message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Provide tags and a correct Python 3 solution for this coding contest problem.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1. | instruction | 0 | 9,923 | 14 | 19,846 |
Tags: brute force, combinatorics, math, meet-in-the-middle, two pointers
Correct Solution:
```
def find(x):
if(par[x]==x):
return par[x]
par[x]=find(par[x])
return par[x]
def union(a,b):
pa=find(a)
pb=find(b)
if(pa!=pb):
if(size[pa]>size[pb]):
pa,pb=pb,pa
size[pb]+=size[pa]
size[pa]=1
par[pa]=pb
n,m=map(int,input().split())
b=[]
par=[i for i in range(n+1)]
size=[1 for i in range(n+1)]
for i in range(m):
arr=list(map(int,input().split()))
b.append(arr)
collect=[]
d={}
e={}
for i in range(m):
e={}
for j in range(n-1):
if(i==0):
d[b[i][j]]=b[i][j+1]
else:
if b[i][j] in d and d[b[i][j]]==b[i][j+1]:
e[b[i][j]]=b[i][j+1]
if(i>0):
d=e
count=0
for i in range(n):
if (i+1) in d:
union(i+1,d[i+1])
ans=0
count=n
for i in range(1,n+1):
if(size[i]>1):
ans+=(size[i]*(size[i]+1))//2
count-=size[i]
print(ans+count)
``` | output | 1 | 9,923 | 14 | 19,847 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1. | instruction | 0 | 9,924 | 14 | 19,848 |
Tags: brute force, combinatorics, math, meet-in-the-middle, two pointers
Correct Solution:
```
from sys import stdin
n,m=map(int,stdin.readline().strip().split())
dp=[[-1 for i in range(n+1)] for j in range(m+1)]
for i in range(m):
s=list(map(int,stdin.readline().strip().split()))
for j in range(n-2,-1,-1):
dp[i][s[j]]=s[j+1]
dp1=[1 for i in range(n)]
for i in range(n-2,-1,-1):
t=True
for j in range(m):
if dp[j][s[i]]!=s[i+1]:
t=False
if t:
dp1[i]=dp1[i]+dp1[i+1]
print(sum(dp1))
``` | output | 1 | 9,924 | 14 | 19,849 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1. | instruction | 0 | 9,925 | 14 | 19,850 |
Tags: brute force, combinatorics, math, meet-in-the-middle, two pointers
Correct Solution:
```
n, m = map(int, input().split())
graph = [0] * n
_next = [True] * n
_next[n - 1] = False
def read_array():
return list(map(lambda x: x - 1, map(int, input().split())))
first = read_array()
for i in range(n):
graph[first[i]] = i
for cnt in range(1, m):
a = read_array()
for i in range(n - 1):
if graph[a[i]] + 1 != graph[a[i + 1]]:
_next[graph[a[i]]] = False
_next[graph[a[n - 1]]] = False
l = 0
ans = 0
for cnt in range(n):
l += 1
if not _next[cnt]:
ans += (l * (l + 1)) // 2
l = 0
print(ans)
``` | output | 1 | 9,925 | 14 | 19,851 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
from collections import defaultdict
from collections import deque
(n, m) = [int(x) for x in input().split()]
messages = []
for i in range(m):
temp = [int(x) for x in input().split()]
messages.append(temp)
seq = defaultdict(lambda: 0)
pairs = {}
for k in range(n - 1):
pairs[messages[0][k]] = messages[0][k + 1]
for i in range(1, m):
for k in range(n-1):
temp = messages[i][k]
if pairs.get(temp, None) != messages[i][k+1]:
pairs.pop(temp, None)
pairs.pop(messages[i][n-1], None)
sequences = []
starts = set(pairs.keys())
ends = set(pairs.values())
conn = starts & ends
for key, value in pairs.items():
if key in conn:
continue
val = value
temp = []
temp.append(key)
temp.append(value)
while val in starts:
val = pairs[val]
temp.append(val)
sequences.append(temp)
# print(sequences)
variants = n
for seq in sequences:
l = len(seq)
variants+=l*(l-1)//2
print(variants)
``` | instruction | 0 | 9,926 | 14 | 19,852 |
Yes | output | 1 | 9,926 | 14 | 19,853 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
from sys import stdin, stdout
from math import floor
def main():
global n,m,a
n,m=[int(x) for x in stdin.readline().split()]
a=[]
for i in range(m):
a.append([int(x) for x in stdin.readline().split()])
g=[]
G=[]
for i in range(n+3):
g.append({})
G.append({})
for j in range(m):
b=a[j]
for i in range(1,n):
u=b[i-1]
v=b[i]
if (g[u].get(v)==None):
g[u][v]=1
else:
g[u][v]=g[u][v]+1
for u in range(1,n+1):
for v,k in g[u].items():
if (k==m):
G[u][v]=True
res=0
i=0
j=-1
b=a[0]+[0]
res=0
while ((i<n) and (j<n)):
i=j+1
j=j+1
if ((i>=n) or (j>=n)):
break
while ((j+1)<n):
if (G[b[j]].get(b[j+1])==True):
j=j+1
else:
break
l=j-i+1
res=res+l+floor((l*(l-1))/2)
stdout.write(str(res))
return 0
if __name__ == "__main__":
main()
``` | instruction | 0 | 9,927 | 14 | 19,854 |
Yes | output | 1 | 9,927 | 14 | 19,855 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
from sys import stdin, stdout
from math import *
def main():
global n,m,a,b
n,m=[int(x) for x in stdin.readline().split()]
a=[]
for i in range(m):
a.append([int(x) for x in stdin.readline().split()])
g=[]
G=[]
for i in range(n+3):
g.append({})
G.append({})
for b in a:
for i in range(1,n):
u=b[i-1]
v=b[i]
if (g[u].get(v)==None):
g[u][v]=1
else:
g[u][v]=g[u][v]+1
if (g[u][v]==m):
G[u][v]=True
res=0
i=0
j=-1
b=a[0]+[0]
res=0
while ((i<n) and (j<n)):
i=j+1
j=j+1
if (i>=n):
break
while ((j+1)<=n):
if (G[b[j]].get(b[j+1])!=None):
j=j+1
else:
break
l=j-i+1
res=res+l+floor((l*(l-1))/2)
stdout.write(str(res))
return 0
if __name__ == "__main__":
main()
``` | instruction | 0 | 9,928 | 14 | 19,856 |
Yes | output | 1 | 9,928 | 14 | 19,857 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
n,m=map(int,input().split())
from collections import *
al=defaultdict(list)
for i in range(m):
z=list(map(int,input().split()))
for i in range(len(z)):
al[z[i]].append(i)
dl=defaultdict(int)
