進んでいく
def _query_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
# i番目の要素にxを足す
# 上に登っていく
def _add(self, i, x):
while i <= self.max:
self.data[i] += x
i += i & -i
# 値xを挿入
def push(self, x):
if not x in self.co_to_ind:
raise KeyError("The pushing number didnt initialized")
self._add(self.co_to_ind[x], 1)
# 値xを削除
def delete(self, x):
if not x in self.co_to_ind:
raise KeyError("The deleting number didnt initialized")
if self.count(x) <= 0:
raise ValueError("The deleting number doesnt exist")
self._add(self.co_to_ind[x], -1)
# 要素xの個数
def count(self, x):
return self._query_sum(self.co_to_ind[x]) - self._query_sum(self.co_to_ind[x]-1)
# 値xを超える最低ind
def bisect_right(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_right(self.ind_to_co, x) - 1
return self._query_sum(i)
# 値xを下回る最低ind
def bisect_left(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_left(self.ind_to_co, x)
if i == 1:
return 0
return self._query_sum(i-1)
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
Ind = [0]*(N+1)
for i, a in enumerate(A):
Ind[a] = i+1
Bit = BITbisect(list(range(N+1)))
ans = [0]
Bit.push(Ind[1])
a = 0
for n in range(2, N+1):
ind = Ind[n]
f = Bit.bisect_left(ind)
#print(Bit)
l = len(Bit)
if l%2 == 0:
if f == l//2:
a += l//2-l//2
elif f < l//2:
p1 = Bit[l//2-1]
a += (p1-ind-1) - (l//2-1) + l-f
else:
p2 = Bit[l//2]
a += (ind-p2-1) - (l//2-1) + l-f
else:
p1 = Bit[l//2]
#print(f, p1, ind, l)
if f <= l//2:
a += (p1-ind-1) - l//2 + l-f
else:
a += (ind-p1-1) - l//2 + l-f
ans.append(a)
Bit.push(ind)
print(*ans, sep=" ")
``` | output | 1 | 85,809 | 12 | 171,619 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
Let f(k) be the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation.
You need to find f(1), f(2), …, f(n).
Input
The first line of input contains one integer n (1 ≤ n ≤ 200 000): the number of elements in the permutation.
The next line of input contains n integers p_1, p_2, …, p_n: given permutation (1 ≤ p_i ≤ n).
Output
Print n integers, the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation, for k=1, 2, …, n.
Examples
Input
5
5 4 3 2 1
Output
0 1 3 6 10
Input
3
1 2 3
Output
0 0 0
Submitted Solution:
```
n = int(input())
a = [0] + list(map(int, input().split()))
pos, bit = [[0] * (n+1), [0] * (n+1)]
def add(i):
while i <= n:
bit[i] += 1
i += i & -i
def sum(i):
res = 0
while i > 0:
res += bit[i]
i -= i & -i
return res
for i in range(1, n+1):
pos[a[i]] = i
Max, Min, invSum = [0, n, 0]
for i in range(1, n+1):
Max = max(Max, pos[i])
Min = min(Min, pos[i])
invSum += i - sum(pos[i]) - 1
print(Max - Min - i + invSum + 1, end=' ')
add(pos[i])
``` | instruction | 0 | 85,810 | 12 | 171,620 |
No | output | 1 | 85,810 | 12 | 171,621 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a permutation p_1, p_2, …, p_n.
In one move you can swap two adjacent values.
You want to perform a minimum number of moves, such that in the end there will exist a subsegment 1,2,…, k, in other words in the end there should be an integer i, 1 ≤ i ≤ n-k+1 such that p_i = 1, p_{i+1} = 2, …, p_{i+k-1}=k.
Let f(k) be the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation.
You need to find f(1), f(2), …, f(n).
Input
The first line of input contains one integer n (1 ≤ n ≤ 200 000): the number of elements in the permutation.
The next line of input contains n integers p_1, p_2, …, p_n: given permutation (1 ≤ p_i ≤ n).
Output
Print n integers, the minimum number of moves that you need to make a subsegment with values 1,2,…,k appear in the permutation, for k=1, 2, …, n.
