text stringlengths 198 433k | conversation_id int64 0 109k |
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Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
Input
3
BRB
Output
1
GRB
Input
7
RGBGRBB
Output
3
RGBRGBR
Submitted Solution:
```
input()
s = input()
orders = ['RGB', 'RBG', 'GRB', 'GBR', 'BRG', 'BGR']
garlands = [''.join(o[i % 3] for i in range(len(s))) for o in orders]
with_changes = [(g, sum(int(c1 != c2) for c1, c2 in zip(g, s))) for g in garlands]
g, c = min(with_changes, key=lambda t: t[1])
print(c)
print(g)
```
Yes
| 12,500 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
Input
3
BRB
Output
1
GRB
Input
7
RGBGRBB
Output
3
RGBRGBR
Submitted Solution:
```
n=int(input())
s1=input()
s=list(s1)
c=0
a=[0]*n
d=['R','G','B']
if(len(s)==3):
for i in range(len(s)-2):
if(s[i]==s[i+2]):
c=1
if(s[i]=='B' and s[i+1]=='R'):
s[i+2]='G'
if(s[i]=='B' and s[i+1]=='G'):
s[i+2]='R'
if(s[i]=='G' and s[i+1]=='R'):
s[i+2]='B'
if(s[i]=='G' and s[i+1]=='B'):
s[i+2]='R'
if(s[i]=='R' and s[i+1]=='G'):
s[i+2]='B'
if(s[i]=='R' and s[i+1]=='B'):
s[i+2]='G'
elif(s[i+1]==s[i+2]):
c=1
if(s[i]=='B' and s[i+1]=='R'):
s[i+2]='G'
if(s[i]=='B' and s[i+1]=='G'):
s[i+2]='R'
if(s[i]=='G' and s[i+1]=='R'):
s[i+2]='B'
if(s[i]=='G' and s[i+1]=='B'):
s[i+2]='R'
if(s[i]=='R' and s[i+1]=='G'):
s[i]='B'
if(s[i+2]=='R' and s[i+1]=='B'):
s[i]='G'
elif(s[i]==s[i+1]==s[i+2]):
c=2
if(s[i]=='B'):
s[i+1]='G'
s[i+2]='R'
if(s[i]=='G'):
s[i+1]='B'
s[i+2]='R'
if(s[i]=='R'):
s[i+1]='G'
s[i+2]='B'
if(len(s)==2):
if(s[0]==s[1]):
c=1
if(s[1]=='R'):
s[0]='G'
if(s[1]=='B'):
s[0]='G'
if(s[1]=='G'):
s[0]='B'
if(len(s)==1):
c=0
else:
for i in range(0,len(s)-3,1):
if(s[i]!=s[i+3]):
s[i+3]=s[i]
c+=1
print(c)
for i in s:
print(i, end="")
```
No
| 12,501 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
Input
3
BRB
Output
1
GRB
Input
7
RGBGRBB
Output
3
RGBRGBR
Submitted Solution:
```
n=int(input())
sa=[0,0,0]
la=[0,0,0]
ka=[0,0,0]
p=input()
x=0
y=1
z=2
while x<n:
if p[x]=="B":
sa[0]+=1
elif p[x]=="G":
sa[1]+=1
else:
sa[2]+=1
if y<n:
if p[y]=="B":
la[0]+=1
elif p[y]=="G":
la[1]+=1
else:
la[2]+=1
if z<n:
if p[z]=="B":
ka[0]+=1
elif p[z]=="G":
ka[1]+=1
else:
ka[2]+=1
x+=3
y+=3
z+=3
aa=sa.index(max(sa))
ba=la.index(max(la))
ca=ka.index(max(ka))
fam=[0,0,0]
if aa==0:
fam[0]="B"
if aa==1:
fam[0]="G"
if aa==2:
fam[0]="R"
if ba==0:
fam[1]="B"
if ba==1:
fam[1]="G"
if ba==2:
fam[1]="R"
if ca==0:
fam[2]="B"
if ca==1:
fam[2]="G"
if ca==2:
fam[2]="R"
jam=fam*n
jam=jam[0:n]
s=0
for x in range(n):
if p[x]==jam[x]:
pass
else:
s+=1
if n==2:
if p[0]!=p[1]:
print(0)
print(p)
else:
print(1)
if p[0]!="B":
print(p[0]+"B")
else:
print(p[0]+"R")
elif n==3:
if len(set(p))==n:
print(0)
print(p)
elif len(set(p))==n-1:
p=list(p)
if p[0]==p[1]:
if p[2]=="R":
p[0]="B"
p[1]="G"
elif p[2]=="G":
p[0]="B"
p[1]="R"
else:
p[0]="G"
p[1]="R"
if p[0]==p[2]:
if p[1]=="R":
p[0]="B"
p[2]="G"
elif p[1]=="G":
p[0]="B"
p[2]="R"
else:
p[0]="G"
p[2]="R"
if p[1]==p[2]:
if p[0]=="R":
p[2]="B"
p[1]="G"
elif p[0]=="G":
p[2]="B"
p[1]="R"
else:
p[2]="G"
p[1]="R"
print(1)
print("".join(p))
else:
p=list(p)
print(2)
if p[0]=="B":
print(p[0]+"G"+"R")
elif p[0]=="G":
print(p[0]+"B"+"R")
else:
print(p[0]+"B"+"G")
elif n==1:
print(0)
print(p)
else:
print(s)
print("".join(jam))
```
No
| 12,502 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
Input
3
BRB
Output
1
GRB
Input
7
RGBGRBB
Output
3
RGBRGBR
Submitted Solution:
```
n = int(input())
s = input()
b = ['R','G','B']
count = 0
k = len(s)
if len(s) == 1:
print(count)
print(s)
else:
for i in range(k-3):
if s[i] == s[i+1]:
for j in b:
if j != s[i] and j != s[i+2]:
s = s[:i+1]+j+s[i+2:]
count+=1
break
if s[k-2] == s[k-1]:
for j in b:
if j != s[k-2]:
s = s[:k-2]+j+s[k-1:]
count+=1
break
print(count)
print(s)
```
No
| 12,503 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a garland consisting of n lamps. Each lamp is colored red, green or blue. The color of the i-th lamp is s_i ('R', 'G' and 'B' — colors of lamps in the garland).
You have to recolor some lamps in this garland (recoloring a lamp means changing its initial color to another) in such a way that the obtained garland is nice.
A garland is called nice if any two lamps of the same color have distance divisible by three between them. I.e. if the obtained garland is t, then for each i, j such that t_i = t_j should be satisfied |i-j|~ mod~ 3 = 0. The value |x| means absolute value of x, the operation x~ mod~ y means remainder of x when divided by y.
For example, the following garlands are nice: "RGBRGBRG", "GB", "R", "GRBGRBG", "BRGBRGB". The following garlands are not nice: "RR", "RGBG".
Among all ways to recolor the initial garland to make it nice you have to choose one with the minimum number of recolored lamps. If there are multiple optimal solutions, print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of lamps.
The second line of the input contains the string s consisting of n characters 'R', 'G' and 'B' — colors of lamps in the garland.
Output
In the first line of the output print one integer r — the minimum number of recolors needed to obtain a nice garland from the given one.
In the second line of the output print one string t of length n — a nice garland obtained from the initial one with minimum number of recolors. If there are multiple optimal solutions, print any of them.