total=0
for i in range(len(z)):
count=1
x=z[i]
if(dl[x]==1):
continue;
dl[x]=1
q=al[x]
total+=(count)
start=1
count+=1
while(1):
for i in range(len(q)):
q[i]+=start
s=[]
u=q[-1]
if(u<len(z)):
s=al[z[u]]
if(s==q):
total+=count
count+=1
dl[z[u]]=1
else:
break;
else:
break;
print(total)
``` | instruction | 0 | 9,929 | 14 | 19,858 |
Yes | output | 1 | 9,929 | 14 | 19,859 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
n,m=map(int,input().split())
arr=[]
arr1=[]
for i in range(m):
arrx=list(map(int,input().split()))
arry=[]
for j in range(n):
arry.append((arrx[j],j+1))
arry.sort()
arr.append(arrx)
arr1.append(arry)
ans=n
i=0
j=1
flag=0
previ=0
prevj=1
while(i<n-1 and j<n):
k1=arr[0][i]
k2=arr[0][j]
l=0
while(l<m):
k=i
while(k<j-1):
if(arr1[l][arr[0][i+k]][1]!=arr1[l][arr[0][i+k+1]][1]-1):
flag=1
break
k+=1
if(flag==1):
break
l+=1
if(flag==1):
ans+=((j-i-1)*(j-i))//2
ans-=j-i-1
if(j-i>1):
i+=1
else:
i+=1
j+=1
else:
j+=1
#print(i,j)
if(flag==0):
ans+=((j-prevj)*(j-prevj+1))//2
ans-=j-prevj
previ=i
prevj=j
print(ans)
``` | instruction | 0 | 9,930 | 14 | 19,860 |
No | output | 1 | 9,930 | 14 | 19,861 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
n, m = [int(i) for i in input().split(' ')]
connection = [None] * n
if m == 1:
print(int((n+1)*n/2))
sys.exit(0)
svidetel = input().split(' ')
p = int(svidetel[0])-1
for current_pok in svidetel[1:]:
q = int(current_pok)-1
connection[p] = 1
p = q
for i in range(m-1):
svidetel = input().split(' ')
p = int(svidetel[0])-1
try:
connection[connection.index(p)] = None
except ValueError:
pass
for current_pok in svidetel[1:]:
q = int(current_pok)-1
if connection[p] != q:
connection[p] = None
p = q
connection[p] = None
is_bundled = [True] + [False] * (n - 1)
unbundled_count = n
current_bundle = []
number_of_variants = 0
research_deck = [0]
while unbundled_count >= 0:
#print(research_deck, current_bundle)
if research_deck:
c = research_deck.pop()
# check _to connection
if (connection[c] != None) and not is_bundled[connection[c]]:
research_deck.append(connection[c])
# check _from connection
try:
fr = connection.index(c)
if not is_bundled[fr]:
research_deck.append(fr)
except ValueError:
pass
current_bundle.append(c)
is_bundled[c] = True
else:
curlen = len(current_bundle)
number_of_variants += int((curlen + 1)*curlen/2)
unbundled_count -= curlen
current_bundle = []
if unbundled_count > 0:
research_deck = [is_bundled.index(False)]
else:
break
print(number_of_variants)
``` | instruction | 0 | 9,931 | 14 | 19,862 |
No | output | 1 | 9,931 | 14 | 19,863 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
n, m = map(int, input().split())
ll = [[*map(lambda s: int(s) - 1, input().split())] for _ in range(m)]
f = [0] * n
if n == 100000 and m > 1:
print(n)
exit()
for l in ll:
for i in range(n - 1):
if f[l[i]] == 0:
f[l[i]] = l[i + 1]
elif f[l[i]] != l[i + 1]:
f[l[i]] = -1
f[l[-1]] = -1
seen = set()
res = [0] * n
for i in ll[0]:
# print(i, res)
if i in seen:
continue
curr = i
while curr != -1:
res[i] += 1
seen.add(curr)
curr = f[curr]
res[i] = (res[i] * (res[i] + 1))//2
print(sum(res))
``` | instruction | 0 | 9,932 | 14 | 19,864 |
No | output | 1 | 9,932 | 14 | 19,865 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Acingel is a small town. There was only one doctor here — Miss Ada. She was very friendly and nobody has ever said something bad about her, so who could've expected that Ada will be found dead in her house? Mr Gawry, world-famous detective, is appointed to find the criminal. He asked m neighbours of Ada about clients who have visited her in that unlucky day. Let's number the clients from 1 to n. Each neighbour's testimony is a permutation of these numbers, which describes the order in which clients have been seen by the asked neighbour.
However, some facts are very suspicious – how it is that, according to some of given permutations, some client has been seen in the morning, while in others he has been seen in the evening? "In the morning some of neighbours must have been sleeping!" — thinks Gawry — "and in the evening there's been too dark to see somebody's face...". Now he wants to delete some prefix and some suffix (both prefix and suffix can be empty) in each permutation, so that they'll be non-empty and equal to each other after that — some of the potential criminals may disappear, but the testimony won't stand in contradiction to each other.
In how many ways he can do it? Two ways are called different if the remaining common part is different.
Input
The first line contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ 10) — the number of suspects and the number of asked neighbors.
Each of the next m lines contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). It is guaranteed that these integers form a correct permutation (that is, each number from 1 to n appears exactly once).
Output
Output a single integer denoting the number of ways to delete some prefix and some suffix of each permutation (possibly empty), such that the remaining parts will be equal and non-empty.
Examples
Input
3 2
1 2 3
2 3 1
Output
4
Input
5 6
1 2 3 4 5
2 3 1 4 5
3 4 5 1 2
3 5 4 2 1
2 3 5 4 1
1 2 3 4 5
Output
5
Input
2 2
1 2
2 1
Output
2
Note
In the first example, all possible common parts are [1], [2], [3] and [2, 3].
In the second and third examples, you can only leave common parts of length 1.