Examples
Input
5
5 4 3 2 1
Output
0 1 3 6 10
Input
3
1 2 3
Output
0 0 0
Submitted Solution:
```
from bisect import bisect_right, bisect_left
# instead of AVLTree
class BITbisect():
def __init__(self, InputProbNumbers):
# 座圧
self.ind_to_co = [-10**18]
self.co_to_ind = {}
for ind, num in enumerate(sorted(list(set(InputProbNumbers)))):
self.ind_to_co.append(num)
self.co_to_ind[num] = ind+1
self.max = len(self.co_to_ind)
self.data = [0]*(self.max+1)
def __str__(self):
retList = []
for i in range(1, self.max+1):
x = self.ind_to_co[i]
if self.count(x):
c = self.count(x)
for _ in range(c):
retList.append(x)
return "[" + ", ".join([str(a) for a in retList]) + "]"
def __getitem__(self, key):
key += 1
s = 0
ind = 0
l = self.max.bit_length()
for i in reversed(range(l)):
if ind + (1<<i) <= self.max:
if s + self.data[ind+(1<<i)] < key:
s += self.data[ind+(1<<i)]
ind += (1<<i)
if ind == self.max or key < 0:
raise IndexError("BIT index out of range")
return self.ind_to_co[ind+1]
def __len__(self):
return self._query_sum(self.max)
def __contains__(self, num):
if not num in self.co_to_ind:
return False
return self.count(num) > 0
# 0からiまでの区間和
# 左に進んでいく
def _query_sum(self, i):
s = 0
while i > 0:
s += self.data[i]
i -= i & -i
return s
# i番目の要素にxを足す
# 上に登っていく
def _add(self, i, x):
while i <= self.max:
self.data[i] += x
i += i & -i
# 値xを挿入
def push(self, x):
if not x in self.co_to_ind:
raise KeyError("The pushing number didnt initialized")
self._add(self.co_to_ind[x], 1)
# 値xを削除
def delete(self, x):
if not x in self.co_to_ind:
raise KeyError("The deleting number didnt initialized")
if self.count(x) <= 0:
raise ValueError("The deleting number doesnt exist")
self._add(self.co_to_ind[x], -1)
# 要素xの個数
def count(self, x):
return self._query_sum(self.co_to_ind[x]) - self._query_sum(self.co_to_ind[x]-1)
# 値xを超える最低ind
def bisect_right(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_right(self.ind_to_co, x) - 1
return self._query_sum(i)
# 値xを下回る最低ind
def bisect_left(self, x):
if x in self.co_to_ind:
i = self.co_to_ind[x]
else:
i = bisect_left(self.ind_to_co, x)
if i == 1:
return 0
return self._query_sum(i-1)
import sys
input = sys.stdin.readline
N = int(input())
A = list(map(int, input().split()))
Ind = [0]*(N+1)
for i, a in enumerate(A):
Ind[a] = i+1
Bit = BITbisect(list(range(N+1)))
ans = [0]
Bit.push(Ind[1])
a = 0
for n in range(2, N+1):
ind = Ind[n]
f = Bit.bisect_left(ind)
#print(Bit)
l = len(Bit)
if l%2 == 0:
if f == l//2:
a += l//2
elif f < l//2:
p1 = Bit[l//2-1]
a += (p1-ind-1) - (l//2-1) + l-f
else:
p2 = Bit[l//2]
a += (ind-p2-1) - (l//2-1) + l-f
else:
p1 = Bit[l//2]
#print(f, p1, ind, l)
if f <= l//2:
a += (p1-ind-1) - l//2 + l-f
else:
a += (ind-p1-1) - l//2 + l-f
ans.append(a)
Bit.push(ind)
print(*ans, sep=" ")
``` | instruction | 0 | 85,811 | 12 | 171,622 |
No | output | 1 | 85,811 | 12 | 171,623 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,871 | 12 | 171,742 |
Tags: geometry, math
Correct Solution:
```
import sys, math
import io, os
#data = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from bisect import bisect_left as bl, bisect_right as br, insort
from heapq import heapify, heappush, heappop
from collections import defaultdict as dd, deque, Counter
#from itertools import permutations,combinations
def data(): return sys.stdin.readline().strip()
def mdata(): return list(map(int, data().split()))
def outl(var) : sys.stdout.write('\n'.join(map(str, var))+'\n')
def out(var) : sys.stdout.write(str(var)+'\n')
from decimal import Decimal
from fractions import Fraction
#sys.setrecursionlimit(100000)
INF = float('inf')
mod = int(1e9)+7
for t in range(int(data())):
n=int(data())
a=mdata()
if a[0]+a[1]<=a[-1]:
print(1,2,n)
else:
out(-1)
``` | output | 1 | 85,871 | 12 | 171,743 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,872 | 12 | 171,744 |
Tags: geometry, math
Correct Solution:
```
import sys
from os import path
if(path.exists('input.txt')):
sys.stdin = open("input.txt","r")
sys.stdout = open("output.txt","w")
