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 |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others. | instruction | 0 | 52,335 | 12 | 104,670 |
Tags: constructive algorithms
Correct Solution:
```
N = int(input())
O = list(map(int, input().split()))
strings = [["0" for i in range(N)] for i in range(N+1)]
scores = sorted([(O[i], i) for i in range(N)], reverse=True)
T = [None for i in range(N)]
for i, (num, ind) in enumerate(scores): T[ind] = i
for i, x in enumerate(O):
for j in range(x):
strings[(T[i] + j) % (N + 1)][i] = "1"
strings = [s for s in strings if "1" in s]
print(len(strings))
for s in strings:
print("".join(s))
``` | output | 1 | 52,335 | 12 | 104,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others. | instruction | 0 | 52,336 | 12 | 104,672 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
v = []
q = 0
z = ['0' for i in range(n)]
for i in input().split() :
v.append([int(i), q])
q += 1
v.sort(reverse=True)
c = []
def predictor(v, z, index_off, index_on, n) :
global c
if n == 0 :
for i in range(len(v)) :
if v[i][0] > 0 :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
#print(''.join(z))
c.append(''.join(z))
return v
for i in range(len(v)) :
if (index_off != i and (v[i][0] > 1 or (i == index_on and v[i][0] == 1))) :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
c.append(''.join(z))
return v
def generator(v, z, n) :
y, x = -1, n-1
q = 0
while v[0][0] != 0 :
if v[x-1][0] == 1 or v[x][0] == 0 : y = -2
while v[x][0] == 0 : x -= 1
if y >= 0 :
while v[y][0]-1 == n :
y += 1
q += 1
predictor(v, z, y, x, n)
if y == -2 : y = q
else : y += 1
n -= 1
return n
generator(v, z, n)
print(len(c))
for i in c : print(i)
``` | output | 1 | 52,336 | 12 | 104,673 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others. | instruction | 0 | 52,337 | 12 | 104,674 |
Tags: constructive algorithms
Correct Solution:
```
n = int(input())
v, c = list(), list()
z = ['0' for i in range(n)]
for i, x in enumerate(input().split()) : v.append([int(x), i])
v.sort(reverse=True)
def predictor(v, z, index_off, index_on, n) :
global c
if n == 0 :
for i in range(len(v)) :
if v[i][0] > 0 :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
c.append(''.join(z))
return
for i in range(len(v)) :
if (index_off != i and (v[i][0] > 1 or (i == index_on and v[i][0] == 1))) :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
c.append(''.join(z))
def generator(v, z, n) :
y, x = -1, n-1
q = 0
while v[0][0] != 0 :
if v[x-1][0] == 1 or v[x][0] == 0 : y = -2
while v[x][0] == 0 : x -= 1
if y >= 0 :
while v[y][0]-1 == n :
y += 1
q += 1
predictor(v, z, y, x, n)
if y == -2 : y = q
else : y += 1
n -= 1
return len(c)
print(generator(v, z, n))
print('\n'.join(c))
``` | output | 1 | 52,337 | 12 | 104,675 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
n = int(input())
v, c = list([[0, 0]]), list()
z = ['0' for i in range(n)]
for i, x in enumerate(input().split()) : v.append([int(x), i])
v.sort(reverse=True, key=lambda x : [x[0], -x[1]])
#v.sort()
def S_N_U_F_F(z, v, c) :
y, k, x, count = -1, 0, len(v)-1, len(v)-1
while v[0][0] != 0 :
#print(v, 'y:', y, 'k:', k, 'x:', x, 'count:', count)
if k > 0 and k < len(v)-1 : z[v[y+k][1]] = '0'
elif v[y+k][0] > 1 or v[y+k+1][0] == 0 :
v[y+k][0], z[v[y+k][1]], x = v[y+k][0]-1, '1', i
else : z[v[y+k][1]] = '0'
for i in range(0, y+k) : v[i][0], z[v[i][1]] = v[i][0]-1, '1'
for i in range(y+k+1, len(v)-1) :
if v[i][0] > 1 or (v[i][0] != 0 and v[i+1][0] == 0) :
v[i][0], z[v[i][1]], x = v[i][0]-1, '1', i
else : z[v[i][1]] = '0'
while v[y+1][0] == count and count != 0 : y, k = y+1, k-1
if v[x-1][0] == 1 or v[x][0] == 0 : k = -1
k += 1
count -= 1
#print(''.join(z), 'y:', y)
c.append(''.join(z))
return len(c)
def S_N_U_F_F_(z, v, c) :
x, count = len(v)-1, len(v)
while v[len(v)-2][0] > 0 :
count, k, s = count-1, 0, 0
print('x =', x)
for i in range(1, len(v)-1) :
value = v[i][0]
if value == count+1 or (v[i-1][0] == -1+s and value == 1) or (value > 1 and i != x) :
if value == count : k += 1
v[i][0], z[v[i][1]] = v[i][0]-1, '1'
if v[i-1][0] == -1+s and v[i][0] == 0 :
print('v[i] =', v[i])
v[i][0], x, s = -1, len(v)-1-k, -1
v[x][0] -= 1
z[v[x][1]] = '1'
else : z[v[i][1]] = '0'
x -= 1
print(v, 'x:', x, 'count', count, 's:', s, 'k:', k)
c.append(''.join(z))
print(''.join(z))
return len(c)
def predictor(v, z, index_off, index_on, n) :
global c
if n == 0 :
for i in range(len(v)) :
if v[i][0] > 0 :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
c.append(''.join(z))
return
for i in range(len(v)) :
if (index_off != i and (v[i][0] > 1 or (i == index_on and v[i][0] == 1))) :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
c.append(''.join(z))
def generator(v, z, n) :
y, x, q = -1, n-1, 0
while v[0][0] != 0 :
if v[x-1][0] == 1 or v[x][0] == 0 : y = -2
if v[x][0] == 0 : x -= 1
if y >= 0 :
while v[y][0]-1 == n :
y += 1
q += 1
predictor(v, z, y, x, n)
if y == -2 : y = q
else : y += 1
n -= 1
return len(c)
#print(generator(v, z, n))
print(S_N_U_F_F(z, v, c))
print('\n'.join(c))
``` | instruction | 0 | 52,338 | 12 | 104,676 |
Yes | output | 1 | 52,338 | 12 | 104,677 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
n = int(input())
v, c = list([[0, 0]]), list()
z = ['0' for i in range(n)]
for i, x in enumerate(input().split()) : v.append([int(x), i])
v.sort(reverse=True)
def S_N_U_F_F(z, v, c) :
y, k, x, count = -1, 0, len(v)-1, len(v)-1
while v[0][0] != 0 :
if (k > 0 and k < len(v)-1) or (v[y+k][0] <= 1 and v[y+k+1][0] != 0) : z[v[y+k][1]] = '0'
else : v[y+k][0], z[v[y+k][1]], x = v[y+k][0]-1, '1', i
for i in range(0, y+k) : v[i][0], z[v[i][1]] = v[i][0]-1, '1'
for i in range(y+k+1, len(v)-1) :
if v[i][0] > 1 or (v[i][0] != 0 and v[i+1][0] == 0) : v[i][0], z[v[i][1]], x = v[i][0]-1, '1', i
else : z[v[i][1]] = '0'
while v[y+1][0] == count and count != 0 : y, k = y+1, k-1
if v[x-1][0] == 1 or v[x][0] == 0 : k = -1
count, k = count-1, k+1
c.append(''.join(z))
return len(c)
print(S_N_U_F_F(z, v, c))
