message stringlengths 2 48.6k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 318 108k | cluster float64 8 8 | __index_level_0__ int64 636 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5 | instruction | 0 | 26,207 | 8 | 52,414 |
Tags: implementation
Correct Solution:
```
n,k=map(int,input().split())
s=list(map(int,input().split()))
dic = { n%s[i] : i+1 for i in range(k) }
l=[ n%s[i] for i in range(k) ]
l.sort()
print(dic[l[0]],n//s[dic[l[0]]-1])
``` | output | 1 | 26,207 | 8 | 52,415 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5 | instruction | 0 | 26,208 | 8 | 52,416 |
Tags: implementation
Correct Solution:
```
#n=int(input())
#k=int(input())
n,k=map(int,input().split())
min=10**18
ans=0
ansi=1
al=input().split()
for i in range(1,k+1):
a=int(al[i-1])
r=n%a
if r<min:
min=r
ans=n//a
ansi=i
print("%d %d\n" % (ansi, ans))
``` | output | 1 | 26,208 | 8 | 52,417 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5 | instruction | 0 | 26,209 | 8 | 52,418 |
Tags: implementation
Correct Solution:
```
n,k=[int(x) for x in input().split()]
s=input()
v=[int(x) for x in s.split()]
d=[n-n//x*x for x in v]
print(d.index(min(d))+1,n//v[d.index(min(d))])
``` | output | 1 | 26,209 | 8 | 52,419 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5 | instruction | 0 | 26,210 | 8 | 52,420 |
Tags: implementation
Correct Solution:
```
n,k=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]
j=k*[0]
for i in range(k):
j[i]=n%a[i]
print(j.index(min(j))+1,n//(a[j.index(min(j))]))
``` | output | 1 | 26,210 | 8 | 52,421 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5
Submitted Solution:
```
def farm(n,k,l):
res = []
for i in range(0,len(l)):
res.append(n%l[i])
v = min(res)
for i in range(0,len(res)):
if v == res[i]:
print((i+1),end=" ")
print(n//l[i])
break
n,k = list(map(int,input().split()))
l = list(map(int,input().split()))
farm(n,k,l)
``` | instruction | 0 | 26,212 | 8 | 52,424 |
Yes | output | 1 | 26,212 | 8 | 52,425 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.
Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.
Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.
Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.
Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.
Input
The first line contains two integers N and K (0 ≤ N ≤ 1018, 1 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.
The second line contains K integers a1, a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.
Output
Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.
If there are many correct answers, output any of them.
Examples
Input
19 3
5 4 10
Output
2 4
Input
28 3
5 6 30
Output
1 5
Submitted Solution:
```
import sys
N, K = map(int, input().split())
arr = list(map(int, input().split()))
Min = int(sys.maxsize)
ans =int(0)
ans1 = int(0)
for i in range(0, K):
x = int(N % arr[i])
if x < Min:
ans = i+1
Min = x
ans1 = int(N/arr[i])
print(ans, ans1)
``` | instruction | 0 | 26,218 | 8 | 52,436 |
No | output | 1 | 26,218 | 8 | 52,437 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,293 | 8 | 52,586 |
"Correct Solution:
```
N , K = map(int, input().split())
h = list(map(int, input().split()))
Dp = [None]*N
Dp[0] = 0
for i in range(1,N):
Dp[i] =min([Dp[j]+abs(h[i]-h[j]) for j in range(max(0,i-K),i)])
print(Dp[-1])
``` | output | 1 | 26,293 | 8 | 52,587 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,294 | 8 | 52,588 |
"Correct Solution:
```
n,k,*h=map(int,open(0).read().split());p=[0]
for i in range(1,n):p+=[min(p[j]+abs(h[j]-h[i])for j in range(max(i-k,0),i))]
print(p[-1])
``` | output | 1 | 26,294 | 8 | 52,589 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,295 | 8 | 52,590 |
"Correct Solution:
```
n,k=map(int,input().split());h=list(map(int,input().split()))
dp=[10**9]*n;dp[0]=0
for i in range(1,n):
kk=max(0,i-k)
for j in range(kk,i):dp[i]=min(dp[i],dp[j]+abs(h[j]-h[i]))
print(dp[-1])
``` | output | 1 | 26,295 | 8 | 52,591 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,296 | 8 | 52,592 |
"Correct Solution:
```
n,k = map(int, input().split())
h = list(map(int, input().split()))
dp = [float('inf')]*n
dp[0] = 0
for i in range(1,n):
dp[i] = min([dp[j]+abs(h[i]-h[j]) for j in range(max(0,i-k),i)])
print(dp[-1])
``` | output | 1 | 26,296 | 8 | 52,593 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,297 | 8 | 52,594 |
"Correct Solution:
```
N,K=map(int,input().split())
H=list(map(int,input().split()))
dp=[0]*N
for i in range(N-1):
A=[]
for j in range(K):
if i-j>=0:
A.append(dp[i-j]+abs(H[i-j]-H[i+1]))
dp[i+1]=min(A)
print(dp[N-1])
``` | output | 1 | 26,297 | 8 | 52,595 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,298 | 8 | 52,596 |
"Correct Solution:
```
n, k = [int(e) for e in input().split()]
arr = [int(e) for e in input().split()]
dp = [0]*len(arr)
for i in range(1,len(arr)):
dp[i] = min(dp[j] + abs(arr[i]-arr[j]) for j in range(max(i-k,0),i))
print(dp[-1])
``` | output | 1 | 26,298 | 8 | 52,597 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,299 | 8 | 52,598 |
"Correct Solution:
```
N,K=map(int, input().split())
a = list(map(int, input().split()))
dp = [0]+[float("inf")]*N
for i,h1 in enumerate(a):
for j in range(i+1,min(i+K+1,N)):
dp[j] = min(dp[j],dp[i]+abs(h1-a[j]))
print(dp[N-1])
``` | output | 1 | 26,299 | 8 | 52,599 |
Provide a correct Python 3 solution for this coding contest problem.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40 | instruction | 0 | 26,300 | 8 | 52,600 |
"Correct Solution:
```
n,k=map(int,input().split())
h=list(map(int,input().split()))
d=[float('inf')]*n
d[0]=0
for i in range(n-1):
for j in range(1,min(k+1,n-i)):
d[i+j]=min(d[i+j],d[i]+abs(h[i+j]-h[i]))
print(d[-1])
``` | output | 1 | 26,300 | 8 | 52,601 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
n,k,*h=map(int,open(0).read().split());d=[0]
for i in range(1,n):d+=[min(d[j]+abs(h[j]-h[i])for j in range(max(0,i-k),i))]
print(d[-1])
``` | instruction | 0 | 26,301 | 8 | 52,602 |
Yes | output | 1 | 26,301 | 8 | 52,603 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
n,k=map(int,input().split())
h=list(map(int,input().split()))
dp=[0]
for i in range(1,n):
p=10**12
for j in range(k):
if j>=i:
break
else:
p=min(p,dp[i-j-1]+abs(h[i]-h[i-j-1]))
dp.append(p)
print(dp[-1])
``` | instruction | 0 | 26,302 | 8 | 52,604 |
Yes | output | 1 | 26,302 | 8 | 52,605 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
n,k = list(map(int, input().split()))
h = list(map(int, input().split()))
dp = [0] * n
dp[0]=0
for i in range(1, n):
dp[i] = min([dp[j]+abs(h[i]-h[j]) for j in range(max(i-k, 0), i)])
print(dp[n-1])
``` | instruction | 0 | 26,303 | 8 | 52,606 |
Yes | output | 1 | 26,303 | 8 | 52,607 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
n, k = map(int,input().split())
h = tuple(map(int, input().split()))
dp = [0]+[float('inf')]*(n-1)
for i in range(1,n):
for j in range(max(0, i-k), i):
dp[i] = min(dp[i], dp[j]+abs(h[j]-h[i]))
print(dp[-1])
``` | instruction | 0 | 26,304 | 8 | 52,608 |
Yes | output | 1 | 26,304 | 8 | 52,609 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
from bisect import bisect_left as bl, bisect_right as br, insort
import sys
import heapq
#from math import *
from collections import defaultdict as dd, deque
def data(): return sys.stdin.readline().strip()
def mdata(): return map(int, data().split())
out=sys.stdout.write
#sys.setrecursionlimit(100000)
n,k=mdata()
H=list(mdata())
d=[1e9]*n
d[0]=0
for i in range(1,n):
for j in range(i-1,max(i-k-1,-1),-1):
d[i]=min(d[i],d[j]+abs(H[i]-H[j]))
print(d[-1])
``` | instruction | 0 | 26,305 | 8 | 52,610 |
No | output | 1 | 26,305 | 8 | 52,611 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
N, K = map(int, input().split())
H = list(map(int, input().split())) + [float('inf')] * K
dp = [float('inf')] * (N + K)
dp[0] = 0
for i in range(1, N):
for j in range(1, K + 1):
dp[i] = min(dp[i], dp[i - j] + abs(H[i - j] - H[i]))
print(dp[N - 1])
``` | instruction | 0 | 26,306 | 8 | 52,612 |
No | output | 1 | 26,306 | 8 | 52,613 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
N, K = map(int, input().split())
h = list(map(int, input().split()))
dp = [10000000]*N
dp[0] = 0
# 配るDP作戦 dp[i]の値が分かっている場合にdp[i]を用いてdp[i+1],dp[i+2]...を更新
# for i in range(N):
# for k in range(1,min(K+1,N-i)):
# dp[i+k]=min(dp[i+k],dp[i]+abs(h[i]-h[i+k]))
# # print("i=",i," dp=",dp)
# print(dp[-1])
for i in range(N):
if i == 1:
dp[i] = dp[i-1]+abs(h[i]-h[i-1])
else:
for k in range(1,K+1):
dp[i] = min(dp[i], dp[i-k]+abs(h[i]-h[i-k]))
print(dp[-1])
``` | instruction | 0 | 26,307 | 8 | 52,614 |
No | output | 1 | 26,307 | 8 | 52,615 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are N stones, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), the height of Stone i is h_i.
There is a frog who is initially on Stone 1. He will repeat the following action some number of times to reach Stone N:
* If the frog is currently on Stone i, jump to one of the following: Stone i + 1, i + 2, \ldots, i + K. Here, a cost of |h_i - h_j| is incurred, where j is the stone to land on.
Find the minimum possible total cost incurred before the frog reaches Stone N.
Constraints
* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq K \leq 100
* 1 \leq h_i \leq 10^4
Input
Input is given from Standard Input in the following format:
N K
h_1 h_2 \ldots h_N
Output
Print the minimum possible total cost incurred.