Submitted Solution:
```
n,m=map(int,input().split())
l=[[0]*(n+1) for i in range(m)]
pos=[[0]*(n+1) for i in range(m)]
dp=[1 for i in range(n)]
for i in range(m):
k=0
for j in list(map(int,input().split())):
l[i][k]=j
pos[i][j]=k
k+=1
ans=0
value=0
if m==1:
ans=n*(n+1)//2
else:
for i in range(n-1,-1,-1):
val=l[0][i]
value+=1
for j in range(1,m):
if l[j][pos[j][val]+1]!=l[0][i+1]:
ans+=value*(value+1)//2
value=0
break
print(ans)
``` | instruction | 0 | 9,933 | 14 | 19,866 |
No | output | 1 | 9,933 | 14 | 19,867 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,989 | 14 | 19,978 |
Tags: greedy
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
s=sum(l)
c=l[0]
k=1
d=[1]
for i in range(1,n):
if(k<=n-2):
if(l[i]<=(l[0]//2)):
c=c+l[i]
k=k+1
d.append(i+1)
if(c>s//2):
print(k)
print(*d)
else:
print(0)
``` | output | 1 | 9,989 | 14 | 19,979 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,990 | 14 | 19,980 |
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(int(x) for x in input().split())
p = a.pop(0)
pc = 0
c = []
for ai in range(len(a)):
if a[ai] * 2 <= p:
c.append(ai)
pc += a[ai]
if pc + p > (sum(a)+p)//2:
print(len(c)+1)
print(1, end=' ')
for ce in c:
print(ce+2, end=' ')
print()
else:
print(0)
``` | output | 1 | 9,990 | 14 | 19,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,991 | 14 | 19,982 |
Tags: greedy
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
sm=sum(a)
b=[(a[i],i) for i in range(1,n)]
b.sort()
ans=[1]
x=a[0]
fl=0
for i in b:
if x>sm/2:
fl=1
break
if a[0]>=2*i[0]:
ans.append(i[1]+1)
x+=i[0]
if(fl):
print(len(ans))
print(*ans)
else:print(0)
``` | output | 1 | 9,991 | 14 | 19,983 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,992 | 14 | 19,984 |
Tags: greedy
Correct Solution:
```
n = int(input())
arr = list(map(int, input().split(' ')))
tot = sum(arr)
curr = 0
ans = []
ac = arr[0]
for idx, num in enumerate(arr):
if idx == 0:
tot -= num
curr += num
ans.append(idx + 1)
if ac >= num * 2:
tot -= num
curr += num
ans.append(idx + 1)
if curr > tot:
print(len(ans))
print(*ans)
exit(0)
print(0)
``` | output | 1 | 9,992 | 14 | 19,985 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,993 | 14 | 19,986 |
Tags: greedy
Correct Solution:
```
n = int(input())
a = list(map(int, input().split()))
our = a[0]
ans = [1]
for i in range(1, n):
if a[i] * 2 <= a[0]:
our += a[i]
ans.append(i + 1)
if our > sum(a) - our:
print(len(ans))
print(*ans)
else:
print(0)
``` | output | 1 | 9,993 | 14 | 19,987 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,994 | 14 | 19,988 |
Tags: greedy
Correct Solution:
```
import math
from collections import deque, defaultdict
from sys import stdin, stdout
input = stdin.readline
# print = stdout.write
listin = lambda : list(map(int, input().split()))
mapin = lambda : map(int, input().split())
n = int(input())
a = listin()
ans = [1]
s = sum(a)
for i in range(1, n):
if a[0] >= 2*a[i]:
ans.append(i+1)
k = 0
for i in ans:
k+=a[i-1]
if k > s//2:
print(len(ans))
print(*ans)
else:
print(0)
``` | output | 1 | 9,994 | 14 | 19,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,995 | 14 | 19,990 |
Tags: greedy
Correct Solution:
```
def solve():
n=int(input())
#=map(int, input().split())
a=list(map(int, input().split()))
nd=(sum(a)//2)+1
par=a[0]
d=[1]
for i in range(1,n):
if par>=nd:
break
if 2*a[i]<=a[0]:
par+=a[i]
d.append(i+1)
if par>=nd:
print(len(d))
for i in range(len(d)):
print(d[i],end=' ')
else:
print(0)
def main():
t=0
if(t==0):
test = 1
else:
test = int(input())
for _ in range(test):
solve()
main()
``` | output | 1 | 9,995 | 14 | 19,991 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement. | instruction | 0 | 9,996 | 14 | 19,992 |
Tags: greedy
Correct Solution:
```
from collections import defaultdict
n = int(input())
a = list(map(int, input().split()))
record =defaultdict(list)
for i, v in enumerate(a):
record[v].append(i)
totalSeats = sum(a)
aliceSeats = a[0]
temp = a[0]
a[0], a[n - 1] = a[n - 1], a[0]
a.pop()
a.sort()
i = 0
ans = [0]
c = 1
while i < n - 1 and aliceSeats <= totalSeats//2:
# print(a[i])
if a[i] <= temp//2:
aliceSeats += a[i]
ans.append(record[a[i]].pop())
# print(ans)
c += 1
else:
break
i += 1
if aliceSeats > totalSeats//2:
print(c)
for i in ans:
print(i + 1, end=' ')
print('')
else:
print('0')
``` | output | 1 | 9,996 | 14 | 19,993 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
n = int(input())
arr = list(map(int,input().strip().split(' ')))
total_seats = sum(arr)
f = [tuple([0,arr[0]])] + list(filter(lambda x: arr[0] >= x[1]*2,enumerate(arr)))
if sum(map(lambda x:x[1],f)) > total_seats // 2:
print(str(len(f)) + "\n" + " ".join(map(lambda x: str(x[0]+1),f)))
else:
print(0)
``` | instruction | 0 | 9,997 | 14 | 19,994 |
Yes | output | 1 | 9,997 | 14 | 19,995 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
n=int(input())
p=input().rstrip().split(' ')
a=int(p[0])
B=0;
for i in range(0,len(p)):
B+=int(p[i])
L=p[1:len(p)]
S=a;
w=[]
w.append(list('1'))
q=[]
for i in range(0,len(L)):
if a >= (int(L[i])*2):
S+=int(L[i])
q.append(i+2)
if S>(B//2):
w.append(q)
q=[]
if len(w)==1:
if a > (B//2):
print(1)
print(1)
else:
print(0)
else:
D=[]
for i in range(0,len(w)):
for j in range(0,len(w[i])):
D.append(int(w[i][j]))
print(len(D))
print(*D)
``` | instruction | 0 | 9,998 | 14 | 19,996 |
Yes | output | 1 | 9,998 | 14 | 19,997 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
line1 = input()
line = input()
tmp = line.split(' ')
a = int(tmp[0])
num = 0
num_list, ans_list = [], []
for idx, i in enumerate(tmp):
num += int(i)
if idx:
num_list.append((int(i), idx + 1))
num_list.sort()
cnt = a
for item in num_list:
c, d = item
if cnt * 2 > num:
break
elif a >= 2 * c:
cnt += c
ans_list.append(item)
else:
break
if cnt * 2 > num:
print(len(ans_list) + 1)
print(1, end=' ')
for item in ans_list:
print(item[1], end=' ')
else:
print(0)
``` | instruction | 0 | 9,999 | 14 | 19,998 |
Yes | output | 1 | 9,999 | 14 | 19,999 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
GI = lambda: int(input()); GIS = lambda: map(int, input().split()); LGIS = lambda: list(GIS())
def main():
GI()
ps = LGIS()
a = ps[0]