print('\n'.join(c))
``` | instruction | 0 | 52,339 | 12 | 104,678 |
Yes | output | 1 | 52,339 | 12 | 104,679 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
n = int(input())
v = []
q = 0
z = ['0' for i in range(n)]
for i in input().split() :
v.append([int(i), q])
q += 1
v.sort(reverse=True, key=lambda x : [x[0], -x[1]])
print(n+1)
def predictor(v, z, index_off, index_on, n) :
if n == 0 :
for i in range(len(v)) :
if v[i][0] > 0 :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
return
for i in range(len(v)) :
#print(index_off, i, n, index_on)
if (index_off != i and (v[i][0] > 1 or (i == index_on and v[i][0] == 1))) :
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
print(''.join(z))
return v
def generator(v, z, n) :
y, x = -1, n-1
q = -1
while v[0][0] != 0 :
if v[x-1][0] == 1 or v[x][0] == 0 : y = q
while v[x][0] == 0 : x -= 1
#print(v)
while v[y][0]-1 == n :
y += 1
q += 1
predictor(v, z, y, x, n)
y += 1
n -= 1
return n
q = generator(v, z, n)
c = '0'*n
for i in range(q+1) : print(c)
``` | instruction | 0 | 52,340 | 12 | 104,680 |
No | output | 1 | 52,340 | 12 | 104,681 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
n = int(input())
v = []
q = 0
z = ['0' for i in range(n)]
for i in input().split() :
v.append([int(i), q])
q += 1
v.sort(reverse=True, key=lambda x : [x[0], -x[1]])
print(n+1)
def predictor(v, z, index_off, index_on, n) :
for i in range(len(v)) :
#print(index_off, i, n, index_on)
#if index_off == i and v[i][0]+1 == n : index_off += 1
if v[i][0]-1 == n or (index_off != i and (v[i][0] > 1 or (i == index_on and v[i][0] == 1))) :
if v[i][0]-1 == n and index_off == i : index_off += 1
v[i][0] -= 1
z[v[i][1]] = '1'
else : z[v[i][1]] = '0'
print(''.join(z))
return v
def generator(v, z, n) :
y, x = -1, n-1
while v[0][0] != 0 :
if v[x-1][0] == 1 or v[x][0] == 0 : y = -1
while v[x][0] == 0 : x -= 1
#print(v)
predictor(v, z, y, x, n)
y += 1
n -= 1
return n
q = generator(v, z, n)
c = '0'*n
for i in range(q) : print(c)
``` | instruction | 0 | 52,341 | 12 | 104,682 |
No | output | 1 | 52,341 | 12 | 104,683 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
print("6\n11111\n01111\n10111\n11011\n11101\n11110")
``` | instruction | 0 | 52,342 | 12 | 104,684 |
No | output | 1 | 52,342 | 12 | 104,685 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given an integer array a_1, a_2, ..., a_n, where a_i represents the number of blocks at the i-th position. It is guaranteed that 1 ≤ a_i ≤ n.
In one operation you can choose a subset of indices of the given array and remove one block in each of these indices. You can't remove a block from a position without blocks.
All subsets that you choose should be different (unique).
You need to remove all blocks in the array using at most n+1 operations. It can be proved that the answer always exists.
Input
The first line contains a single integer n (1 ≤ n ≤ 10^3) — length of the given array.
The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n) — numbers of blocks at positions 1, 2, ..., n.
Output
In the first line print an integer op (0 ≤ op ≤ n+1).
In each of the following op lines, print a binary string s of length n. If s_i='0', it means that the position i is not in the chosen subset. Otherwise, s_i should be equal to '1' and the position i is in the chosen subset.
All binary strings should be distinct (unique) and a_i should be equal to the sum of s_i among all chosen binary strings.
If there are multiple possible answers, you can print any.
It can be proved that an answer always exists.
Examples
Input
5
5 5 5 5 5
Output
6
11111
01111
10111
11011
11101
11110
Input
5
5 1 1 1 1
Output
5
11000
10000
10100
10010
10001
Input
5
4 1 5 3 4
Output
5
11111
10111
10101
00111
10100
Note
In the first example, the number of blocks decrease like that:
{ 5,5,5,5,5 } → { 4,4,4,4,4 } → { 4,3,3,3,3 } → { 3,3,2,2,2 } → { 2,2,2,1,1 } → { 1,1,1,1,0 } → { 0,0,0,0,0 }. And we can note that each operation differs from others.
Submitted Solution:
```
n = int(input())
a = list(map(int, input().split()))
d = []
res = []
for i in range(n+1):
res.append([])
for j in range(n):
res[i].append(0)
for i in range(len(a)):
a[i] = (a[i], i)
a.sort(reverse=True)
f = 0
for i in range(n):
if a[i][0] in d:
f += 1
else:
d.append(a[i][0])
for j in range(a[i][0]):
if j >= a[i][0] - f:
res[(j+1) % (n+1)][a[i][1]] = 1
else:
res[(j) % (n+1)][a[i][1]] = 1
ans = 0
for line in res:
if sum(line) > 0:
ans += 1
print(ans)
for line in res:
if sum(line) > 0:
print(' '.join(map(lambda x: str(x), line)))
``` | instruction | 0 | 52,343 | 12 | 104,686 |
No | output | 1 | 52,343 | 12 | 104,687 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,344 | 12 | 104,688 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
n = int(input())
s = []
for i in range(n):
s.append(list(map(int, input().split()))[1:])
# List of sequence that are non-increasing sorted by first number
f = sorted([e for e in s if e == sorted(e, reverse=True)], key=lambda x: -x[0])
# print(f)
# List of sequence that are non-increasing sorted by last number
l = sorted(f[::], key=lambda x: -x[-1])
# print(l)
q = ans = 0
for e in l:
while q < len(f) and e[-1] < f[q][0]:
q += 1
ans += len(f) - q
print(n*n - ans)
``` | output | 1 | 52,344 | 12 | 104,689 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,345 | 12 | 104,690 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
n = int(input())
lis = []
allok = 0
maxlis = []
minlis = []
for i in range(n):
s = list(map(int,input().split()))
flag = True
for j in range(s[0]):
j += 1
if j == 1:
nmin = s[j]
nmax = s[j]
elif nmin < s[j]:
flag = False
break
else:
nmin = min(nmin,s[j])
nmax = max(nmax,s[j])
if flag:
minlis.append(nmin)
maxlis.append(nmax)
else:
allok += 1
ans = 0
ans += allok * allok + 2 * allok * (n-allok)
minlis.sort()
maxlis.sort()
minlis.reverse()
maxlis.reverse()
maxlis.append(float("-inf"))
#print (ans,minlis,maxlis)
maxind = 0
for i in minlis:
while i < maxlis[maxind]:
maxind += 1
ans += maxind
#print (ans , i ,maxlis[maxind])
print (ans)
``` | output | 1 | 52,345 | 12 | 104,691 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,346 | 12 | 104,692 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
import sys
input = sys.stdin.readline
n=int(input())
A=[list(map(int,input().split())) for i in range(n)]
B=[0]*n
for i in range(n):
k=A[i][0]
for j in range(2,k+1):
if A[i][j]>A[i][j-1]:
B[i]=(1<<31,-1)
break
else:
B[i]=(A[i][1],A[i][-1])
from operator import itemgetter
C=sorted(B,key=itemgetter(0))
D=sorted(B,key=itemgetter(1))
ANS=0
ind=0
for _,d in D:
while ind<n and C[ind][0]<=d:
ind+=1
ANS+=n-ind
print(ANS)
``` | output | 1 | 52,346 | 12 | 104,693 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,347 | 12 | 104,694 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
import sys
from bisect import bisect_right
n = int(input())
s = []
has_ascents=[False for _ in range(n)]
firsts=[
]
lasts=[]
for k in range(n):
a=list(map(lambda x: int(x), sys.stdin.readline().split()[1:]))
for i in range(0,len(a)-1):
if a[i]<a[i+1]:
has_ascents[k]=True
break
a.sort()
firsts.append(a[0])
if has_ascents[k]==False:
lasts.append(a[-1])
lasts.sort()
c=0
for i in range(0,n):
# print(i)
if has_ascents[i]==True:
continue
else:
# print(firsts[i],lasts)
ind=bisect_right(lasts,firsts[i])
# print(ind,"index")
if ind>0:
c+=ind
print(n*n-c)
``` | output | 1 | 52,347 | 12 | 104,695 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,348 | 12 | 104,696 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
import sys
input=sys.stdin.readline
n=int(input())
mi=[]
ma=[]
cnt=0
ans=0
for _ in range(n):
a=list(map(int,input().split()))
b=min(a[1:])
c=max(a[1:])
saisho=10**10
f=0
for j in range(a[0]):
saisho=min(saisho,a[j+1])
if saisho<a[j+1]:
f=1
ans+=(n-cnt)*2-1
cnt+=1
break
if not f:
mi.append(b)
ma.append(c)
mi.sort()
ma.sort()
import bisect
for i in range(n-cnt):
p=bisect.bisect_right(ma,mi[i])
ans+=n-cnt-p
print(ans)
``` | output | 1 | 52,348 | 12 | 104,697 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,349 | 12 | 104,698 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
n = int(input())
minis, maxis, casc, cdes = [], [], 0, 0
for i in range(n):
s = list(map(int, input().split()))
s.pop(0)
for j in range(len(s)):
if j > 0 and s[j] > s[j - 1]:
casc += 1
break
elif j == len(s) - 1:
cdes += 1
minis.append(min(s))
maxis.append(max(s))
total = (casc * casc) + (2 * casc * cdes)
minis.sort()
maxis.sort()
i = 0
for j in maxis:
while i < cdes and minis[i] < j:
i += 1
total += i
print(total)
``` | output | 1 | 52,349 | 12 | 104,699 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,350 | 12 | 104,700 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
from bisect import bisect, insort
from sys import stdin, stdout
n=int(stdin.readline())
a,b=[],[]
for _ in [0]*n:
l,*s=map(int,stdin.readline().split())
if all(s[i]>=s[i+1] for i in range(len(s) -1)):insort(a,s[0]);b.append(s[-1]),
stdout.write(str(n*n-sum(bisect(a,x)for x in b)))
``` | output | 1 | 52,350 | 12 | 104,701 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences. | instruction | 0 | 52,351 | 12 | 104,702 |
Tags: binary search, combinatorics, data structures, dp, implementation, sortings
Correct Solution:
```
import sys
import math
import heapq
import bisect
import re
from collections import deque
sys.setrecursionlimit(1000000000)
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def finput():
return float(input())
def tinput():
return input().split()
def rinput():
return map(int, tinput())
def rlinput():
return list(rinput())
def modst(w, s):
res = 1
while s:
if s % 2:
res *= w
w *= w
s //= 2
return res
def error(n):
print("YES")
exit(n)
def main():
n = iinput()
res = 0
q = []
t = 0
for i in range(n):
w = rlinput()
mi = 1e6
ma = 0
fl = 1
for i in range(1, w[0] + 1):
if mi > w[i]:
mi = w[i]
if ma < w[i]:
ma = w[i]
for i in range(1, w[0]):
if w[i + 1] > w[i]:
res += n + n - 1
fl = 0
t += 1
break
if fl:
q.append([mi, ma])
res -= (t * (t - 1))
q = sorted(q)
w, e = [], []
for i in q:
w.append(i[0])
e.append(i[1])
for i in range(len(w)):
res += bisect.bisect_left(w, e[i])
print(res)
for i in range(1):
main()
``` | output | 1 | 52,351 | 12 | 104,703 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
import bisect
def is_ascent(sequence):
minimum = float('inf')
for x in sequence:
if x > minimum:
return True
minimum = min(x, minimum)
return False
def find_numbers_greater_than(x, maximas):
return len(maximas) - bisect.bisect_left(maximas, x+1)
n = int(input())
sequences = []
for i in range(0, n):
ls = input().split()
l = int(ls[0])
sequence = [int(v) for v in ls[1:]]
sequences.append(sequence)
non_ascent_seq = []
ans = 0
asc = 0
non_asc = 0
for sequence in sequences:
if is_ascent(sequence):
asc += 1
else:
non_asc += 1
non_ascent_seq.append(sequence)
ans += 2*asc*non_asc + asc*asc
maximas = []
for sequence in non_ascent_seq:
maximas.append(max(sequence))
maximas.sort()
for sequence in non_ascent_seq:
x = min(sequence)
ans += find_numbers_greater_than(x, maximas)
print(ans)
``` | instruction | 0 | 52,352 | 12 | 104,704 |
Yes | output | 1 | 52,352 | 12 | 104,705 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
t = int(input())
yes=0;
lno=[]
M=[]
m=[]
for i in range(t):
inp=list(map(int,input().split()))[1:]
if inp[::-1] != sorted(inp):
yes+=1;
else:
M.append(inp[0])
m.append(inp[-1])
ans = 2*yes*t-yes*yes
M.sort()
lM=len(M)
m.sort()
j=0
for x in M:
if j<lM:
while(m[j]<x):
j+=1
if j==lM: #creating a mistake
break
ans+=j
i+=1
print(ans)
``` | instruction | 0 | 52,353 | 12 | 104,706 |
Yes | output | 1 | 52,353 | 12 | 104,707 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
N = int(readline())
M = 10**6+5
table1 = [0]*M
table2 = [0]*M
ans = 0
cnt = 0
for _ in range(N):
l, *k = list(map(int, readline().split()))
if all(k2-k1 <= 0 for k1, k2 in zip(k, k[1:])):
table1[k[-1]] += 1
table2[k[0]] += 1
else:
cnt += 1