Examples
Input
3
BRB
Output
1
GRB
Input
7
RGBGRBB
Output
3
RGBRGBR
Submitted Solution:
```
n = int(input())
s = input()
if n == 1:
print(n)
print(s)
else:
A = [[0]*3 for i in range(n)]
Aid = [[0]*3 for i in range(n)]
M = {'R':0,'G':1,'B':2}
M_ = {0:'R',1:'G',2:'B'}
v = M[s[0]]
A[0][(v+1)%3]= 1
A[0][(v+2)%3] = 1
def whichmax(A,id):
if A[id][0] <= A[id][1] and A[id][0] <= A[id][2]:
return 0
if A[id][1] <= A[id][0] and A[id][1] <= A[id][2]:
return 1
return 2
def min_and_id(a,ida,b,idb):
if a<b:
return a,ida
else:
return b,idb
for i in range(1,n):
v = M[s[i]]
vo0 = A[i-1][v]
vo1 = A[i-1][(v+1)%3]
vo2 = A[i-1][(v+2)%3]
val, vid = min_and_id(vo1,(v+1)%3,vo2,(v+2)%3)
#self
A[i][v] = val
Aid[i][v] = vid
val2,vid2 = min_and_id(vo0,v,vo2,(v+2)%3)
#N-1
A[i][(v+1)%3] = val2 + 1
Aid[i][(v + 1) % 3] = vid2
val3, vid3 = min_and_id(vo0, v, vo1, (v + 1) % 3)
#N-2
A[i][(v + 2) % 3] = val3 + 1
Aid[i][(v + 2) % 3] = vid3
ans = ""
i = n-1
midx = whichmax(A,n-1)
ans += M_[midx]
pid = Aid[n-1][midx]
while i>0:
ans += M_[pid]
pid = Aid[i][pid]
i-=1
print(A[n-1][midx])
print(ans[::-1])
```
No
| 12,504 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
"""for p in range(int(input())):
n,k=map(int,input().split(" "))
number=input().split(" ")
chances=[k for i in range(n)]
prev=-1
prev_updated=-1
last_used=False
toSub=0
start=0
prevSub=0
if(number[0]=='1'):
prev=0
prev_updated=0
start=1
for i in range(start,n):
if(number[i]=='1'):
# print("\ni",i,"\ntoSub",toSub,"\nprevUpadted",prev_updated,"\nprev",prev,"\nlast_used",last_used)
f1=False
# toSub+=1
toSub=0
zeros=i - prev_updated - 1
if(last_used):
zeros-=1
#chances[i]-=toSub
#print(prevSub,(i - prev - 1 ) +1)
if(i - prev - 1 <= prevSub):
chances[i]-= prevSub - (i - prev - 1 ) +1
if(chances[i]<zeros):
chances[i]=zeros
toSub+= prevSub - (i - prev - 1 ) +1
f1=True
if(zeros==0 or chances[i]==0):
prev_updated=i
prev=i
last_used=False
prevSub=toSub
continue
# print("\nchances: ",chances[i],"\t\tzeroes : ",zeros,"\t\tprevSub :",prevSub)
if(chances[i]>zeros):
# print("\t\t\t\t1")
number[i-zeros]='1'
number[i]='0'
prev_updated=i-zeros
last_used=False
elif(chances[i]==zeros):
# print("\t\t\t\t2")
number[i]='0'
number[i-chances[i]]='1'
prev_updated=i-chances[i]
last_used=True
else:
# print("\t\t\t\t3")
number[i]='0'
number[i-chances[i]]='1'
prev_updated=i-chances[i]
last_used=True
prev=i
prevSub=toSub
if(prev_updated>2 and f1):
if(number[prev_updated]=='1' and number[prev_updated-1]=='0' and number[prev_updated-2]=='1'):
last_used=False
#if()
# print("\ni",i,"\ntoSub",toSub,"\nprevUpadted",prev_updated,"\nprev",prev,"\nlast_used",last_used)
# print(number)
else:
toSub=0
print(*number)
# print(chances)"""
"""class offer:
def __init__(self, n, fre):
self.num = n
self.free = fre
self.delta= n-fre
n,m,k=map(int,input().split(" "))
shovel=list(map(int,input().split(" ")))
#dicti={}
offers=[]