s = a
l = [1]
for i, p in enumerate(ps[1:], 2):
if p <= a/2:
s += p
l.append(i)
if s > sum(ps) / 2:
print(len(l))
print(' '.join(map(str, l)))
else:
print(0)
main()
``` | instruction | 0 | 10,000 | 14 | 20,000 |
Yes | output | 1 | 10,000 | 14 | 20,001 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
n = int(input())
m= list(map(int,input().split()))
r=0
for i in m:r+=(i//2)
gd=["1"]
for x in range(1,n):
if m[0]>=(m[x]*2):
m[0]+=m[x]
gd.append(str(x+1))
if m[0]>r:
print(len(gd))
print(' '.join(gd))
else:
print(0)
``` | instruction | 0 | 10,001 | 14 | 20,002 |
No | output | 1 | 10,001 | 14 | 20,003 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,copy,functools
# import time,random,resource
# sys.setrecursionlimit(10**6)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
mod2 = 998244353
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def pe(s): return print(str(s), file=sys.stderr)
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)
def IF(c, t, f): return t if c else f
def YES(c): return IF(c, "YES", "NO")
def Yes(c): return IF(c, "Yes", "No")
class Prime():
def __init__(self, n):
self.M = m = int(math.sqrt(n)) + 10
self.A = a = [True] * m
a[0] = a[1] = False
self.T = t = [2]
for j in range(4, m, 2):
a[j] = False
for i in range(3, m, 2):
if not a[i]:
continue
t.append(i)
for j in range(i*i,m,i):
a[j] = False
self.ds_memo = {}
self.ds_memo[1] = set([1])
def is_prime(self, n):
return self.A[n]
def division(self, n):
d = collections.defaultdict(int)
for c in self.T:
while n % c == 0:
d[c] += 1
n //= c
if n < 2:
break
if n > 1:
d[n] += 1
return d.items()
# memo
def divisions(self, n):
if n in self.ds_memo:
return self.ds_memo[n]
for c in self.T:
if n % c == 0:
rs = set([c])
for cc in self.divisions(n // c):
rs.add(cc)
rs.add(cc * c)
self.ds_memo[n] = rs
return rs
rs = set([1, n])
self.ds_memo[n] = rs
return rs
def main():
t = 1
rr = []
for _ in range(t):
n = I()
a = LI()
s = sum(a)
t = a[0]
k = t
r = [1]
for i,c in enumerate(a[1:],2):
if c * 2 <= t:
k += c
r.append(i)
if k > s//2:
rr.append(len(r))
rr.append(JA(r, " "))
continue
m = max(a)
if m < t*2:
rr.append(0)
continue
mi = a.index(m)
k = 0
r = []
for i,c in enumerate(a, 1):
if i == mi or m >= c*2:
k += c
r.append(i)
if k > s//2:
rr.append(len(r))
rr.append(JA(r, " "))
else:
rr.append(0)
return JA(rr, "\n")
print(main())
``` | instruction | 0 | 10,002 | 14 | 20,004 |
No | output | 1 | 10,002 | 14 | 20,005 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
n=int(input())
a=list(map(int,input().split()))
b=[1]
c=[a[0]]
for i in range(1,n):
if(a[i]*2<=a[0]):
b.append(i+1)
c.append(a[i])
x=sum(a)
if(a[0]>=x//2):
print(len(b))
print(*b)
else:
print(0)
``` | instruction | 0 | 10,003 | 14 | 20,006 |
No | output | 1 | 10,003 | 14 | 20,007 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Alice is the leader of the State Refactoring Party, and she is about to become the prime minister.
The elections have just taken place. There are n parties, numbered from 1 to n. The i-th party has received a_i seats in the parliament.
Alice's party has number 1. In order to become the prime minister, she needs to build a coalition, consisting of her party and possibly some other parties. There are two conditions she needs to fulfil:
* The total number of seats of all parties in the coalition must be a strict majority of all the seats, i.e. it must have strictly more than half of the seats. For example, if the parliament has 200 (or 201) seats, then the majority is 101 or more seats.
* Alice's party must have at least 2 times more seats than any other party in the coalition. For example, to invite a party with 50 seats, Alice's party must have at least 100 seats.
For example, if n=4 and a=[51, 25, 99, 25] (note that Alice'a party has 51 seats), then the following set [a_1=51, a_2=25, a_4=25] can create a coalition since both conditions will be satisfied. However, the following sets will not create a coalition:
* [a_2=25, a_3=99, a_4=25] since Alice's party is not there;
* [a_1=51, a_2=25] since coalition should have a strict majority;
* [a_1=51, a_2=25, a_3=99] since Alice's party should have at least 2 times more seats than any other party in the coalition.
Alice does not have to minimise the number of parties in a coalition. If she wants, she can invite as many parties as she wants (as long as the conditions are satisfied). If Alice's party has enough people to create a coalition on her own, she can invite no parties.
Note that Alice can either invite a party as a whole or not at all. It is not possible to invite only some of the deputies (seats) from another party. In other words, if Alice invites a party, she invites all its deputies.
Find and print any suitable coalition.
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of parties.
The second line contains n space separated integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the number of seats the i-th party has.
Output
If no coalition satisfying both conditions is possible, output a single line with an integer 0.
Otherwise, suppose there are k (1 ≤ k ≤ n) parties in the coalition (Alice does not have to minimise the number of parties in a coalition), and their indices are c_1, c_2, ..., c_k (1 ≤ c_i ≤ n). Output two lines, first containing the integer k, and the second the space-separated indices c_1, c_2, ..., c_k.
You may print the parties in any order. Alice's party (number 1) must be on that list. If there are multiple solutions, you may print any of them.
Examples
Input
3
100 50 50
Output
2
1 2
Input
3
80 60 60
Output
0
Input
2
6 5
Output
1
1
Input
4
51 25 99 25
Output
3
1 2 4
Note
In the first example, Alice picks the second party. Note that she can also pick the third party or both of them. However, she cannot become prime minister without any of them, because 100 is not a strict majority out of 200.
In the second example, there is no way of building a majority, as both other parties are too large to become a coalition partner.
In the third example, Alice already has the majority.
The fourth example is described in the problem statement.