ans = N**2 - (N-cnt)**2
for i in range(M-2, -1, -1):
table2[i] += table2[i+1]
for i in range(M):
if not table1[i]:
continue
ans += table1[i]*table2[i+1]
print(ans)
``` | instruction | 0 | 52,354 | 12 | 104,708 |
Yes | output | 1 | 52,354 | 12 | 104,709 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
def bin_search(a,x):
if x < a[0]:
return 0
if a[len(a) - 1] <= x:
return len(a)
left = 0
right = len(a)
while left + 1 < right:
mid = (left+right)//2
if x < a[mid]:
right = mid
else:
left = mid
return right
n = int(input())
amin = []
amax = []
for i in range(n):
s = [int(s) for s in input().split()]
r = False
minimal = s[1]
maximal = s[1]
last = s[1]
for j in range(2,s[0]+1):
minimal = min(minimal,s[j])
maximal = max(maximal,s[j])
if not r:
if s[j] > last:
r = True
else:
last = s[j]
if not r:
amin.append(minimal)
amax.append(maximal)
res = n**2-len(amax)**2
amax.sort()
for i in amin:
res += len(amax)-bin_search(amax,i)
print(res)
``` | instruction | 0 | 52,355 | 12 | 104,710 |
Yes | output | 1 | 52,355 | 12 | 104,711 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
numberlist=[]
lengthofnumber=int(input())
for i in range(lengthofnumber):
inputel=(list(map(int,input().split( ))))
numberlist.append(inputel[1:])
ascents=0
def ascentexist1(inputlist):
countervariable=0
if len(inputlist)==1:
return False
countervariable=1
while countervariable<=(len(inputlist)-1) and inputlist[countervariable-1]>=inputlist[countervariable]:
countervariable+=1
if countervariable<=(len(inputlist)-1):
return True
else:
return False
def ascentexist2(a,b):
if min(a)<max(b):
return True
else:
return False
nonascents=[]
for a in numberlist:
if ascentexist1(a):
ascents+=lengthofnumber
else:
nonascents.append(a)
ascents+=len(nonascents)*(abs(lengthofnumber-len(nonascents)))
minnonascents=[]
maxnonascents=[]
for b in nonascents:
minnonascents.append(min(b))
maxnonascents.append(max(b))
minnonascents.sort()
maxnonascents.sort()
def indexmin(item,array):
if item>=array[-1]:
return None
start=0
end=len(array)-1
while start<=end:
midpoint=(start+end)//2
if abs(start-end)<=1:
if item==array[midpoint]:
return midpoint+1
else:
return midpoint
elif item<array[midpoint]:
end=midpoint
elif item>=array[midpoint]:
start=midpoint
return None
for k in minnonascents:
alphavalue=indexmin(k,maxnonascents)
if alphavalue!=None:
ascents+=len(maxnonascents)-(alphavalue)
print(ascents)
``` | instruction | 0 | 52,356 | 12 | 104,712 |
No | output | 1 | 52,356 | 12 | 104,713 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
n = int(input())
s = n * n
fi = []
la = []
def binp(seq, num):
l = 0
r = len(seq) - 1
m = (l + r) // 2
while l < r:
m = (l + r) // 2
if l == r - 1:
if seq[r] <= num:
return r
else:
if seq[l] <= num:
return l
else:
return l - 1
if seq[m] > num:
r = m
else:
l = m
return m
for i in range(n):
g, *L = [int(i) for i in input().split()]
t = 1
for j in range(1, g):
if L[j] > L[j - 1]:
t = 0
break
if t:
fi.append(L[0])
la.append(L[-1])
fi.sort()
la.sort()
for i in la:
g = binp(fi, i) + 1
s -= g
print(s)
``` | instruction | 0 | 52,357 | 12 | 104,714 |
No | output | 1 | 52,357 | 12 | 104,715 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
import sys
n = int(input())
def inp(n):
L = []
for i in range(n):
L.append([int(i) for i in input().split()[1:]])
return L
def binary(l, M):
m = min(l)
start = 0
end = len(M)
while start<end:
mid = (start + end)//2
if m == M[mid]:
return len(M)-mid-1
elif m>M[mid]:
start = mid+1
else:
end = mid
return len(M)-mid-1
def has_ascent(l):
L = [sys.maxsize]
for i in l:
if i <= L[-1]:
L.append(i)
else:
return True
return False
d = 0
k = 0
K = inp(n)
L = []
for i in K:
if has_ascent(i):
d = d+(2*n-1)
k = k+1
else:
L.append(i)
M = [max(L[i]) for i in range(len(L))]
M.sort()
for l in L:
d += binary(l, M)
d=d-(k*k)+k
print(d)
``` | instruction | 0 | 52,358 | 12 | 104,716 |
No | output | 1 | 52,358 | 12 | 104,717 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A sequence a = [a_1, a_2, …, a_l] of length l has an ascent if there exists a pair of indices (i, j) such that 1 ≤ i < j ≤ l and a_i < a_j. For example, the sequence [0, 2, 0, 2, 0] has an ascent because of the pair (1, 4), but the sequence [4, 3, 3, 3, 1] doesn't have an ascent.
Let's call a concatenation of sequences p and q the sequence that is obtained by writing down sequences p and q one right after another without changing the order. For example, the concatenation of the [0, 2, 0, 2, 0] and [4, 3, 3, 3, 1] is the sequence [0, 2, 0, 2, 0, 4, 3, 3, 3, 1]. The concatenation of sequences p and q is denoted as p+q.
Gyeonggeun thinks that sequences with ascents bring luck. Therefore, he wants to make many such sequences for the new year. Gyeonggeun has n sequences s_1, s_2, …, s_n which may have different lengths.
Gyeonggeun will consider all n^2 pairs of sequences s_x and s_y (1 ≤ x, y ≤ n), and will check if its concatenation s_x + s_y has an ascent. Note that he may select the same sequence twice, and the order of selection matters.
Please count the number of pairs (x, y) of sequences s_1, s_2, …, s_n whose concatenation s_x + s_y contains an ascent.
Input
The first line contains the number n (1 ≤ n ≤ 100 000) denoting the number of sequences.
The next n lines contain the number l_i (1 ≤ l_i) denoting the length of s_i, followed by l_i integers s_{i, 1}, s_{i, 2}, …, s_{i, l_i} (0 ≤ s_{i, j} ≤ 10^6) denoting the sequence s_i.
It is guaranteed that the sum of all l_i does not exceed 100 000.
Output
Print a single integer, the number of pairs of sequences whose concatenation has an ascent.
Examples
Input
5
1 1
1 1
1 2
1 4
1 3
Output
9
Input
3
4 2 0 2 0
6 9 9 8 8 7 7
1 6
Output
7
Input
10
3 62 24 39
1 17
1 99
1 60
1 64
1 30
2 79 29
2 20 73
2 85 37
1 100
Output
72
Note
For the first example, the following 9 arrays have an ascent: [1, 2], [1, 2], [1, 3], [1, 3], [1, 4], [1, 4], [2, 3], [2, 4], [3, 4]. Arrays with the same contents are counted as their occurences.