Examples
Input
5 3
10 30 40 50 20
Output
30
Input
3 1
10 20 10
Output
20
Input
2 100
10 10
Output
0
Input
10 4
40 10 20 70 80 10 20 70 80 60
Output
40
Submitted Solution:
```
n,k = map(int,input().split())
h = list(map(int,input().split()))
dp = [0]
for i in range(n-1):
cnt = float("inf")
for j in range(k):
if i-j < 0: continue
cnt = min(dp[-1-j] + abs(h[i-j]-h[i+1]), cnt)
#print("i =",i,"j =",j,"i-j =",i-j,"i+j =",i+j,"cnt =",cnt)
dp.append(cnt)
print(dp[-1])
``` | instruction | 0 | 26,308 | 8 | 52,616 |
No | output | 1 | 26,308 | 8 | 52,617 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,463 | 8 | 52,926 |
"Correct Solution:
```
def solve():
n = int(input())
dp = [0] * (n + 1)
def update(i, x):
while i <= n:
if dp[i] < x:
dp[i] = x
i += i & -i
def get_sum(i):
s = 0
while i > 0:
t = dp[i]
if s < t:
s = t
i -= i & -i
return s
A = map(int, input().split())
for a_i in A:
update(a_i, get_sum(a_i - 1) + a_i)
ans = n * (n + 1) // 2 - get_sum(n)
print(ans)
solve()
``` | output | 1 | 26,463 | 8 | 52,927 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,464 | 8 | 52,928 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# segment tree
class SegmentTree:
def _max(self,a,b):
if a < b:
return b
else:
return a
def __init__(self,n):
self.N = 1
while(self.N < n):
self.N *= 2
self.seg = [0] * (self.N * 2 -1)
def max_update(self,k,a):
k += self.N - 1
self.seg[k] = a
while(0 < k):
k = (k-1)//2;
self.seg[k] = self._max(self.seg[2*k +1], self.seg[2*k + 2])
def _max_query(self,a, b, k, l, r):
if r <= a or b <= l:
return 0
if a <= l and r <= b:
return self.seg[k]
else:
vl = self._max_query(a, b, k * 2 + 1, l, (l + r)//2)
vr = self._max_query(a, b, k * 2 + 2, (l + r)//2, r)
return self._max(vl,vr)
def max_query(self,a, b):
return self._max_query(a, b, 0, 0, self.N)
n = int(input())
X = list(map(int,input().split()))
st = SegmentTree(n+1)
for x in X:
st.max_update(x, st.max_query(0,x) + x)
print((n * (n + 1))//2 - st.max_query(0,n+1))
``` | output | 1 | 26,464 | 8 | 52,929 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,465 | 8 | 52,930 |
"Correct Solution:
```
class RangeMaximumQuery:
def __init__(self, n):
self.size = n
self.dat = [0] * (2 * n - 1)
def update(self, i, x):
i += self.size - 1
self.dat[i] = x
while i > 0:
i = (i - 1) // 2
d1 = self.dat[i * 2 + 1]
d2 = self.dat[i * 2 + 2]
if d1 > d2:
self.dat[i] = d1
else:
self.dat[i] = d2
def getmax(self, a, b, k, l, r):
if r <= a or b <= l:
return 0
elif a <= l and r <= b:
return self.dat[k]
else:
vl = self.getmax(a, b, k * 2 + 1, l, (l + r) // 2)
vr = self.getmax(a, b, k * 2 + 2, (l + r) // 2, r)
if vl > vr:
return vl
else:
return vr
def solve():
from math import ceil, log
n = int(input())
A = map(int, input().split())
s = 2 ** ceil(log(n, 2))
W = RangeMaximumQuery(s)
for a_i in A:
cost = W.getmax(0, a_i - 1, 0, 0, s) + a_i
W.update(a_i - 1, cost)
ans = n * (n + 1) // 2 - W.getmax(0, n, 0, 0, s)
print(ans)
solve()
``` | output | 1 | 26,465 | 8 | 52,931 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,466 | 8 | 52,932 |
"Correct Solution:
```
import math
class SegmentTree:
"""
Segment Tree
query:
1. update(i, val): update i-th value to val
2. query(low, high): find f(value) in [low, high)
time complexity: O(log n)
space complexity: O(2n)
"""
def __init__(self, N, f, default):
self.N = 1 << math.ceil(math.log(N, 2))
self.default = default
self.f = f
self.segtree = [self.default] * (self.N * 2 - 1)
def update(self, i, val):
i += self.N - 1
self.segtree[i] = val
while i > 0:
i = (i - 1) // 2
self.segtree[i] = self.f(self.segtree[2*i+1], self.segtree[2*i+2])
def query(self, low, high, k=0, left=0, right=-1):
if right == -1:
right = self.N
if right <= low or high <= left:
return self.default
if low <= left and right <= high:
return self.segtree[k]
else:
mid = (left + right) // 2
return self.f(self.query(low, high, 2*k+1, left, mid),
self.query(low, high, 2*k+2, mid, right))
N = int(input())
X = [int(x) for x in input().split()]
dp = SegmentTree(N, lambda x, y: max(x, y), 0)
for x, i in sorted((x, i) for i, x in enumerate(X)):
dp.update(i, dp.query(0, i) + x)
print(N * (N + 1) // 2 - dp.query(0, N))
``` | output | 1 | 26,466 | 8 | 52,933 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,467 | 8 | 52,934 |
"Correct Solution:
```
import math
class FenwickTree:
def __init__(self, a_list, f, default):
# 0-indexed
self.N = len(a_list)
self.bit = a_list[:]
self.f = f
self.default = default
for _ in range(self.N, 1 << (math.ceil(math.log(self.N, 2)))):
self.bit.append(self.default)
for i in range(self.N - 1):
self.bit[i | (i + 1)] = self.f(self.bit[i | (i + 1)], self.bit[i])
def update(self, i, val):
while i < self.N:
self.bit[i] = self.f(self.bit[i], val)
i |= i + 1
def query(self, n):
# [0, n]
ret = 0
while n >= 0:
ret = self.f(ret, self.bit[n])
n = (n & (n + 1)) - 1
return ret
N = int(input())
X = [int(x) for x in input().split()]
dp = FenwickTree([0] * N, lambda x, y: max(x, y), 0)
for x, i in sorted((x, i) for i, x in enumerate(X)):
dp.update(i, dp.query(i) + x)
print(N * (N + 1) // 2 - dp.query(N - 1))
``` | output | 1 | 26,467 | 8 | 52,935 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,468 | 8 | 52,936 |
"Correct Solution:
```
import math
import sys
if sys.version[0] == '2':
range, input = xrange, raw_input
class FenwickTree:
def __init__(self, a_list, f, default):
# 0-indexed
self.N = len(a_list)
self.bit = a_list[:]
self.f = f
self.default = default
for _ in range(self.N, 1 << int(math.ceil(math.log(self.N, 2)))):
self.bit.append(self.default)
for i in range(self.N - 1):
self.bit[i | (i + 1)] = self.f(self.bit[i | (i + 1)], self.bit[i])
def update(self, i, val):
while i < self.N:
self.bit[i] = self.f(self.bit[i], val)
i |= i + 1
def query(self, n):
# [0, n]
ret = 0
while n >= 0:
ret = self.f(ret, self.bit[n])
n = (n & (n + 1)) - 1
return ret
N = int(input())
X = [int(x) for x in input().split()]
dp = FenwickTree([0] * N, lambda x, y: max(x, y), 0)
for x in X:
dp.update(x - 1, dp.query(x - 1) + x)