Submitted Solution:
```
n = int(input())
l = list(map(int, input().split()))
total = sum(l)
seats = l[0]
maj = total//2 + 1
ans = []
ans.append(1)
least = l[0]//2 + 1 if l[0]%2 == 1 else l[0]//2
i = 1;
flag = 0
while (seats < maj):
if (i >= n):
flag = 1
break
if (l[i] <= least):
seats += l[i]
ans.append(i+1)
i+=1;
final = [str(x) for x in ans]
if (flag == 1):
print(0)
else:
print(len(final))
print(" ".join(final))
``` | instruction | 0 | 10,004 | 14 | 20,008 |
No | output | 1 | 10,004 | 14 | 20,009 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,328 | 14 | 20,656 |
Tags: sortings
Correct Solution:
```
n = int(input())
a = []
b = []
c = 0
for i in range(n):
x,y = map(int, input().split())
a.append(x)
b.append(y)
for i in range(n):
if a[i] != b[i]:
c += 1
if c > 0:
print("Happy Alex")
else:
print("Poor Alex")
``` | output | 1 | 10,328 | 14 | 20,657 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,329 | 14 | 20,658 |
Tags: sortings
Correct Solution:
```
# with open("input.txt","r") as f:
laptops = []
for _ in range(int(input())):
laptop = list(map(int,input().split()))
laptops.append(laptop)
laptops.sort()
truth = False
for i in range(len(laptops)-1):
if laptops[i][0]<laptops[i+1][0] and laptops[i][1]>laptops[i+1][1]:
truth = True
break
if(truth):
print("Happy Alex")
else:
print("Poor Alex")
``` | output | 1 | 10,329 | 14 | 20,659 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,330 | 14 | 20,660 |
Tags: sortings
Correct Solution:
```
n=int(input())
s='Poor Alex'
for i in range(0,n):
x, y = input().split()
if int(x) > int(y):
s='Happy Alex'
print(s)
``` | output | 1 | 10,330 | 14 | 20,661 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,331 | 14 | 20,662 |
Tags: sortings
Correct Solution:
```
n = int(input())
temp1 = 0
for x in range(n):
a, b = input().split()
a = int(a)
b = int(b)
if a < b:
temp1 += 1
if temp1 == 0:
print("Poor Alex")
else:
print("Happy Alex")
``` | output | 1 | 10,331 | 14 | 20,663 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,332 | 14 | 20,664 |
Tags: sortings
Correct Solution:
```
def CF456A():
N = int(input())
laptops = [tuple(map(int, input().split())) for _ in range(N)]
laptops = sorted(laptops, key=lambda x:(x[0], x[1]))
for i in range(N-1):
if laptops[i][1] > laptops[i+1][1]: return "Happy Alex"
return "Poor Alex"
if __name__ == '__main__':
res = CF456A()
print(res)
``` | output | 1 | 10,332 | 14 | 20,665 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,333 | 14 | 20,666 |
Tags: sortings
Correct Solution:
```
N=int(input())
q=[[0,0] for i in range(N)]
x=False
for i in range (N):
q[i][0],q[i][1]=map(int,input().split())
s=sorted(q, key=lambda x: x[0])
for k in range(1,N):
if s[k-1][1]-s[k][1]>0:
x=True
if x:
print('Happy Alex')
else:
print('Poor Alex')
``` | output | 1 | 10,333 | 14 | 20,667 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,334 | 14 | 20,668 |
Tags: sortings
Correct Solution:
```
import sys
input=sys.stdin.readline
l=[]
for i in range(int(input())):l.append(list(map(int,input().split())))
l.sort()
m=[i[1]for i in l]
print(["Happy","Poor"][m==sorted(m)],"Alex")
``` | output | 1 | 10,334 | 14 | 20,669 |
Provide tags and a correct Python 3 solution for this coding contest problem.
One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.
Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.
Input
The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.
Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).
All ai are distinct. All bi are distinct.
Output
If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).
Examples
Input
2
1 2
2 1
Output
Happy Alex | instruction | 0 | 10,335 | 14 | 20,670 |
Tags: sortings
Correct Solution:
```
entrada = int(input())
sub = 0
ind = 0
pos = 0
tuplas = []
flag = False
for i in range(entrada):
tupla = list(int(x) for x in input().split())
tuplas.append(tupla)
s = tupla[0] - tupla[1]
if s < 0:
if sub == 0 and pos == 0:
sub = s
ind = i
elif s < sub or s == sub:
if tuplas[ind][0] > tupla[0] and tuplas[ind][1] < tupla[1]:
flag = True
break
else:
flag = True
break
elif s > 0:
if sub != 0:
flag = True
break
pos+=1
if flag:
print("Happy Alex")
else:
print("Poor Alex")
``` | output | 1 | 10,335 | 14 | 20,671 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is known that passages in Singer house are complex and intertwined. Let's define a Singer k-house as a graph built by the following process: take complete binary tree of height k and add edges from each vertex to all its successors, if they are not yet present.
<image> Singer 4-house
Count the number of non-empty paths in Singer k-house which do not pass the same vertex twice. Two paths are distinct if the sets or the orders of visited vertices are different. Since the answer can be large, output it modulo 109 + 7.
Input
The only line contains single integer k (1 ≤ k ≤ 400).
Output
Print single integer — the answer for the task modulo 109 + 7.
Examples
Input
2
Output
9
Input
3
Output
245
Input
20
Output
550384565
Note
There are 9 paths in the first example (the vertices are numbered on the picture below): 1, 2, 3, 1-2, 2-1, 1-3, 3-1, 2-1-3, 3-1-2.
<image> Singer 2-house | instruction | 0 | 10,477 | 14 | 20,954 |
Tags: combinatorics, dp, graphs, trees
Correct Solution:
```
print([0,1,9,245,126565,54326037,321837880,323252721,754868154,328083248,838314395,220816781,893672292,166441208,251255697,114256285,118775501,482714697,11784725,460862131,550384565,106742050,425241115,626692854,674266678,320014275,345949512,527320049,897822749,137190263,491039182,810384961,482023334,658099864,886790989,845381174,371433224,278969124,420088324,696766322,388302635,141033366,46387851,932125021,278342766,371131134,922501918,110778457,506223573,806353719,391845991,923507761,780307355,109951115,830090230,605558495,344686604,988110893,944684429,715019947,799898820,384672708,907325090,758952329,550672104,368337206,394915145,401744167,923781939,831857516,407845661,329267374,927004007,891609656,897919613,481297880,737337940,651873737,287246681,973133651,679864988,784719328,820504764,875613823,806512665,164851642,500228957,951814419,447763649,273141670,979349615,964027956,809510400,276634497,116631976,426739449,175282420,885948162,62270880,974395255,675165056,759589968,837957573,931897605,152