Submitted Solution:
```
n = int(input())
mins = []
maxs = []
ascents = 0
def isAscent(s):
return list(reversed(s))!=sorted(s)
ans = 0
for i in range(n):
l,*s = list(map(int,input().split()))
if isAscent(s):
ascents+=1
else:
mins.append(min(s))
maxs.append(max(s))
for minn in mins:
for maxx in maxs:
if maxx>minn:
ans+=1
nnn = 2*n-1
for i in range(ascents):
ans+=nnn
nnn-=(i+1)*2
if nnn<=0:
break
print(ans)
``` | instruction | 0 | 52,359 | 12 | 104,718 |
No | output | 1 | 52,359 | 12 | 104,719 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,360 | 12 | 104,720 |
Tags: bitmasks, greedy
Correct Solution:
```
from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
import itertools
from functools import *
from itertools import permutations,combinations,groupby
import sys
import bisect
import string
import math
import time
import random
def Golf():
*a,=map(int,open(0))
def S_():
return input()
def IS():
return input().split()
def LS():
return [i for i in input().split()]
def I():
return int(input())
def MI():
return map(int,input().split())
def LI():
return [int(i) for i in input().split()]
def LI_():
return [int(i)-1 for i in input().split()]
def NI(n):
return [int(input()) for i in range(n)]
def NI_(n):
return [int(input())-1 for i in range(n)]
def StoI():
return [ord(i)-97 for i in input()]
def ItoS(nn):
return chr(nn+97)
def LtoS(ls):
return ''.join([chr(i+97) for i in ls])
def GI(V,E,Directed=False,index=0):
org_inp=[]
g=[[] for i in range(n)]
for i in range(E):
inp=LI()
org_inp.append(inp)
if index==0:
inp[0]-=1
inp[1]-=1
if len(inp)==2:
a,b=inp
g[a].append(b)
if not Directed:
g[b].append(a)
elif len(inp)==3:
a,b,c=inp
aa=(inp[0],inp[2])
bb=(inp[1],inp[2])
g[a].append(bb)
if not Directed:
g[b].append(aa)
return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0}):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0}) # sample usage
mp=[1]*(w+2)
found={}
for i in range(h):
s=input()
for char in search:
if char in s:
found[char]=((i+1)*(w+2)+s.index(char)+1)
mp_def[char]=mp_def[replacement_of_found]
mp+=[1]+[mp_def[j] for j in s]+[1]
mp+=[1]*(w+2)
return h+2,w+2,mp,found
def bit_combination(k,n=2):
rt=[]
for tb in range(n**k):
s=[tb//(n**bt)%n for bt in range(k)]
rt+=[s]
return rt
def show(*inp,end='\n'):
if show_flg:
print(*inp,end=end)
YN=['YES','NO']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
u_alp=string.ascii_uppercase
ts=time.time()
#sys.setrecursionlimit(10**7)
input=lambda: sys.stdin.readline().rstrip()
def ran_input():
import random
n=random.randint(4,16)
rmin,rmax=1,10
a=[random.randint(rmin,rmax) for _ in range(n)]
return n,a
show_flg=False
show_flg=True
ans=0
t=I()
for _ in range(t):
ans=0
N=70
n,m=LI()
a=LI()
a.sort()
dif=sum(a)-n
if dif<0:
print(-1)
continue
D=[0]*N
B=[0]*N
for j in a:
B[j.bit_length()-1]+=1
for i in range(N):
if (n>>i)&1:
D[i]=1
j=2*N
for i in range(N):
if i>j and B[i]>0:
ans+=i-j
B[i]-=1
j=2*N
if D[i]>0:
if B[i]>0:
D[i]-=1
B[i]-=1
else:
j=min(j,i)
if B[i]>1:
B[i+1]+=B[i]//2
B[i]-=(B[i]//2)*2
print(ans)
``` | output | 1 | 52,360 | 12 | 104,721 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,361 | 12 | 104,722 |
Tags: bitmasks, greedy
Correct Solution:
```
for t in range(int(input())):
n, m = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
pows = [0]*50
for i in a:
pows[i.bit_length()-1] += 1
p = 1
i = 0
tot = 0
make = 100
#print(n,i,p,make)
#print(*pows)
while (n>0 or make<100) and i<len(pows)-1:
if n%(2*p) != 0:
if pows[i] > 0:
pows[i] -= 1
else:
make = min(make,i)
n -= p
if pows[i]>0 and make<i:
tot += i-make
make = 100
pows[i] -= 1
pows[i+1] += pows[i]//2
pows[i] %= 2
i += 1
p *= 2
# print(n,i,p,make)
# print(*pows)
if n==0 and make==100:
print(tot)
else:
print(-1)
``` | output | 1 | 52,361 | 12 | 104,723 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,362 | 12 | 104,724 |
Tags: bitmasks, greedy
Correct Solution:
```
import math as mt
for tc in range(int(input())):
n,m=map(int,input().split())
l=list(map(int,input().split()))
if sum(l)<n:
print(-1)
continue
co=[0]*(10**5+1)
for i in l:
co[int(mt.log2(i))]+=1
i=0
ans=0
while i<60:
if (1<<i)&n != 0:
if co[i]:
co[i]-=1
else:
while i<60 and co[i]==0:
ans+=1
i+=1
co[i]-=1
continue
co[i+1]+=co[i]//2
i+=1
print(ans)
``` | output | 1 | 52,362 | 12 | 104,725 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,363 | 12 | 104,726 |
Tags: bitmasks, greedy
Correct Solution:
```
from math import ceil,gcd,sqrt
def ii():return int(input())
def mi():return map(int,input().split())
def li():return list(mi())
def si():return input()
for _ in range(ii()):
n,m=mi()
a=li()
dp=[0]*61
if(n>sum(a)):
print('-1')
continue
for i in a:
s=bin(i)[2:]
dp[len(s)-1]+=1
i,res=0,0
while i<60:
if(1<<i & n):
if(dp[i]>0):
dp[i]-=1
else:
while(dp[i]==0 and i<60):
i+=1
res+=1
dp[i]-=1
continue
dp[i+1]+=dp[i]//2
i+=1
print(res)
``` | output | 1 | 52,363 | 12 | 104,727 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,364 | 12 | 104,728 |
Tags: bitmasks, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n, m = map(int, input().split(' '))
a = list(map(int, input().strip().split(' ')))
if sum(a) < n: print("-1"); continue
b = [0 for _ in range(64)]
for z in a:
for q in range(64):
b[q] += (z>>q)&1
tar = [0 for _ in range(64)]
for q in range(64):
tar[q] += (n>>q)&1
# clear directs
for i in range(64):
if tar[i] == 1 and b[i] >= 1:
tar[i] -= 1
b[i] -= 1
# combine up
splits = 0
for i in range(63):
if tar[i] == 0: continue
for j in range(i):
if b[j] > 1:
b[j+1] += b[j]//2
b[j] = b[j]%2
if tar[i] == 1 and b[i] >= 1:
tar[i] -= 1
b[i] -= 1
continue
for j in range(i+1, 64):
if b[j] >= 1:
splits += 1
b[j] -= 1
b[i] -= 1
break
else:
b[j] += 1
splits += 1
print(splits)
``` | output | 1 | 52,364 | 12 | 104,729 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,365 | 12 | 104,730 |
Tags: bitmasks, greedy
Correct Solution:
```
def log2(n):
i = 0
while n >> i != 1:
i += 1
return i
def solve(n, po2s):
if sum(po2s) < n:
return -1
po_cnt = [0] * 60
for x in po2s:
po_cnt[log2(x)] += 1
ans = 0
for i in range(60):
if n & (1<<i):
if po_cnt[i]:
po_cnt[i] -= 1
else:
j = i
while po_cnt[j] == 0:
j += 1
while j > i:
ans += 1
po_cnt[j] -= 1
po_cnt[j-1] += 2
j -= 1
po_cnt[i] -= 1
if i < 59:
po_cnt[i+1] += po_cnt[i] // 2
return ans
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
po2s = list(map(int, input().split()))
print(solve(n, po2s))
``` | output | 1 | 52,365 | 12 | 104,731 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,366 | 12 | 104,732 |
Tags: bitmasks, greedy
Correct Solution:
```
from math import log2
for _ in range(int(input())) :
n, m = map(int,input().split())
bit = [0]*61
sum = 0
for x in map(int,input().split()) :
bit[int(log2(x))] +=1
sum+=x
if sum < n :
print(-1)
continue
else :
index,res = 0,0
while index < 60 :
if (1<<index)&n != 0 :
if bit[index] > 0 :
bit[index]-=1
else :
while index<60 and bit[index] == 0 :
index+=1
res+=1
bit[index]-=1
continue
bit[index+1]+= bit[index]//2
index+=1
print(res)
``` | output | 1 | 52,366 | 12 | 104,733 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0 | instruction | 0 | 52,367 | 12 | 104,734 |
Tags: bitmasks, greedy
Correct Solution:
```
def main():
import sys
input = sys.stdin.readline
T = int(input())
for t in range(T):
n, m = map(int, input().split())
a = list(map(int, input().split()))
cnt = [0] * 64
for i in range(m):
cnt[a[i].bit_length() - 1] += 1
sum = 0
ans = 0
for j in range(64):
bit = 1 << j
sum += cnt[j] * bit
if bit & n:
if sum >= bit:
sum -= bit
else:
found = False
for k in range(j + 1, 64):
if cnt[k] > 0:
found = True
cnt[k] -= 1
break
if not found:
ans = -1
break
sum += (1 << k)
ans += k - j
sum -= bit
print(ans)
main()
``` | output | 1 | 52,367 | 12 | 104,735 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a bag of size n. Also you have m boxes. The size of i-th box is a_i, where each a_i is an integer non-negative power of two.