print(N * (N + 1) // 2 - dp.query(N - 1))
``` | output | 1 | 26,468 | 8 | 52,937 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,469 | 8 | 52,938 |
"Correct Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# st = SegmentTree(n, lambda a,b: a if a > b else b)
class SegmentTree:
def __init__(self,n,f):
self.N = 1
self.f = f
while(self.N < n):
self.N *= 2
self.seg = [0] * (self.N * 2 -1)
def _update(self,k,a,f):
k += self.N - 1
self.seg[k] = a
while(0 < k):
k = (k-1)//2;
self.seg[k] = self.f(self.seg[2*k +1], self.seg[2*k + 2])
def _query(self,a, b, k, l, r,f):
if r <= a or b <= l:
return 0
if a <= l and r <= b:
return self.seg[k]
else:
vl = self._query(a, b, k * 2 + 1, l, (l + r)//2,f)
vr = self._query(a, b, k * 2 + 2, (l + r)//2, r,f)
return self.f(vl,vr)
def query(self, a, b):
return self._query(a, b, 0, 0, self.N, self.f)
def update(self, k ,a):
return self._update(k, a, self.f)
n = int(input())
X = list(map(int,input().split()))
st = SegmentTree(n+1, lambda a,b: a if b < a else b)
for x in X:
st.update(x, st.query(0,x) + x)
print((n * (n + 1))//2 - st.query(0,n+1))
``` | output | 1 | 26,469 | 8 | 52,939 |
Provide a correct Python 3 solution for this coding contest problem.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19 | instruction | 0 | 26,470 | 8 | 52,940 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**13
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
class Seg():
def __init__(self, na, default, func):
if isinstance(na, list):
n = len(na)
else:
n = na
i = 1
while 2**i <= n:
i += 1
self.D = default
self.H = i
self.N = 2**i
if isinstance(na, list):
self.A = [default] * (self.N) + na + [default] * (self.N-n)
for i in range(self.N-1,0,-1):
self.A[i] = func(self.A[i*2], self.A[i*2+1])
else:
self.A = [default] * (self.N*2)
self.F = func
def find(self, i):
return self.A[i + self.N]
def update(self, i, x):
i += self.N
self.A[i] = x
while i > 1:
i = i // 2
self.A[i] = self.merge(self.A[i*2], self.A[i*2+1])
def merge(self, a, b):
return self.F(a, b)
def total(self):
return self.A[1]
def query(self, a, b):
A = self.A
l = a + self.N
r = b + self.N
res = self.D
while l < r:
if l % 2 == 1:
res = self.merge(res, A[l])
l += 1
if r % 2 == 1:
r -= 1
res = self.merge(res, A[r])
l >>= 1
r >>= 1
return res
def main():
rr = []
def f(n):
a = LI()
seg = Seg(n, 0, max)
for c in a:
t = seg.query(0,c)
seg.update(c, t+c)
r = sum(a) - seg.total()
return r
while 1:
n = I()
if n == 0:
break
rr.append(f(n))
break
return '\n'.join(map(str,rr))
print(main())
``` | output | 1 | 26,470 | 8 | 52,941 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19
Submitted Solution:
```
def solve():
N = int(input())
*A, = map(int, input().split())
data = [0]*(N+1)
def update(k, x):
while k <= N:
data[k] = max(data[k], x)
k += k & -k
def get(k):
s = 0
while k:
s = max(s, data[k])
k -= k & -k
return s
B = [0]*N
for i in range(N):
B[A[i]-1] = i
ans = 0
for i in range(N):
j = B[i]
v = get(j+1)
update(j+1, v+i+1)
print(N*(N+1)//2 - get(N))
solve()
``` | instruction | 0 | 26,471 | 8 | 52,942 |
Yes | output | 1 | 26,471 | 8 | 52,943 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19
Submitted Solution:
```
import math
class RangeMaximumQuery:
"""
RangeMaximumQuery by Segment Tree
query:
1. update(i, val): update i-th value to val
2. query(low, high): find maximun value in [low, high)
time complexity: O(log n)
space complexity: O(2n)
"""
def __init__(self, N, zero=0):
self.N = 1 << math.ceil(math.log(N, 2))
self.zero = zero
self.segtree = [self.zero] * (self.N * 2 - 1)
def update(self, i, val):
i += self.N - 1
self.segtree[i] = val
while i > 0:
i = (i - 1) // 2
self.segtree[i] = max(self.segtree[2*i+1], self.segtree[2*i+2])
def query(self, low, high, k=0, left=0, right=-1):
if right == -1:
right = self.N
if right <= low or high <= left:
return self.zero
if low <= left and right <= high:
return self.segtree[k]
else:
mid = (left + right) // 2
return max(self.query(low, high, 2*k+1, left, mid),
self.query(low, high, 2*k+2, mid, right))
N = int(input())
X = [int(x) for x in input().split()]
dp = RangeMaximumQuery(N)
for x, i in sorted((x, i) for i, x in enumerate(X)):
dp.update(i, dp.query(0, i) + x)
print(N * (N + 1) // 2 - dp.query(0, N))
``` | instruction | 0 | 26,472 | 8 | 52,944 |
Yes | output | 1 | 26,472 | 8 | 52,945 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19
Submitted Solution:
```
class RMQ:
inf = 0
def __init__(self,n_):
self.n_ = n_
self.n = 1
while self.n < n_: self.n*=2
self.st = [self.inf]*(2*self.n-1)
def update(self,k,x):
k+=(self.n-1)
self.st[k] = x
while k >0:
k = (k-1)//2
self.st[k] = max(self.st[2*k+1],self.st[2*k+2])
def search(self,a,b,k,l,r):
if r<=a or b<=l :return self.inf
if a<=l and r<=b:return self.st[k]
L = self.search(a,b,k*2+1,l,(l+r)//2)
R = self.search(a,b,k*2+2,(l+r)//2,r)
return max(L,R)
def query(self,a,b):
return self.search(a,b,0,0,self.n)
n = int(input())
x = list(map(int,input().split()))
rmq = RMQ(n+1)
for i in x:
res = rmq.query(1,i)
rmq.update(i,i+res)
print (sum(x)-rmq.query(1,n+1))
``` | instruction | 0 | 26,473 | 8 | 52,946 |
Yes | output | 1 | 26,473 | 8 | 52,947 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Taro has decided to move. Taro has a lot of luggage, so I decided to ask a moving company to carry the luggage. Since there are various weights of luggage, I asked them to arrange them in order from the lightest one for easy understanding, but the mover left the luggage in a different order. So Taro tried to sort the luggage, but the luggage is heavy and requires physical strength to carry. Each piece of luggage can be carried from where it is now to any place you like, such as between other pieces of luggage or at the edge of the piece of luggage, but carrying one piece of luggage uses as much physical strength as the weight of that piece of luggage. Taro doesn't have much physical strength, so I decided to think of a way to arrange the luggage in order from the lightest one without using as much physical strength as possible.