352780,585420109,1772087,333401718,898833639,745874265,786209423,691982338,498790927,473374639,274302623,971280670,241671319,13070005,302088807,550276351,436592588,631667314,548656698,730626984,146295220,674398632,400383348,454138904,786220712,118620797,233440672,217349271,274853536,310607544,105221205,769566615,853585061,800665807,695377419,924327065,388199705,551624811,721435546,501720515,308465454,825369234,396065729,451899519,295058424,142088952,473485086,378771634,734511215,462404399,959198328,337668263,794122911,38911400,951992982,472696081,373904752,105884826,630251717,28980684,845136347,353665773,691661192,19922354,231463797,757917231,242739918,979036950,713722080,234689388,2243164,209872853,240808787,539523346,425797848,913772061,224613100,421742777,222232478,92712941,215137570,949901408,274827432,15162482,593145989,274574232,239282092,762720192,804146934,500629424,565985054,81127381,671811155,655565571,890331075,237994348,743647404,667160634,713914299,668506729,741341289,277636808,762781382,14272789,902864131,567443405,149113383,648844381,825489976,933016723,192288078,734493315,240985733,861817693,762711459,525904609,532463481,377133989,620711079,772561562,980733194,227599811,162774370,209512798,787116594,3509258,748795368,378035466,612938915,802091952,857679599,481748937,493370392,358420805,48301629,412001241,463126722,509578422,967799131,994766554,687287243,863623583,771554899,690911527,855314994,923686429,246862514,192479791,133487041,703444043,295281758,801816257,920762934,749306433,973004841,848644684,560026478,952127278,616654635,839390326,975154012,409583672,635350249,343228425,335331602,223826406,952341037,589677800,249747234,555694261,137143500,628190328,461598392,431912756,29349807,759199489,783281228,781971312,915823407,388508707,718062705,27424111,309999451,963383322,831185229,132910888,347028136,850484840,223055285,142335980,144754000,772005560,81796039,167696020,79454283,172772542,201056991,484957644,716630285,763194701,211505841,903448791,926964672,257752668,482951716,411539070,620249847,592476107,170473128,814662613,898000271,57354872,361106091,488697643,889007954,138725767,684860983,36248116,304610143,137633385,413715776,99010024,779653665,100387568,286328069,564731826,621740468,943513219,506666491,249987886,553719884,769853086,337485319,702455584,809637762,755400257,892290368,502180086,364275817,118162370,873374339,261271695,970132574,744105500,434447173,117975095,383088393,625447969,180281249,545367713,133236931,360175662,148087453,806871297,498529036,886076476,65645000,465138299,967109895,331362616,472283705,796894900,199697765,503759892,472807906,187586706,941198065,782234442,57693411,18678611,82626204,395317191,570588915,152519440,449852456,63696518,763741345,878748386,494317541,444782633,93316211,929164666,529288371,165769871,730546850,955877127,994202767,492009567,275683011,415902127,95725776,718047399,786963365,73091278,986172399,174591541,913259286][int(input())])
``` | output | 1 | 10,477 | 14 | 20,955 |
Provide tags and a correct Python 3 solution for this coding contest problem.
It is known that passages in Singer house are complex and intertwined. Let's define a Singer k-house as a graph built by the following process: take complete binary tree of height k and add edges from each vertex to all its successors, if they are not yet present.
<image> Singer 4-house
Count the number of non-empty paths in Singer k-house which do not pass the same vertex twice. Two paths are distinct if the sets or the orders of visited vertices are different. Since the answer can be large, output it modulo 109 + 7.
Input
The only line contains single integer k (1 ≤ k ≤ 400).
Output
Print single integer — the answer for the task modulo 109 + 7.
Examples
Input
2
Output
9
Input
3
Output
245
Input
20
Output
550384565
Note
There are 9 paths in the first example (the vertices are numbered on the picture below): 1, 2, 3, 1-2, 2-1, 1-3, 3-1, 2-1-3, 3-1-2.
<image> Singer 2-house | instruction | 0 | 10,478 | 14 | 20,956 |
Tags: combinatorics, dp, graphs, trees
Correct Solution:
```
l=[0,1,9,245,126565,54326037,321837880,323252721,754868154,328083248,838314395,220816781,893672292,166441208,251255697,114256285,118775501,482714697,11784725,460862131,550384565,106742050,425241115,626692854,674266678,320014275,345949512,527320049,897822749,137190263,491039182,810384961,482023334,658099864,886790989,845381174,371433224,278969124,420088324,696766322,388302635,141033366,46387851,932125021,278342766,371131134,922501918,110778457,506223573,806353719,391845991,923507761,780307355,109951115,830090230,605558495,344686604,988110893,944684429,715019947,799898820,384672708,907325090,758952329,550672104,368337206,394915145,401744167,923781939,831857516,407845661,329267374,927004007,891609656,897919613,481297880,737337940,651873737,287246681,973133651,679864988,784719328,820504764,875613823,806512665,164851642,500228957,951814419,447763649,273141670,979349615,964027956,809510400,276634497,116631976,426739449,175282420,885948162,62270880,974395255,675165056,759589968,837957573,931897605,152352780,585420109,1772087,333401718,898833639,745874265,786209423,691982338,498790927,473374639,274302623,971280670,241671319,13070005,302088807,550276351,436592588,631667314,548656698,730626984,146295220,674398632,400383348,454138904,786220712,118620797,233440672,217349271,274853536,310607544,105221205,769566615,853585061,800665807,695377419,924327065,388199705,551624811,721435546,501720515,308465454,825369234,396065729,451899519,295058424,142088952,473485086,378771634,734511215,462404399,959198328,337668263,794122911,38911400,951992982,472696081,373904752,105884826,630251717,28980684,845136347,353665773,691661192,19922354,231463797,757917231,242739918,979036950,713722080,234689388,2243164,209872853,240808787,539523346,425797848,913772061,224613100,421742777,222232478,92712941,215137570,949901408,274827432,15162482,593145989,274574232,239282092,762720192,804146934,500629424,565985054,81127381,671811155,655565571,890331075,237994348,743647404,667160634,713914299,668506729,741341289,277636808,762781382,14272789,902864131,567443405,149113383,648844381,825489976,933016723,192288078,734493315,240985733,861817693,762711459,525904609,532463481,377133989