You can divide boxes into two parts of equal size. Your goal is to fill the bag completely.
For example, if n = 10 and a = [1, 1, 32] then you have to divide the box of size 32 into two parts of size 16, and then divide the box of size 16. So you can fill the bag with boxes of size 1, 1 and 8.
Calculate the minimum number of divisions required to fill the bag of size n.
Input
The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and m (1 ≤ n ≤ 10^{18}, 1 ≤ m ≤ 10^5) — the size of bag and the number of boxes, respectively.
The second line of each test case contains m integers a_1, a_2, ... , a_m (1 ≤ a_i ≤ 10^9) — the sizes of boxes. It is guaranteed that each a_i is a power of two.
It is also guaranteed that sum of all m over all test cases does not exceed 10^5.
Output
For each test case print one integer — the minimum number of divisions required to fill the bag of size n (or -1, if it is impossible).
Example
Input
3
10 3
1 32 1
23 4
16 1 4 1
20 5
2 1 16 1 8
Output
2
-1
0
Submitted Solution:
```
from functools import *
from itertools import permutations,combinations,groupby
import sys
import bisect
import string
import math
import time
import random
def Golf():
*a,=map(int,open(0))
def S_():
return input()
def IS():
return input().split()
def LS():
return [i for i in input().split()]
def I():
return int(input())
def MI():
return map(int,input().split())
def LI():
return [int(i) for i in input().split()]
def LI_():
return [int(i)-1 for i in input().split()]
def NI(n):
return [int(input()) for i in range(n)]
def NI_(n):
return [int(input())-1 for i in range(n)]
def StoI():
return [ord(i)-97 for i in input()]
def ItoS(nn):
return chr(nn+97)
def LtoS(ls):
return ''.join([chr(i+97) for i in ls])
def GI(V,E,Directed=False,index=0):
org_inp=[]
g=[[] for i in range(n)]
for i in range(E):
inp=LI()
org_inp.append(inp)
if index==0:
inp[0]-=1
inp[1]-=1
if len(inp)==2:
a,b=inp
g[a].append(b)
if not Directed:
g[b].append(a)
elif len(inp)==3:
a,b,c=inp
aa=(inp[0],inp[2])
bb=(inp[1],inp[2])
g[a].append(bb)
if not Directed:
g[b].append(aa)
return g,org_inp
def GGI(h,w,search=None,replacement_of_found='.',mp_def={'#':1,'.':0}):
#h,w,g,sg=GGI(h,w,search=['S','G'],replacement_of_found='.',mp_def={'#':1,'.':0}) # sample usage
mp=[1]*(w+2)
found={}
for i in range(h):
s=input()
for char in search:
if char in s:
found[char]=((i+1)*(w+2)+s.index(char)+1)
mp_def[char]=mp_def[replacement_of_found]
mp+=[1]+[mp_def[j] for j in s]+[1]
mp+=[1]*(w+2)
return h+2,w+2,mp,found
def bit_combination(k,n=2):
rt=[]
for tb in range(n**k):
s=[tb//(n**bt)%n for bt in range(k)]
rt+=[s]
return rt
def show(*inp,end='\n'):
if show_flg:
print(*inp,end=end)
YN=['YES','NO']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
u_alp=string.ascii_uppercase
ts=time.time()
#sys.setrecursionlimit(10**7)
input=lambda: sys.stdin.readline().rstrip()
def ran_input():
import random
n=random.randint(4,16)
rmin,rmax=1,10
a=[random.randint(rmin,rmax) for _ in range(n)]
return n,a
show_flg=False
show_flg=True
ans=0
t=I()
for _ in range(t):
ans=0
N=70
n,m=LI()
a=LI()
a.sort()
dif=sum(a)-n
if dif<0:
print(-1)
continue
D=[0]*N
B=[0]*N
for j in a:
B[j.bit_length()-1]+=1
for i in range(N):
if (dif>>i)&1:
D[i]=1
#show(n,dif,bin(n)[2:],bin(dif)[2:])
#print('B',B[::-1],'D',D[::-1],a)
for i in range(N):
if D[i]>0 and B[i]>0:
D[i]-=1
B[i]-=1
if B[i]>1:
B[i+1]+=B[i]//2
B[i]-=(B[i]//2)*2
#print('B',B[::-1],'D',D[::-1],a)
if sum(D)==0:
print(0)
continue
j=N*2
for i in range(N):
if D[i]==1:
j=min(j,i)
if i>j and B[i]==1:
ans+=i-j
j=2*N
print(ans)
#A=[bin(i)[2:] for i in a]
#show(n,dif,bin(n),bin(dif))
#show(n,m,A,a)
``` | instruction | 0 | 52,375 | 12 | 104,750 |
No | output | 1 | 52,375 | 12 | 104,751 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,554 | 12 | 105,108 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
import math
def primeDec(n):
res = []
p = 2
while p * p <= n:
if n % p == 0:
count = 0
while n % p == 0:
n //= p
count += 1
res.append((p, count))
p += 1
if n > 1:
res.append((n, 1))
return res
def comb(k, n):
res = 1
for i in range(n - k + 1, n + 1):
res *= i
for i in range(1, k + 1):
res //= i
return res
def ev(n, k):
d = primeDec(n)
res = 1
for p, i in d:
#print(k - 1, k + i - 1)
res = (res * (comb(i, k + i - 1) % N)) % N
#print(n, k, res)
return res
N = 10 ** 9 + 7
n, k = map(int, input().split())
d = primeDec(n)
res = 0
for i in range(1, n + 1):
res = (res + ev(i, k)) % N
print(res)
``` | output | 1 | 52,554 | 12 | 105,109 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,555 | 12 | 105,110 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
# Author : nitish420 --------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
mod=10**9+7
# sys.setrecursionlimit(10**6)
def main():
n,k=map(int,input().split())
ans=0
dp=[[0 for _ in range(n+1)]for _ in range(k+1)]
for i in range(1,n+1):
dp[0][i]=1
dp[1][i]=1
for r in range(2,k+1):
for c in range(1,n+1):
for z in range(c,n+1,c):
dp[r][c]=(dp[r][c]+dp[r-1][z])%mod
ans=0
for i in range(1,n+1):
ans+=(dp[k][i])%mod
print(ans%mod)
# region fastio
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')
# endregion
if __name__ == '__main__':
main()
``` | output | 1 | 52,555 | 12 | 105,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,556 | 12 | 105,112 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
MOD=10**9+7
n,k=map(int,input().split())
dd={}
for i in range(1,n+1):
dd[i]=[]
for a in range(i,n+1,i):
dd[i].append(a)
dp=[[0]*(n+1) for i in range(k+1)]
dp[1]=[0]+[1]*(n)
for i in range(2,k+1):
for a in dd:
h=dp[i-1][a]
for c in dd[a]:
dp[i][c]+=h
dp[i][c]%=MOD
print(sum(dp[-1])%(MOD))
``` | output | 1 | 52,556 | 12 | 105,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,557 | 12 | 105,114 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
ans = [[0]*2001 for i in range(2001)]
n, k = [int(x) for x in input().split(' ')] ; p = 0;
for end in range(1, n + 1):
ans[1][end] = 1
for le in range(2, k + 1):
for end in range(1, n + 1):