def solve():
n = int(input().strip())
a = [int(i) for i in input().rstrip().split()]
if(a[-1] - a[0] - a[1]) >= 0 :
print(f"1 2 {n}" )
else:
print(-1)
def main():
tc = int(input().strip())
for case in range(tc):
solve()
if __name__ == '__main__':
main()
``` | output | 1 | 85,872 | 12 | 171,745 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,873 | 12 | 171,746 |
Tags: geometry, math
Correct Solution:
```
t=int(input())
while(t):
n=int(input())
a=list(map(int,input().split()))
if n>2:
flag=0
for i in range(len(a)-2):
if(a[i]+a[i+1]>a[n-1]):
continue
else:
flag=1
print(i+1,end=" ")
print(i+2,end=" ")
print(n)
break
if(flag!=1):
print("-1")
t=t-1
``` | output | 1 | 85,873 | 12 | 171,747 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,874 | 12 | 171,748 |
Tags: geometry, math
Correct Solution:
```
t = int(input())
for _ in range (t):
n = int(input())
A = list(map(int, input().split()))
if A[0]+A[1] <= A[-1]:
print (1, 2, n)
else:
print (-1)
``` | output | 1 | 85,874 | 12 | 171,749 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,875 | 12 | 171,750 |
Tags: geometry, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
a=list(map(int,input().split()))
if(a[0]+a[1]>a[-1]):
print("-1")
else:
print(1,2,n)
``` | output | 1 | 85,875 | 12 | 171,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,876 | 12 | 171,752 |
Tags: geometry, math
Correct Solution:
```
t=int(input())
for i in range(t):
n=int(input())
a=[int(v) for v in input().split()]
p=a[0]
q=a[1]
r=a[-1]
if p+q<=r:
print(1,2,n)
else:
print(-1)
``` | output | 1 | 85,876 | 12 | 171,753 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,877 | 12 | 171,754 |
Tags: geometry, math
Correct Solution:
```
def solve(n, a):
a, b, c = a[0], a[1], a[-1]
if a + b > c:
print(-1)
else:
print(1, 2, n)
return
def main():
inp = lambda: [int(x) for x in input().split()]
tc = int(input())
for _ in range(tc):
n, a = int(input()), inp()
solve(n, a)
if __name__ == '__main__':
main()
``` | output | 1 | 85,877 | 12 | 171,755 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an array a_1, a_2, ... , a_n, which is sorted in non-decreasing order (a_i ≤ a_{i + 1}).
Find three indices i, j, k such that 1 ≤ i < j < k ≤ n and it is impossible to construct a non-degenerate triangle (a triangle with nonzero area) having sides equal to a_i, a_j and a_k (for example it is possible to construct a non-degenerate triangle with sides 3, 4 and 5 but impossible with sides 3, 4 and 7). If it is impossible to find such triple, report it.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains one integer n (3 ≤ n ≤ 5 ⋅ 10^4) — the length of the array a.
The second line of each test case contains n integers a_1, a_2, ... , a_n (1 ≤ a_i ≤ 10^9; a_{i - 1} ≤ a_i) — the array a.
It is guaranteed that the sum of n over all test cases does not exceed 10^5.
Output
For each test case print the answer to it in one line.
If there is a triple of indices i, j, k (i < j < k) such that it is impossible to construct a non-degenerate triangle having sides equal to a_i, a_j and a_k, print that three indices in ascending order. If there are multiple answers, print any of them.
Otherwise, print -1.
Example
Input
3
7
4 6 11 11 15 18 20
4
10 10 10 11
3
1 1 1000000000
Output
2 3 6
-1
1 2 3
Note
In the first test case it is impossible with sides 6, 11 and 18. Note, that this is not the only correct answer.
In the second test case you always can construct a non-degenerate triangle. | instruction | 0 | 85,878 | 12 | 171,756 |
Tags: geometry, math
Correct Solution:
```
for _ in range(int(input())):
n=int(input())
lst=list(map(int,input().split()))
if lst[0]+lst[1]<=lst[-1]:
print(1,2,n)
else:
print(-1)
``` | output | 1 | 85,878 | 12 | 171,757 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.
There is a process that takes place on arrays a and b of length n and length n-1 respectively.
The process is an infinite sequence of operations. Each operation is as follows:
* First, choose a random integer i (1 ≤ i ≤ n-1).
* Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer).
See notes for an example of an operation.
It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.
You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.
Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.
Input
The first line contains a single integer n (2 ≤ n ≤ 100).
The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).
The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).
The fourth line contains a single integer q (q=1).
The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).
Output
Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.
Example
Input
3
2 3 4
2 1
1
-1
Output
56
Note
The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).
Examples of arrays a that are not good:
* a = [3, 2, 3] is not good because a_1 > c_1;
* a = [0, -1, 3] is not good because a_2 < 0.
One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.
Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2. | instruction | 0 | 85,967 | 12 | 171,934 |
Tags: dp, math
Correct Solution:
```
def putin():
return map(int, input().split())
def sol():
n = int(input())
C = list(putin())
B = list(putin())
q = int(input())
x = int(input())
min_arr = [x]
min_part_sums = [x]
part_sums = [C[0]]
for i in range(1, n):
part_sums.append(part_sums[-1] + C[i])
for elem in B:
min_arr.append(min_arr[-1] + elem)
min_part_sums.append(min_arr[-1] + min_part_sums[-1])
for i in range(n):
if min_part_sums[i] > part_sums[i]:
return 0
if min_part_sums[0] > C[0]:
return 0
answer = [1] * (part_sums[0] - max(0, min_part_sums[0]) + 1)
for k in range(1, n):
new_answer = [0] * (part_sums[k] - max(0, min_part_sums[k]) + 1)
cnt = 1
window = answer[-1]
new_answer[-1] = window
while cnt <= len(new_answer) - 1:
cnt += 1
if cnt <= len(answer):
window += answer[-cnt]
if C[k] + 1 < cnt:
window -= answer[C[k] + 1 - cnt]
new_answer[-cnt] = window
answer = new_answer.copy()
m = 10 ** 9 + 7
return sum(answer) % m
print(sol())
``` | output | 1 | 85,967 | 12 | 171,935 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.
There is a process that takes place on arrays a and b of length n and length n-1 respectively.
The process is an infinite sequence of operations. Each operation is as follows:
* First, choose a random integer i (1 ≤ i ≤ n-1).
* Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer).
See notes for an example of an operation.
It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.
You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.
Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.
Input
The first line contains a single integer n (2 ≤ n ≤ 100).
The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).
The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).
The fourth line contains a single integer q (q=1).
The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).
Output
Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.
Example
Input
3
2 3 4
2 1
1
-1
Output
56
Note
The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).
Examples of arrays a that are not good:
* a = [3, 2, 3] is not good because a_1 > c_1;
* a = [0, -1, 3] is not good because a_2 < 0.
One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.
Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2. | instruction | 0 | 85,968 | 12 | 171,936 |
Tags: dp, math
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input());C = [int(a) for a in input().split()];B = [int(a) for a in input().split()];Q = int(input());x = int(input());P = 10 ** 9 + 7;
dp = [[0] * 20100 for _ in range(N + 1)];dp[0][0] = 1;ans = 0;s = x;t = s
for i in range(N):
for j in range(20050, t - 1, -1):
if j < 0: break
dp[i+1][j] = (dp[i+1][j+1] + dp[i][max(j-C[i], 0)] - dp[i][j+1]) % P
for j in range(min(t - 1, 20050), -1, -1):dp[i+1][j] = dp[i+1][j+1]
if i < N - 1:s += B[i];t += s
print(dp[-1][0] % P)
``` | output | 1 | 85,968 | 12 | 171,937 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.
There is a process that takes place on arrays a and b of length n and length n-1 respectively.
The process is an infinite sequence of operations. Each operation is as follows:
* First, choose a random integer i (1 ≤ i ≤ n-1).
* Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer).
See notes for an example of an operation.
It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.
You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.
Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.
Input
The first line contains a single integer n (2 ≤ n ≤ 100).
The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).
The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).
The fourth line contains a single integer q (q=1).
The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).
Output
Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.
Example
Input
3
2 3 4
2 1
1
-1
Output
56
Note
The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).
Examples of arrays a that are not good:
* a = [3, 2, 3] is not good because a_1 > c_1;
* a = [0, -1, 3] is not good because a_2 < 0.
One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.
Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2. | instruction | 0 | 85,969 | 12 | 171,938 |
Tags: dp, math
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().rstrip()
N = int(input())
C = [int(a) for a in input().split()]
B = [int(a) for a in input().split()]
Q = int(input())
x = int(input())
P = 10 ** 9 + 7
dp = [[0] * 20100 for _ in range(N + 1)]
dp[0][0] = 1
ans = 0
s = x
t = s
for i in range(N):
for j in range(20050, t - 1, -1):
if j < 0: break
dp[i+1][j] = (dp[i+1][j+1] + dp[i][max(j-C[i], 0)] - dp[i][j+1]) % P
for j in range(min(t - 1, 20050), -1, -1):
dp[i+1][j] = dp[i+1][j+1]
if i < N - 1:
s += B[i]
t += s
print(dp[-1][0] % P)
``` | output | 1 | 85,969 | 12 | 171,939 |
Provide tags and a correct Python 3 solution for this coding contest problem.
This is the easy version of the problem. The only difference is that in this version q = 1. You can make hacks only if both versions of the problem are solved.
There is a process that takes place on arrays a and b of length n and length n-1 respectively.
The process is an infinite sequence of operations. Each operation is as follows:
* First, choose a random integer i (1 ≤ i ≤ n-1).
* Then, simultaneously set a_i = min\left(a_i, \frac{a_i+a_{i+1}-b_i}{2}\right) and a_{i+1} = max\left(a_{i+1}, \frac{a_i+a_{i+1}+b_i}{2}\right) without any rounding (so values may become non-integer).
See notes for an example of an operation.
It can be proven that array a converges, i. e. for each i there exists a limit a_i converges to. Let function F(a, b) return the value a_1 converges to after a process on a and b.
You are given array b, but not array a. However, you are given a third array c. Array a is good if it contains only integers and satisfies 0 ≤ a_i ≤ c_i for 1 ≤ i ≤ n.
Your task is to count the number of good arrays a where F(a, b) ≥ x for q values of x. Since the number of arrays can be very large, print it modulo 10^9+7.
Input
The first line contains a single integer n (2 ≤ n ≤ 100).
The second line contains n integers c_1, c_2 …, c_n (0 ≤ c_i ≤ 100).
The third line contains n-1 integers b_1, b_2, …, b_{n-1} (0 ≤ b_i ≤ 100).
The fourth line contains a single integer q (q=1).
The fifth line contains q space separated integers x_1, x_2, …, x_q (-10^5 ≤ x_i ≤ 10^5).
Output
Output q integers, where the i-th integer is the answer to the i-th query, i. e. the number of good arrays a where F(a, b) ≥ x_i modulo 10^9+7.
Example
Input
3
2 3 4
2 1
1
-1
Output
56
Note
The following explanation assumes b = [2, 1] and c=[2, 3, 4] (as in the sample).
Examples of arrays a that are not good:
* a = [3, 2, 3] is not good because a_1 > c_1;
* a = [0, -1, 3] is not good because a_2 < 0.
One possible good array a is [0, 2, 4]. We can show that no operation has any effect on this array, so F(a, b) = a_1 = 0.
Another possible good array a is [0, 1, 4]. In a single operation with i = 1, we set a_1 = min((0+1-2)/(2), 0) and a_2 = max((0+1+2)/(2), 1). So, after a single operation with i = 1, a becomes equal to [-1/2, 3/2, 4]. We can show that no operation has any effect on this array, so F(a, b) = -1/2. | instruction | 0 | 85,970 | 12 | 171,940 |
Tags: dp, math
Correct Solution:
```
def solve():
MOD = 10**9+7
n = int(input())
c = list(map(int, input().split()))
b = [0] + list(map(int, input().split()))
q = int(input())
queries = list(map(int, input().split()))
maxans = 1
for c1 in c:
maxans = maxans * (c1+1) % MOD
ans = {}
for i in range(1, n):
b[i] += b[i-1]
s = lb = 0
for i in range(1, n):
s -= b[i]
lb = min(lb, s//(i+1))
s = ub = c[0]
for i in range(n):
s += c[i] - b[i]
ub = min(ub, s//(i+1))
for x in queries:
if x <= lb:
print(maxans)
elif x > ub:
print(0)
elif x in ans:
print(ans[x])
else:
dp0 = [1] * 10002
dp0[0] = 0
bd = 0
for i in range(n):
dp1 = [0] * 10002
bd += b[i] + x
for j in range(max(bd, 0), 10001):
dp1[j+1] = (dp1[j] + dp0[j+1] - dp0[max(j-c[i], 0)]) % MOD
dp0 = dp1[:]
a = dp0[-1]
ans[x] = a
print(a)
import sys
input = lambda: sys.stdin.readline().rstrip()
solve()
``` | output | 1 | 85,970 | 12 | 171,941 |
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