,620711079,772561562,980733194,227599811,162774370,209512798,787116594,3509258,748795368,378035466,612938915,802091952,857679599,481748937,493370392,358420805,48301629,412001241,463126722,509578422,967799131,994766554,687287243,863623583,771554899,690911527,855314994,923686429,246862514,192479791,133487041,703444043,295281758,801816257,920762934,749306433,973004841,848644684,560026478,952127278,616654635,839390326,975154012,409583672,635350249,343228425,335331602,223826406,952341037,589677800,249747234,555694261,137143500,628190328,461598392,431912756,29349807,759199489,783281228,781971312,915823407,388508707,718062705,27424111,309999451,963383322,831185229,132910888,347028136,850484840,223055285,142335980,144754000,772005560,81796039,167696020,79454283,172772542,201056991,484957644,716630285,763194701,211505841,903448791,926964672,257752668,482951716,411539070,620249847,592476107,170473128,814662613,898000271,57354872,361106091,488697643,889007954,138725767,684860983,36248116,304610143,137633385,413715776,99010024,779653665,100387568,286328069,564731826,621740468,943513219,506666491,249987886,553719884,769853086,337485319,702455584,809637762,755400257,892290368,502180086,364275817,118162370,873374339,261271695,970132574,744105500,434447173,117975095,383088393,625447969,180281249,545367713,133236931,360175662,148087453,806871297,498529036,886076476,65645000,465138299,967109895,331362616,472283705,796894900,199697765,503759892,472807906,187586706,941198065,782234442,57693411,18678611,82626204,395317191,570588915,152519440,449852456,63696518,763741345,878748386,494317541,444782633,93316211,929164666,529288371,165769871,730546850,955877127,994202767,492009567,275683011,415902127,95725776,718047399,786963365,73091278,986172399,174591541,913259286]
p=int(input())
print(l[p])
``` | output | 1 | 10,478 | 14 | 20,957 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
N = int(input())
N -= 1
A = int(input())
B = int(input())
C = int(input())
print(min(N * A, N * B, min(A, B) + (C * (N - 1))) if N != 0 else 0)
``` | instruction | 0 | 10,487 | 14 | 20,974 |
Yes | output | 1 | 10,487 | 14 | 20,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
s = 0
n -= 1
x = 1
for i in range(n):
if x == 1:
if a < b:
s += a
x = 2
else:
s += b
x = 3
elif x == 2:
if a < c:
s += a
x = 1
else:
s += c
x = 3
else:
if b < c:
s += b
x = 1
else:
s += c
x = 2
print(s)
``` | instruction | 0 | 10,488 | 14 | 20,976 |
Yes | output | 1 | 10,488 | 14 | 20,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
if n==1:
print(0)
else:
if a<=b and a<=c:
print(a*(n-1))
elif b<=a and b<=c:
print(b*(n-1))
else:
print(min(a,b)+c*(n-2))
``` | instruction | 0 | 10,489 | 14 | 20,978 |
Yes | output | 1 | 10,489 | 14 | 20,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n=int(input())
e=int(input())
o=int(input())
r=int(input())
meal=1
way=0
pos='R'
dic={'R':[o,e],'E':[o,r],'O':[e,r]}
while meal<n :
way+=min(dic[pos])
if min(dic[pos])==e :
if pos=='R' :
pos='O'
else :
pos='R'
elif min(dic[pos])==o :
if pos=='R' :
pos='E'
else :
pos='R'
else :
if pos=='O' :
pos='E'
else :
pos='O'
meal+=1
print(way)
``` | instruction | 0 | 10,490 | 14 | 20,980 |
Yes | output | 1 | 10,490 | 14 | 20,981 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n=int(input())
a=int(input())
b=int(input())
c=int(input())
re=[a,b,c]
w=0
mi=0
i=1
while n-i!=0:
er=3000
dw=w
for j in range(3):
if w!=j:
if re[j]<er:
er=re[j]
dw=j
w=dw
mi+=er
i+=1
print(mi)
``` | instruction | 0 | 10,491 | 14 | 20,982 |
No | output | 1 | 10,491 | 14 | 20,983 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n=int(input())
a=int(input())
b=int(input())
c=int(input())
m=0
z=a
if min(a,b,c)==a:
print((n-1)*min(a,b,c))
elif min(a,b,c)==b:
print((n-1)*b)
else:
m+=min(a,b)
print(m+(n-2)*c)
``` | instruction | 0 | 10,492 | 14 | 20,984 |
No | output | 1 | 10,492 | 14 | 20,985 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
short = min((a,b,c))
ans = 0
if a == short or b == short:
ans = min(a,b)*(n-1)
else:
ans = min(a,b) + c*(n-2)
print(ans)
``` | instruction | 0 | 10,493 | 14 | 20,986 |
No | output | 1 | 10,493 | 14 | 20,987 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Winnie-the-Pooh likes honey very much! That is why he decided to visit his friends. Winnie has got three best friends: Rabbit, Owl and Eeyore, each of them lives in his own house. There are winding paths between each pair of houses. The length of a path between Rabbit's and Owl's houses is a meters, between Rabbit's and Eeyore's house is b meters, between Owl's and Eeyore's house is c meters.
For enjoying his life and singing merry songs Winnie-the-Pooh should have a meal n times a day. Now he is in the Rabbit's house and has a meal for the first time. Each time when in the friend's house where Winnie is now the supply of honey is about to end, Winnie leaves that house. If Winnie has not had a meal the required amount of times, he comes out from the house and goes to someone else of his two friends. For this he chooses one of two adjacent paths, arrives to the house on the other end and visits his friend. You may assume that when Winnie is eating in one of his friend's house, the supply of honey in other friend's houses recover (most probably, they go to the supply store).
Winnie-the-Pooh does not like physical activity. He wants to have a meal n times, traveling minimum possible distance. Help him to find this distance.
Input
First line contains an integer n (1 ≤ n ≤ 100) — number of visits.
Second line contains an integer a (1 ≤ a ≤ 100) — distance between Rabbit's and Owl's houses.
Third line contains an integer b (1 ≤ b ≤ 100) — distance between Rabbit's and Eeyore's houses.
Fourth line contains an integer c (1 ≤ c ≤ 100) — distance between Owl's and Eeyore's houses.
Output
Output one number — minimum distance in meters Winnie must go through to have a meal n times.
Examples
Input
3
2
3
1
Output
3
Input
1
2
3
5
Output
0
Note
In the first test case the optimal path for Winnie is the following: first have a meal in Rabbit's house, then in Owl's house, then in Eeyore's house. Thus he will pass the distance 2 + 1 = 3.
In the second test case Winnie has a meal in Rabbit's house and that is for him. So he doesn't have to walk anywhere at all.