for nend in range(end, n + 1, end):
ans[le][nend] = (ans[le - 1][end] + ans[le][nend]) % 1000000007
for i in range(n):
p = (p + ans[k][i + 1]) % 1000000007
print(p)
``` | output | 1 | 52,557 | 12 | 105,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,558 | 12 | 105,116 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
import math,sys,bisect,heapq
from collections import defaultdict,Counter,deque
from itertools import groupby,accumulate
#sys.setrecursionlimit(200000000)
int1 = lambda x: int(x) - 1
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
ilele = lambda: map(int,input().split())
alele = lambda: list(map(int, input().split()))
ilelec = lambda: map(int1,input().split())
alelec = lambda: list(map(int1, input().split()))
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
MOD = 1000000000 + 7
def Y(c): print(["NO","YES"][c])
def y(c): print(["no","yes"][c])
def Yy(c): print(["No","Yes"][c])
n,k = ilele()
dp = list2d(k+1,n+1,0)
for i in range(n+1):
dp[0][i] = 1
for i in range(1,k+1):
for l in range(1,n+1):
for j in range(l,n+1,l):
dp[i][l] += dp[i-1][j]
dp[i][l] %= MOD
print(dp[k][1])
``` | output | 1 | 52,558 | 12 | 105,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,559 | 12 | 105,118 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
n,k=map(int,input().split())
dp=[[1 for i in range(k+2)] for j in range(n+2)]
for i in range(n+2):
dp[i][0]=0
mod=1000000007
for i in range(n,0,-1):
for j in range(2,k+1):
for l in range(i,n+1,i):
dp[i][j]=(dp[l][j-1]+dp[i][j])%mod
ans=0
for i in range(1,n+1):
ans=(ans+dp[i][k]-dp[i][k-1])%mod
print(ans%mod)
``` | output | 1 | 52,559 | 12 | 105,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3]. | instruction | 0 | 52,561 | 12 | 105,122 |
Tags: combinatorics, dp, number theory
Correct Solution:
```
def mlt(): return map(int, input().split())
x, y = mlt()
divs = [[] for _ in range(x+1)]
for n in range(1, x+1):
for k in range(n, x+1, n):
divs[k].append(n)
dp = [[0 for n in range(y+1)] for k in range(x+1)]
for n in range(1, y+1):
dp[1][n] = 1
for n in range(1, x+1):
dp[n][1] = 1
mod = int(1e9 + 7)
for n in range(2, x+1):
for k in range(2, y+1):
for dv in divs[n]:
dp[n][k] += dp[dv][k-1]
dp[n][k] %= mod
res = 0
for n in dp:
res += n[-1]
res %= mod
print(res)
``` | output | 1 | 52,561 | 12 | 105,123 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
n,k=map(int,input().split())
m=10**9+7
dp=[[1 if i==1 else 0 for i in range(k+1)]for j in range(n+1)]
for j in range(1,k):
for i in range(1,n+1):
for ii in range(i,n+1,i):
dp[ii][j+1]=(dp[ii][j+1]%m+dp[i][j]%m)%m
sums=0
for i in range(1,n+1):
sums=(sums%m+dp[i][k]%m)%m
#ans = sum([ dp[i][k] for i in range(1, n + 1) ]) % m
print(sums%m)
``` | instruction | 0 | 52,562 | 12 | 105,124 |
Yes | output | 1 | 52,562 | 12 | 105,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
import math
n, k = map(int, input().split())
dp = [[1 for _ in range(k+1)] for _ in range(n+1)]
for i in range(1, n+1):
f = set((1, i))
for m in range(2, int(math.sqrt(i))+1):
if i%m==0:
f.add(m)
if i/m != i:
f.add(i//m)
f = list(f)
for j in range(2, k+1):
dp[i][j] = 0
for x in f:
dp[i][j] = (dp[i][j] + dp[x][j-1])%1000000007
ans = 0
for i in range(1, n+1):
ans += dp[i][k]
ans = ans%1000000007
print(ans%1000000007)
``` | instruction | 0 | 52,563 | 12 | 105,126 |
Yes | output | 1 | 52,563 | 12 | 105,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
# aadiupadhyay
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def solve():
n, k = mp()
dp = [[0]*(k+1) for i in range(n+1)]
for i in range(n+1):
dp[i][1] = 1
for j in range(1, k):
for i in range(1, n+1):
x = 1
while x*i <= n:
dp[x*i][j+1] += dp[i][j]
dp[x*i][j+1] %= mod
x += 1
ans = 0
for i in range(1, n+1):
ans += dp[i][k]
ans %= mod
pr(ans)
for _ in range(1):
solve()
``` | instruction | 0 | 52,564 | 12 | 105,128 |
Yes | output | 1 | 52,564 | 12 | 105,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
mod = 10**9+7
n,k = map(int,input().split())
dp=[[0 for i in range(n+1)] for j in range(k+1)]
for i in range(1,n+1):
dp[1][i]+=1
for length in range(2,k+1):
for j in range(1,n+1):
for end in range(j,n+1,j):
dp[length][end]= (dp[length][end]+ dp[length-1][j])%mod
print(sum(dp[k])%mod)
``` | instruction | 0 | 52,565 | 12 | 105,130 |
Yes | output | 1 | 52,565 | 12 | 105,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
n, k = map(int, input().split())
MOD = 1e9 + 7
divs = []
for i in range(1, n + 1):
divs.append([])
for j in range(1, i + 1):
if i % j == 0:
divs[-1].append(j)
dp = [[1] * n for i in range(k)]
for ki in range(1, k):
for li in range(n):
last = li + 1
dp[ki][li] = sum(map(lambda div: dp[ki - 1][div - 1], divs[li])) % MOD
print(sum(dp[k - 1]) % MOD)
``` | instruction | 0 | 52,566 | 12 | 105,132 |
No | output | 1 | 52,566 | 12 | 105,133 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
mod = 10**9+7
n, k = map(int, input().strip().split())
dp = [[0 for i in range(n)] for i in range(k)]
for i in range(n):
dp[0][i] = 1
for i in range(k-1):
for j in range(n):
for p in range(j+1, n+1, j+1):
dp[i+1][p-1] = (dp[i+1][p-1] + dp[i][j])%mod
# print('----')
# for i in dp:
# print(i)
print(sum(dp[-1]))
``` | instruction | 0 | 52,567 | 12 | 105,134 |
No | output | 1 | 52,567 | 12 | 105,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mashmokh's boss, Bimokh, didn't like Mashmokh. So he fired him. Mashmokh decided to go to university and participate in ACM instead of finding a new job. He wants to become a member of Bamokh's team. In order to join he was given some programming tasks and one week to solve them. Mashmokh is not a very experienced programmer. Actually he is not a programmer at all. So he wasn't able to solve them. That's why he asked you to help him with these tasks. One of these tasks is the following.
A sequence of l integers b1, b2, ..., bl (1 ≤ b1 ≤ b2 ≤ ... ≤ bl ≤ n) is called good if each number divides (without a remainder) by the next number in the sequence. More formally <image> for all i (1 ≤ i ≤ l - 1).