Constraints
> 1 ≤ n ≤ 105
> 1 ≤ xi ≤ n (1 ≤ i ≤ n)
> xi ≠ xj (1 ≤ i, j ≤ n and i ≠ j)
>
* All inputs are given as integers
Input
> n
> x1 x2 ... xn
>
* n represents the number of luggage that Taro has
* x1 to xn represent the weight of each piece of luggage, and are currently arranged in the order of x1, x2, ..., xn.
Output
> S
>
* Output the total S of the minimum physical strength required to arrange the luggage in order from the lightest, but output the line break at the end
Examples
Input
4
1 4 2 3
Output
4
Input
5
1 5 3 2 4
Output
7
Input
7
1 2 3 4 5 6 7
Output
0
Input
8
6 2 1 3 8 5 4 7
Output
19
Submitted Solution:
```
def solve():
n = int(input())
A = map(int, input().split())
W = [0] * (n + 1)
for a_i in A:
W[a_i] = max(W[:a_i]) + a_i
ans = n * (n + 1) // 2 - max(W)
print(ans)
solve()
``` | instruction | 0 | 26,474 | 8 | 52,948 |
No | output | 1 | 26,474 | 8 | 52,949 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,742 | 8 | 53,484 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
from collections import deque
import os
import io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
cases = int(input())
for _ in range(cases):
n, m = map(int, input().split())
adj = [set() for i in range(n + 1)]
visited = [False] * (n + 1)
for i in range(m):
start, end = map(int, input().split())
adj[start].add(end)
adj[end].add(start)
queue = deque()
uncolored = deque()
uncolored.append(1)
solution = []
allE = 0
while uncolored:
index = uncolored.popleft()
if visited[index]:
continue
visited[index] = True
allE += 1
solution.append(str(index))
for i in adj[index]:
if not visited[i]:
queue.append(i)
while queue:
indexQ = queue.popleft()
if visited[indexQ]:
continue
visited[indexQ] = True
allE += 1
for i in adj[indexQ]:
if not visited[i]:
uncolored.append(i)
# print(visited)
if allE == n:
print("YES")
print(len(solution))
print(" ".join(solution))
else:
print("NO")
``` | output | 1 | 26,742 | 8 | 53,485 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,743 | 8 | 53,486 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
from collections import deque
for _ in range(int(input())):
n,m = map(int,input().split())
edge = [[] for i in range(n)]
for i in range(m):
u,v = map(int,input().split())
edge[u-1].append(v-1)
edge[v-1].append(u-1)
color = [-1 for i in range(n)]
used = [False]*n
color[0] = 0
used[0] = True
stack = [0]
for nv in edge[0]:
color[nv] = 1
while stack:
v = stack[-1]
#print(v+1,color[v])
if edge[v]:
nv = edge[v].pop()
if color[v]==0:
if not used[nv]:
used[nv] = True
stack.append(nv)
else:
if not used[nv] and color[nv]==-1:
used[nv] = True
color[nv] = 0
for nnv in edge[nv]:
color[nnv] = 1
stack.append(nv)
else:
stack.pop()
if not all(color[v]!=-1 for v in range(n)):
print("NO")
#print(color)
else:
print("YES")
ans = []
for i in range(n):
if color[i]==0:
ans.append(i+1)
print(len(ans))
print(*ans)
``` | output | 1 | 26,743 | 8 | 53,487 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,745 | 8 | 53,490 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
import sys, io, os
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
out = []
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
adj = [[] for _ in range(n)]
color = [-1]*n
color[0] = 0
for _ in range(m):
u, v= map(int, input().split())
adj[u-1].append(v-1)
adj[v-1].append(u-1)
stack = [0]
while stack:
nex = stack.pop()
for v in adj[nex]:
if color[nex] == 1:
if color[v] == -1:
stack.append(v)
color[v] = 0
elif color[v] == -1:
color[v] = 1 - color[nex]
stack.append(v)
for v in color:
if v == -1:
out.append('NO')
break
else:
out.append('YES')
final = [i + 1 for i in range(n) if color[i]]
out.append(str(len(final)))
out.append(' '.join(map(str,final)))
print('\n'.join(out))
``` | output | 1 | 26,745 | 8 | 53,491 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,746 | 8 | 53,492 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
#!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
def main():
for _ in range(int(input())):
n,m = map(int,input().split())
adjacencyList = []
for _ in range(n):
adjacencyList.append([])
for _ in range(m):
x,y = map(int,input().split())
adjacencyList[x-1].append(y-1)
adjacencyList[y-1].append(x-1)
isCovered = [False] * n
neighborList = [0]
ans = []
while neighborList:
cur = neighborList.pop()
if isCovered[cur]:
continue
ans.append(cur + 1)
isCovered[cur] = True
for elem in adjacencyList[cur]:
if not isCovered[elem]:
isCovered[elem] = True
for elem2 in adjacencyList[elem]:
neighborList.append(elem2)
if sum(isCovered) == n:
print("YES")
print(len(ans))
print(" ".join(map(str,ans)))
else:
print("NO")
# 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 | 26,746 | 8 | 53,493 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,747 | 8 | 53,494 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
import sys,os,io
input = sys.stdin.buffer.readline
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import deque
class Graph:
def __init__(self, N, M=-1):
self.V = N
if M>=0: self.E = M
self.edge = [[] for _ in range(self.V)]
self.color = [-1]*self.V
def add_edges(self, ind=1, bi=True):
for i in range(self.E):
a,b = map(int, input().split())
a -= ind; b -= ind
self.edge[a].append(b)
if bi: self.edge[b].append(a)
def bfs(self, start):
d = deque([start])
self.color[start]=1
while len(d)>0:
v = d.popleft()
for w in self.edge[v]:
if self.color[w]==-1:
self.color[w]=self.color[v]^1
d.append(w)
elif self.color[w]==self.color[v]==1:
self.color[w]=0
d.append(w)
T = int(input())
for t in range(T):
N, M = map(int, input().split())
G = Graph(N,M)
G.add_edges(ind=1, bi=True)
G.bfs(0)
if min(G.color)==-1:
print('NO')
continue
print('YES')
ans = []
for i in range(N):
if G.color[i]==1:
ans.append(i+1)
print(len(ans))
print(*ans)
``` | output | 1 | 26,747 | 8 | 53,495 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image> | instruction | 0 | 26,748 | 8 | 53,496 |
Tags: constructive algorithms, dfs and similar, graph matchings, graphs, greedy
Correct Solution:
```
import sys
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def ngsearch(x):
for to in E[x]:
if NG[to]==0:
NG[to]=1
Q.append(to)
t=int(input())
for tests in range(t):
n,m=map(int,input().split())
E=[[] for i in range(n+1)]
for i in range(m):
x,y=map(int,input().split())
E[x].append(y)
E[y].append(x)
OK=[0]*(n+1)
NG=[0]*(n+1)
Q=[]
ngsearch(1)
while Q:
x=Q.pop()
for to in E[x]:
if NG[to]==0:
OK[to]=1
ngsearch(to)
count=0
ANS=[]
for i in range(1,n+1):
if OK[i]==1 or NG[i]==1:
count+=1
if OK[i]==1:
ANS.append(i)
if count==n:
print("YES")
print(len(ANS))
print(*ANS)
else:
print("NO")
``` | output | 1 | 26,748 | 8 | 53,497 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
import sys,os,io
# input = sys.stdin.readline
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
from collections import deque
class Graph:
def __init__(self, N, M=-1):
self.V = N
if M>=0: self.E = M
self.edge = [[] for _ in range(self.V)]
self.color = [-1]*self.V
def add_edges(self, ind=1, bi=True):
for i in range(self.E):
a,b = map(int, input().split())
a -= ind; b -= ind
self.edge[a].append(b)
if bi: self.edge[b].append(a)
def bfs(self, start):
d = deque([start])
self.color[start]=1
while len(d)>0:
v = d.popleft()
for w in self.edge[v]:
if self.color[w]==-1:
self.color[w]=self.color[v]^1
d.append(w)
elif self.color[w]==self.color[v]==1:
self.color[w]=0
d.append(w)
T = int(input())
for t in range(T):
N, M = map(int, input().split())
G = Graph(N,M)
G.add_edges(ind=1, bi=True)
G.bfs(0)
if min(G.color)==-1:
print('NO')
continue
print('YES')
ans = []
for i in range(N):
if G.color[i]==1:
ans.append(i+1)
print(len(ans))
print(*ans)
``` | instruction | 0 | 26,750 | 8 | 53,500 |
Yes | output | 1 | 26,750 | 8 | 53,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
import io
import os
from collections import deque
def solve(N, M, edges):
graph = [[] for i in range(N)]
for u, v in edges:
graph[u].append(v)
graph[v].append(u)
source = 0
q = deque([source])
isTeacher = {}
while q:
node = q.popleft()
if node in isTeacher:
continue
for nbr in graph[node]:
if nbr in isTeacher and isTeacher[nbr] == 1:
isTeacher[node] = 0
break
else:
isTeacher[node] = 1
for nbr in graph[node]:
if nbr not in isTeacher:
q.append(nbr)
if len(isTeacher) != N:
return "NO"
ans = [str(k + 1) for k, v in isTeacher.items() if v]
return "YES\n" + str(len(ans)) + "\n" + " ".join(ans)
if __name__ == "__main__":
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
TC = int(input())
for tc in range(1, TC + 1):
(N, M) = [int(x) for x in input().split()]
edges = [[int(x) - 1 for x in input().split()] for i in range(M)] # 0 indexed
ans = solve(N, M, edges)
print(ans)
``` | instruction | 0 | 26,751 | 8 | 53,502 |
Yes | output | 1 | 26,751 | 8 | 53,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
from sys import stdin, stdout
def strange_housing(n, m, uv_d):
c_a = [-1] * (n + 1)
st = [1]
while len(st) > 0:
c = st.pop()
if c_a[c] != -1:
continue
c_a[c] = 1
s = set()
if c in uv_d:
for nc in uv_d[c]:
if c_a[nc] == -1:
c_a[nc] = 0
s.add(nc)
for nc in s:
if nc in uv_d:
for nnc in uv_d[nc]:
if c_a[nnc] == -1:
st.append(nnc)
r = []
for i in range(1, n+1):
if c_a[i] == -1:
return ['NO']
elif c_a[i] == 1:
r.append(i)
return ['YES', len(r), r]
t = int(stdin.readline())
for _ in range(t):
n, m = map(int, stdin.readline().split())
uv_d = {}
for _ in range(m):
u, v = map(int, stdin.readline().split())
if u not in uv_d:
uv_d[u] = []
if v not in uv_d:
uv_d[v] = []
uv_d[u].append(v)
uv_d[v].append(u)
r = strange_housing(n, m, uv_d)
stdout.write(r[0] + '\n')
if len(r) > 1:
stdout.write(str(r[1]) + '\n')
stdout.write(' '.join(map(str, r[2])) + '\n')
``` | instruction | 0 | 26,752 | 8 | 53,504 |
Yes | output | 1 | 26,752 | 8 | 53,505 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
import io,os
readline = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
T = int(readline())
Ans = []
for qu in range(T):
N, M = map(int, readline().split())
Edge = [[] for _ in range(N)]
for _ in range(M):
u, v = map(int, readline().split())
u -= 1
v -= 1
Edge[u].append(v)
Edge[v].append(u)
col = [-1]*N
col[0] = 0
stack = [0]
while stack:
vn = stack.pop()
if col[vn] == 2:
col[vn] = 0
if col[vn] == 0:
for vf in Edge[vn]:
if col[vf] == -1:
col[vf] = 1
stack.append(vf)
else:
col[vf] = 1
else:
for vf in Edge[vn]:
if col[vf] == -1:
col[vf] = 2
stack.append(vf)
if -1 in col:
Ans.append('NO')
else:
Ans.append('YES')
Ans.append(str(col.count(0)))
L = [l + 1 for l in range(N) if col[l] == 0]
Ans.append(' '.join(map(str, L)))
print('\n'.join(Ans))
``` | instruction | 0 | 26,753 | 8 | 53,506 |
Yes | output | 1 | 26,753 | 8 | 53,507 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
def dfs(G,u,col):
for v in G[u]:
if col[v] == -1:
col[v] = 0 if col[u]==1 else 1
dfs(G,v,col)
for _ in range(int(input())):
n,m=map(int,input().split())
G = [[]for i in range(n)]
for i in range(m):
u,v = map(int,input().split())
G[u-1].append(v-1)
G[v-1].append(u-1)
col = None
found = False
for c in [0,1]:
col = [-1]*n
col[0] = c
dfs(G, 0, col)
if min(col) == 0:
found = True
break
if not found:
print('NO')
else:
ans = []
for i in range(n):
if col[i]==1:
ans.append(str(i+1))
print('YES')
print(len(ans))
print(' '.join(ans))
``` | instruction | 0 | 26,754 | 8 | 53,508 |
No | output | 1 | 26,754 | 8 | 53,509 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
print("YES")
print(1)
print(2)
``` | instruction | 0 | 26,755 | 8 | 53,510 |
No | output | 1 | 26,755 | 8 | 53,511 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.buffer.readline
from collections import deque
for _ in range(int(input())):
n,m = map(int,input().split())
edge = [[] for i in range(n)]
for i in range(m):
u,v = map(int,input().split())
edge[u-1].append(v-1)
edge[v-1].append(u-1)
color = [-1 for i in range(n)]
used = [False]*n
color[0] = 0
used[0] = True
stack = [0]
for nv in edge[0]:
color[nv] = 1
while stack:
v = stack[-1]
#print(v+1,color[v])
if edge[v]:
nv = edge[v].pop()
if color[v]==0:
if not used[nv]:
used[nv] = True
stack.append(nv)
else:
if not used[nv]:
used[nv] = True
color[nv] = 0
for nnv in edge[nv]:
color[nnv] = 1
stack.append(nv)
else:
stack.pop()
if not all(color[v]!=-1 for v in range(n)):
print("NO")
#print(color)
else:
print("YES")
ans = []
for i in range(n):
if color[i]==0:
ans.append(i+1)
print(len(ans))
print(*ans)
``` | instruction | 0 | 26,756 | 8 | 53,512 |
No | output | 1 | 26,756 | 8 | 53,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Students of Winter Informatics School are going to live in a set of houses connected by underground passages. Teachers are also going to live in some of these houses, but they can not be accommodated randomly. For safety reasons, the following must hold:
* All passages between two houses will be closed, if there are no teachers in both of them. All other passages will stay open.
* It should be possible to travel between any two houses using the underground passages that are open.
* Teachers should not live in houses, directly connected by a passage.
Please help the organizers to choose the houses where teachers will live to satisfy the safety requirements or determine that it is impossible.
Input
The first input line contains a single integer t — the number of test cases (1 ≤ t ≤ 10^5).
Each test case starts with two integers n and m (2 ≤ n ≤ 3 ⋅ 10^5, 0 ≤ m ≤ 3 ⋅ 10^5) — the number of houses and the number of passages.
Then m lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v), describing a passage between the houses u and v. It is guaranteed that there are no two passages connecting the same pair of houses.
The sum of values n over all test cases does not exceed 3 ⋅ 10^5, and the sum of values m over all test cases does not exceed 3 ⋅ 10^5.
Output
For each test case, if there is no way to choose the desired set of houses, output "NO". Otherwise, output "YES", then the total number of houses chosen, and then the indices of the chosen houses in arbitrary order.
Examples
Input
2
3 2
3 2
2 1
4 2
1 4
2 3
Output
YES
2
1 3
NO
Input
1
17 27
1 8
2 9
3 10
4 11
5 12
6 13
7 14
8 9
8 14
8 15
9 10
9 15
10 11
10 15
10 17
11 12
11 17
12 13
12 16
12 17
13 14
13 16
14 16
14 15
15 16
15 17
16 17
Output
YES
8
1 3 4 5 6 9 14 17
Note
The picture below shows the second example test.
<image>
Submitted Solution:
```
import sys
from collections import deque
zz=1
sys.setrecursionlimit(10**5)
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('all.txt','w')
di=[[-1,0],[1,0],[0,1],[0,-1]]
def fori(n):
return [fi() for i in range(n)]
def inc(d,c,x=1):
d[c]=d[c]+x if c in d else x
def ii():
return input().rstrip()
def li():
return [int(xx) for xx in input().split()]
def fli():
return [float(x) for x in input().split()]
def comp(a,b):
if(a>b):
return 2
return 2 if a==b else 0
def gi():
return [xx for xx in input().split()]
def gtc(tc,ans):
print("Case #"+str(tc)+":",ans)
def cil(n,m):
return n//m+int(n%m>0)
def fi():
return int(input())
def pro(a):
return reduce(lambda a,b:a*b,a)
def swap(a,i,j):
a[i],a[j]=a[j],a[i]
def si():
return list(input().rstrip())
def mi():
return map(int,input().split())
def gh():
sys.stdout.flush()
def isvalid(i,j,n,m):
return 0<=i<n and 0<=j<m
def bo(i):
return ord(i)-ord('a')
def graph(n,m):
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
t=fi()
uu=t
def dfs(i):
vis[i]=1
for j in a[i]:
if j not in vis:
dfs(j)
while t>0:
t-=1
n,m=mi()
a=[[] for i in range(n+1)]
col=[-1]*(n+1)
vis={}
d={}
for i in range(m):
x,y=mi()
a[x].append(y)
a[y].append(x)
inc(d,x)
inc(d,y)
c=0
for i in range(1,n+1):
if i not in vis:
dfs(i)
c+=1
if c>1:
print("NO")
else:
print("YES")
vis={}
col=[0]*(n+1)
q=deque()
q.append(1)
while q:
i=q[0]
vis[i]=1
q.popleft()
for j in a[i]:
if j not in vis:
col[j]=col[i]^1
vis[j]=1
q.append(j)
print(col.count(0)-1)
for i in range(1,n+1):
if col[i]==0:
print(i,end=" ")
print()
``` | instruction | 0 | 26,757 | 8 | 53,514 |
No | output | 1 | 26,757 | 8 | 53,515 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Chris is very keen on his toy blocks. His teacher, however, wants Chris to solve more problems, so he decided to play a trick on Chris.