Submitted Solution:
```
n = int(input())
a = int(input())
b = int(input())
c = int(input())
score = 0
r = True
o = False
e = False
for i in range(n - 1):
if r is True:
if a > b:
score += b
o = True
r = False
else:
score += a
e = True
r = False
elif o is True:
if a > c:
score += c
e = True
o = False
else:
score += a
r = True
o = False
elif e is True:
if b > c:
score += c
o = True
e = False
else:
score += a
r = True
e = False
print(score)
``` | instruction | 0 | 10,494 | 14 | 20,988 |
No | output | 1 | 10,494 | 14 | 20,989 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image> | instruction | 0 | 10,958 | 14 | 21,916 |
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
import sys
import operator
import array
#-----------
def solve():
a = [int(x) for x in input().split()]
n = int(input())
b = [int(x) for x in input().split()]
have = [ [] ] * (n*6)
arr_append = have.append
for i in range(0, n):
for j in range(0, 6):
#arr_append([ b[i] - a[j], i ])
have[i*6 + j] = [ b[i] - a[j], i ]
cnt = array.array('L', [0])*n
z = n
sz = len(have)
ans = 999999999999
r = 0
have.sort(key=operator.itemgetter(0))
for i in range(0, sz):
while (r < sz) and (z > 0):
cnt[have[r][1]] += 1
if (cnt[have[r][1]] == 1):
z-=1
r+=1
if (z > 0):
break
ans = min(ans, have[r - 1][0] - have[i][0]);
cnt[have[i][1]] -= 1
if (cnt[have[i][1]] == 0):
z+=1
print(ans)
#-----------
def main(argv):
solve()
if __name__ == "__main__":
main(sys.argv)
``` | output | 1 | 10,958 | 14 | 21,917 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image> | instruction | 0 | 10,959 | 14 | 21,918 |
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
# 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()
# ------------------------------
def RL(): return map(int, sys.stdin.readline().split())
def RLL(): return list(map(int, sys.stdin.readline().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# sys.setrecursionlimit(300000)
# from heapq import *
# from collections import deque as dq
# from math import ceil,floor,sqrt,pow
import bisect as bs
# from collections import Counter
# from collections import defaultdict as dc
# from functools import lru_cache
a = RLL()
a.sort()
n = N()
b = RLL()
data = [(b[i]-a[j],i) for i in range(n) for j in range(6)]
# print(data)
data.sort()
res = float('inf')
now = [0]*n
count = 0
l,r = 0,0
mn = 6*n
while 1:
while count<n and r<mn:
k = data[r][1]
now[k]+=1
if now[k]==1:
count+=1
r+=1
if count<n:
break
while count==n:
k = data[l][1]
now[k]-=1
if now[k]==0:
count-=1
l+=1
res = min(data[r-1][0]-data[l-1][0],res)
print(res)
``` | output | 1 | 10,959 | 14 | 21,919 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image> | instruction | 0 | 10,960 | 14 | 21,920 |
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
from sys import stdin
input = stdin.readline
a = sorted([int(i) for i in input().split()])
n = int(input())
b = sorted([int(i) for i in input().split()])
c = []
for i in range(n):
c += [[b[i] - a[j], i] for j in range(6)]
c.sort()
d = [0] * n
e = 0
ans = 10 ** 10
u = 0
for i in range(len(c)):
while u < len(c) and e < n:
x = c[u][1]
if d[x] == 0:
e += 1
d[x] += 1
u += 1
if e == n:
ans = min(ans, c[u - 1][0] - c[i][0])
x = c[i][1]
d[x] -= 1
if d[x] == 0:
e -= 1
print(ans)
``` | output | 1 | 10,960 | 14 | 21,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image> | instruction | 0 | 10,961 | 14 | 21,922 |
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
a=list(map(int,input().split()));n=int(input());s=list(map(int,input().split()));b=[];i=j=0;ans=10**18;cs=[0]*n;nz=1;z=n*6
for y in range(n):
for x in a:b.append((s[y]-x)*n+y)
b.sort();cs[b[0]%n]+=1
while j+1<z:
while j+1<z and nz<n:j+=1;nz+=cs[b[j]%n]<1;cs[b[j]%n]+=1
while nz==n:ans=min(ans,b[j]//n-b[i]//n);cs[b[i]%n]-=1;nz-=cs[b[i]%n]==0;i+=1
print(ans)
``` | output | 1 | 10,961 | 14 | 21,923 |
Provide tags and a correct Python 3 solution for this coding contest problem.
After battling Shikamaru, Tayuya decided that her flute is too predictable, and replaced it with a guitar. The guitar has 6 strings and an infinite number of frets numbered from 1. Fretting the fret number j on the i-th string produces the note a_{i} + j.
Tayuya wants to play a melody of n notes. Each note can be played on different string-fret combination. The easiness of performance depends on the difference between the maximal and the minimal indices of used frets. The less this difference is, the easier it is to perform the technique. Please determine the minimal possible difference.
For example, if a = [1, 1, 2, 2, 3, 3], and the sequence of notes is 4, 11, 11, 12, 12, 13, 13 (corresponding to the second example), we can play the first note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7, as shown on the picture.
<image>
Input
The first line contains 6 space-separated numbers a_{1}, a_{2}, ..., a_{6} (1 ≤ a_{i} ≤ 10^{9}) which describe the Tayuya's strings.
The second line contains the only integer n (1 ≤ n ≤ 100 000) standing for the number of notes in the melody.
The third line consists of n integers b_{1}, b_{2}, ..., b_{n} (1 ≤ b_{i} ≤ 10^{9}), separated by space. They describe the notes to be played. It's guaranteed that b_i > a_j for all 1≤ i≤ n and 1≤ j≤ 6, in other words, you can play each note on any string.
Output
Print the minimal possible difference of the maximal and the minimal indices of used frets.
Examples
Input
1 4 100 10 30 5
6
101 104 105 110 130 200
Output
0
Input
1 1 2 2 3 3
7
13 4 11 12 11 13 12
Output
7
Note
In the first sample test it is optimal to play the first note on the first string, the second note on the second string, the third note on the sixth string, the fourth note on the fourth string, the fifth note on the fifth string, and the sixth note on the third string. In this case the 100-th fret is used each time, so the difference is 100 - 100 = 0.
<image>
In the second test it's optimal, for example, to play the second note on the first string, and all the other notes on the sixth string. Then the maximal fret will be 10, the minimal one will be 3, and the answer is 10 - 3 = 7.
<image> | instruction | 0 | 10,962 | 14 | 21,924 |
Tags: binary search, brute force, dp, implementation, sortings, two pointers
Correct Solution:
```
#If FastIO not needed, used this and don't forget to strip
import sys
input = sys.stdin.readline
"""
import os
import sys
from io import BytesIO, IOBase
import heapq as h
from bisect import bisect_left, bisect_right
from types import GeneratorType
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
import os
self.os = os
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 = self.os.read(self._fd, max(self.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 = self.os.read(self._fd, max(self.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:
self.os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
"""
#from collections import Counter as cc
#import math, string
def getInts():
return [int(s) for s in input().strip().split()]
def getInt():
return int(input().strip())
"""
Fret = B[j] minus A[i]
There are 6 possible values for each note
1 4 8 13 20 25
1 4 9 13 20 25
0 0 1 0 0 0
Order the notes
1,2,3,4,5,6
(0,-1,-1,-1,-1,-1)
(1,0,-1,-1,-1,-1)
(1,2,3,4,5,6)
Binary search? Is it possible to attain a minimum difference of <= D?
We need to find a range [L,R] within which every note can be played, such that R-L is minimal
"""
def solve():
A = getInts()
M = getInt()
B = getInts()
X = []
P = []
for i, b in enumerate(B):
for a in A:
P.append((b-a,i))
P.sort()
i = 0
j = -1
counts = [0]*M
sset = set()
ans = 2*10**9
MAX = M*6
set_len = 0
while i < MAX:
while set_len < M and j < MAX:
j += 1
try:
if not counts[P[j][1]]: set_len += 1
counts[P[j][1]] += 1
except:
break
if set_len < M:
break
z = P[i][1]
ans = min(ans,P[j][0]-P[i][0])
counts[z] -= 1
if not counts[z]:
set_len -= 1
i += 1
return ans
#for _ in range(getInt()):
print(solve())
``` | output | 1 | 10,962 | 14 | 21,925 |
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