Given n and k find the number of good sequences of length k. As the answer can be rather large print it modulo 1000000007 (109 + 7).
Input
The first line of input contains two space-separated integers n, k (1 ≤ n, k ≤ 2000).
Output
Output a single integer — the number of good sequences of length k modulo 1000000007 (109 + 7).
Examples
Input
3 2
Output
5
Input
6 4
Output
39
Input
2 1
Output
2
Note
In the first sample the good sequences are: [1, 1], [2, 2], [3, 3], [1, 2], [1, 3].
Submitted Solution:
```
C_cache = [[-1]*2002 for _ in range(2002)]
for _N in range(1, 2002):
C_cache[_N][0] = 1
C_cache[_N][1] = _N
C_cache[_N][2] = (_N*(_N+1))//2
for _K in range(2002): C_cache[0][_K] = 0
def C(_n, _k):
if C_cache[_n][_k] < 0:
res = 0
for i in range(1, _n):
res += C(i, _k-1)
C_cache[_n][_k] = res%1000000007
return C_cache[_n][_k]
def get_divisors_amounts(n):
lst = []
i = 2
while i <= n:
amount = 0
while n%i == 0:
n /= i
amount += 1
if amount != 0: lst.append(amount)
i += 1
return lst
n, k = map(int, input().split(' '))
res = 0
for i in range(1, n+1):
r = 1
for el in get_divisors_amounts(i): r *= C(k, el)
res += r
print(res)
``` | instruction | 0 | 52,568 | 12 | 105,136 |
No | output | 1 | 52,568 | 12 | 105,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.
Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.
Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:
1. group of all i such that ai belongs to no longest increasing subsequences.
2. group of all i such that ai belongs to at least one but not every longest increasing subsequence.
3. group of all i such that ai belongs to every longest increasing subsequence.
Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.
Input
The first line contains the single integer n (1 ≤ n ≤ 105) denoting the number of elements of sequence a.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105).
Output
Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index i belongs to.
Examples
Input
1
4
Output
3
Input
4
1 3 2 5
Output
3223
Input
4
1 5 2 3
Output
3133
Note
In the second sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 3, 2, 5}. Sequence a has exactly 2 longest increasing subsequences of length 3, they are {a1, a2, a4} = {1, 3, 5} and {a1, a3, a4} = {1, 2, 5}.
In the third sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 5, 2, 3}. Sequence a have exactly 1 longest increasing subsequence of length 3, that is {a1, a3, a4} = {1, 2, 3}. | instruction | 0 | 52,586 | 12 | 105,172 |
Tags: data structures, dp, greedy, hashing, math
Correct Solution:
```
# a simple parser for python. use get_number() and get_word() to read
def main():
def parser():
while 1:
data = list(input().split(' '))
for number in data:
if len(number) > 0:
yield(number)
input_parser = parser()
gets = lambda: next(input_parser)
def getNum():
data = gets()
try:
return int(data)
except ValueError:
return float(data)
# ---------program---------
from bisect import bisect_left as binsleft
# bisect_left = bisect_left
MAXA = int(9e9)
n = getNum()
RANGN = range(n)
a = [ getNum() for _ in RANGN ]
revlis = []
g = [MAXA]*n
for i in reversed(RANGN):
x = -a[i]
pt = binsleft( g, x )
revlis.append(pt+1)
if( x < g[pt] ):
g[pt] = x
hlis = max( revlis )
lis, inlis = [], []
d = [0]*n
for i in RANGN: g[i] = MAXA
for i in RANGN:
pt = binsleft( g, a[i] )
lis.append( pt+1 )
inlis.append( lis[i] + revlis[n-i-1] > hlis )
d[pt] += inlis[-1]
if( a[i] < g[pt] ):
g[pt] = a[i]
print( ''.join(
[ '32'[d[lis[i]-1] > 1] if inlis[i] else '1'
for i in RANGN ]
) )
if __name__ == "__main__":
main()
``` | output | 1 | 52,586 | 12 | 105,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The next "Data Structures and Algorithms" lesson will be about Longest Increasing Subsequence (LIS for short) of a sequence. For better understanding, Nam decided to learn it a few days before the lesson.
Nam created a sequence a consisting of n (1 ≤ n ≤ 105) elements a1, a2, ..., an (1 ≤ ai ≤ 105). A subsequence ai1, ai2, ..., aik where 1 ≤ i1 < i2 < ... < ik ≤ n is called increasing if ai1 < ai2 < ai3 < ... < aik. An increasing subsequence is called longest if it has maximum length among all increasing subsequences.
Nam realizes that a sequence may have several longest increasing subsequences. Hence, he divides all indexes i (1 ≤ i ≤ n), into three groups:
1. group of all i such that ai belongs to no longest increasing subsequences.
2. group of all i such that ai belongs to at least one but not every longest increasing subsequence.
3. group of all i such that ai belongs to every longest increasing subsequence.
Since the number of longest increasing subsequences of a may be very large, categorizing process is very difficult. Your task is to help him finish this job.
Input
The first line contains the single integer n (1 ≤ n ≤ 105) denoting the number of elements of sequence a.
The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 105).
Output
Print a string consisting of n characters. i-th character should be '1', '2' or '3' depending on which group among listed above index i belongs to.
Examples
Input
1
4
Output
3
Input
4
1 3 2 5
Output
3223
Input
4
1 5 2 3
Output
3133
Note
In the second sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 3, 2, 5}. Sequence a has exactly 2 longest increasing subsequences of length 3, they are {a1, a2, a4} = {1, 3, 5} and {a1, a3, a4} = {1, 2, 5}.
In the third sample, sequence a consists of 4 elements: {a1, a2, a3, a4} = {1, 5, 2, 3}. Sequence a have exactly 1 longest increasing subsequence of length 3, that is {a1, a3, a4} = {1, 2, 3}. | instruction | 0 | 52,587 | 12 | 105,174 |
Tags: data structures, dp, greedy, hashing, math
Correct Solution:
```
N = int( input() )
A = list( map( int, input().split() ) )
maxa = max( A )
def upd( ftree, x, v ):
while x <= maxa:
ftree[ x ] = max( ftree[ x ], v )
x += x & -x
def qry( ftree, x ):
res = 0
while x:
res = max( res, ftree[ x ] )
x -= x & -x
return res
st_len = [ 0 for i in range( N ) ]
ftree = [ 0 for i in range( maxa + 1 ) ]
for i in range( N - 1, -1, -1 ):
st_len[ i ] = qry( ftree, maxa + 1 - A[ i ] - 1 ) + 1
upd( ftree, maxa + 1 - A[ i ], st_len[ i ] )
ed_len = [ 0 for i in range( N ) ]
ftree = [ 0 for i in range( maxa + 1 ) ]
for i in range( N ):
ed_len[ i ] = qry( ftree, A[ i ] - 1 ) + 1
upd( ftree, A[ i ], ed_len[ i ] )
max_len = max( st_len )
st_cnt_len = [ 0 for i in range( N + 1 ) ]
for i in range( N ):
if ed_len[ i ] + st_len[ i ] - 1 == max_len:
st_cnt_len[ st_len[ i ] ] += 1
for i in range( N ):
if ed_len[ i ] + st_len[ i ] - 1 != max_len:
print( 1, end = "" )
elif st_cnt_len[ st_len[ i ] ] > 1:
print( 2, end = "" )
else:
print( 3, end = "" )
print()
``` | output | 1 | 52,587 | 12 | 105,175 |
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