There are exactly s blocks in Chris's set, each block has a unique number from 1 to s. Chris's teacher picks a subset of blocks X and keeps it to himself. He will give them back only if Chris can pick such a non-empty subset Y from the remaining blocks, that the equality holds:
<image> "Are you kidding me?", asks Chris.
For example, consider a case where s = 8 and Chris's teacher took the blocks with numbers 1, 4 and 5. One way for Chris to choose a set is to pick the blocks with numbers 3 and 6, see figure. Then the required sums would be equal: (1 - 1) + (4 - 1) + (5 - 1) = (8 - 3) + (8 - 6) = 7.
<image>
However, now Chris has exactly s = 106 blocks. Given the set X of blocks his teacher chooses, help Chris to find the required set Y!
Input
The first line of input contains a single integer n (1 ≤ n ≤ 5·105), the number of blocks in the set X. The next line contains n distinct space-separated integers x1, x2, ..., xn (1 ≤ xi ≤ 106), the numbers of the blocks in X.
Note: since the size of the input and output could be very large, don't use slow output techniques in your language. For example, do not use input and output streams (cin, cout) in C++.
Output
In the first line of output print a single integer m (1 ≤ m ≤ 106 - n), the number of blocks in the set Y. In the next line output m distinct space-separated integers y1, y2, ..., ym (1 ≤ yi ≤ 106), such that the required equality holds. The sets X and Y should not intersect, i.e. xi ≠ yj for all i, j (1 ≤ i ≤ n; 1 ≤ j ≤ m). It is guaranteed that at least one solution always exists. If there are multiple solutions, output any of them.
Examples
Input
3
1 4 5
Output
2
999993 1000000
Input
1
1
Output
1
1000000 | instruction | 0 | 26,874 | 8 | 53,748 |
Tags: greedy, implementation, math
Correct Solution:
```
import sys
from math import gcd,sqrt,ceil
from collections import defaultdict,Counter,deque
import math
# input=sys.stdin.readline
# def print(x):
# sys.stdout.write(str(x)+"\n")
# sys.stdout=open("CP1/output.txt",'w')
# sys.stdin=open("CP1/input.txt",'r')
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# import sys
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_sum(bit,i):
s = 0
i+=1
while i>0:
s+=bit[i]
i-=i&(-i)
return s
def update(bit,n,i,v):
i+=1
while i<=n:
bit[i]+=v
i+=i&(-i)
def modInverse(b,m):
g = math.gcd(b, m)
if (g != 1):
return -1
else:
return pow(b, m - 2, m)
# print(lo,ha)
n = int(input())
l = list(map(int,input().split()))
s = 10**6
ans = set()
l = set(l)
w = 0
for i in range(1,s+1):
opp = s+1-i
if i in l and opp not in l:
ans.add(opp)
elif i in l and opp in l:
w+=1
w//=2
for i in range(1,s+1):
opp = s+1-i
if w == 0:
break
if opp not in ans and i not in ans and i not in l and opp not in l:
ans.add(opp)
ans.add(i)
w-=1
print(len(ans))
print(*ans)
``` | output | 1 | 26,874 | 8 | 53,749 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Anya loves to fold and stick. Today she decided to do just that.
Anya has n cubes lying in a line and numbered from 1 to n from left to right, with natural numbers written on them. She also has k stickers with exclamation marks. We know that the number of stickers does not exceed the number of cubes.
Anya can stick an exclamation mark on the cube and get the factorial of the number written on the cube. For example, if a cube reads 5, then after the sticking it reads 5!, which equals 120.
You need to help Anya count how many ways there are to choose some of the cubes and stick on some of the chosen cubes at most k exclamation marks so that the sum of the numbers written on the chosen cubes after the sticking becomes equal to S. Anya can stick at most one exclamation mark on each cube. Can you do it?
Two ways are considered the same if they have the same set of chosen cubes and the same set of cubes with exclamation marks.
Input
The first line of the input contains three space-separated integers n, k and S (1 ≤ n ≤ 25, 0 ≤ k ≤ n, 1 ≤ S ≤ 1016) — the number of cubes and the number of stickers that Anya has, and the sum that she needs to get.
The second line contains n positive integers ai (1 ≤ ai ≤ 109) — the numbers, written on the cubes. The cubes in the input are described in the order from left to right, starting from the first one.
Multiple cubes can contain the same numbers.
Output
Output the number of ways to choose some number of cubes and stick exclamation marks on some of them so that the sum of the numbers became equal to the given number S.
Examples
Input
2 2 30
4 3
Output
1
Input
2 2 7
4 3
Output
1
Input
3 1 1
1 1 1
Output
6
Note
In the first sample the only way is to choose both cubes and stick an exclamation mark on each of them.
In the second sample the only way is to choose both cubes but don't stick an exclamation mark on any of them.
In the third sample it is possible to choose any of the cubes in three ways, and also we may choose to stick or not to stick the exclamation mark on it. So, the total number of ways is six.
Submitted Solution:
```
from itertools import combinations
import math
if __name__ == '__main__':
string = input().split(' ')
n, k, G = int(string[0]), int(string[1]), int(string[2])
a = []
for each in input().split(' '):
a.append(int(each))
b = list(range(len(a)))
arr = []
for i in range(k + 1):
com = list(combinations(b, i))
for each in com:
arr1 = a[:]
for value in each:
arr1[value] = math.factorial(arr1[value])
arr.append(arr1)
count = 0
for each in arr:
for i in range(1, len(a)):
for d in combinations(each, i):
if sum(d) == G:
count += 1
print(count)
``` | instruction | 0 | 26,917 | 8 | 53,834 |
No | output | 1 | 26,917 | 8 | 53,835 |
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