temp_arr=[False for i in range(n)]
for i in range(m):
p,q=map(int,input().split(" "))
if(p>k):
continue
offers.append(offer(p,q))
# dicti[p]=q
#for i in dicti:
# dicti[i].sort()
shovel.sort()
shovel=shovel[:k+1]
offers.sort(key=lambda x: x.delta/x.num,reverse=True)
bestoffer=[]
for i in offers:
if(not temp_arr[i.num]):
temp_arr[i.num]=True
bestoffer.append(i)
cost=0
for i in bestoffer:
"""
"""
n=int(input())
arr=list(map(int,input().split(" ")))
ans=0
for i in range(n-1):
print(ans)
if(((arr[i]==2 and arr[i+1]==3) or (arr[i]==3 and arr[i+1]==2))):
print("Infinite")
ans=-100
break
else:
if(((arr[i]==1 and arr[i+1]==3) or (arr[i]==3 and arr[i+1]==1))):
ans+=4
elif(((arr[i]==1 and arr[i+1]==2) or (arr[i]==2 and arr[i+1]==1))):
ans+=3
if(ans>0):
print("Finite")
print(ans)
#for p in range(1):
for p in range(int(input())):
arr=list(input())
n=len(arr)
for i in range(n):
arr[i]=ord(arr[i])-96
arr.sort()
arr1=arr[:n//2]
arr2=arr[n//2:]
arr=[]
#print(arr,arr1,arr2)
i1=n//2-1
i2=n-i1-2
while (i1!=-1 and i2!=-1):
arr.append(arr1[i1])
arr.append(arr2[i2])
i1-=1
i2-=1
if(i1!=-1):
arr.append(arr1[i1])
elif(i2!=-1):
arr.append(arr2[i2])
#print(arr)
s=""
for i in range(n-1):
if(abs(arr[i]-arr[i+1])==1):
s=-1
print("No answer")
break
else:
s+=chr(arr[i]+96)
if(s!=-1):
s+=chr(arr[-1]+96)
print(s)"""
"""
n,m=map(int,input().split(" "))
seti=[]
ans=[1 for i in range(n)]
for i in range(m):
arr=list(map(int,input().split(" ")))
if(arr[0]>1):
seti.append(set(arr[1:]))
else:
m-=1
parent=[-1 for i in range(m)]
#print(seti)
for i in range(m-1):
for j in range(i+1,m):
if(parent[j]==-1):
if(len(seti[i].intersection(seti[j]))>0):
seti[i]=seti[i].union(seti[j])
parent[j]=i
#print(parent)
for i in range(m):
if(parent[i]==-1):
temp=list(seti[i])
store=len(temp)
for j in temp:
ans[j-1]=store
print(*ans)"""
n=int(input())
temp=input()
arr=list(map(int,temp.split(" ")))
wrong=False
for i in range(1,n):
if(arr[i-1]*arr[i]==6):
print("Infinite")
wrong=True
break
ans=0
if(not wrong):
for j in range(1,n):
if(arr[j-1]*arr[j]==2):
ans+=3
elif(arr[j-1]*arr[j]==3):
ans+=4
print("Finite")
print(ans-temp.count("3 1 2"))
```
| 12,505 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
n = int(input())
l = list(map(int,input().split()))
w = 0
c = 0
for i in range(n-1):
if l[i] == 2:
if i!=0 and i!=n-1:
if l[i+1] == 3 or l[i-1] == 3:
w = 1
break
else:
if i == 0:
if l[i+1] == 3:
w = 1
break
else:
if l[i-1] == 3:
w = 1
break
c+=3
if l[i] == 1:
if l[i+1] == 2:
if l[i-1] == 3 and i!=0:
c+=2
else:
c+=3
if l[i+1] == 3:
c+=4
if l[i] == 3:
if l[i+1] == 2:
w = 1
break
c+=4
if w == 1:
print("Infinite")
else:
print("Finite")
print(c)
```
| 12,506 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
cnt,f = 0,0
for i in range(1,n):
if (a[i-1]-1)*(a[i]-1) > 0:
f = 1
cnt += a[i-1]+a[i]
if i > 1 and a[i-2] == 3 and a[i-1] == 1 and a[i] == 2:
cnt -= 1
if f == 1:
print("Infinite")
else:
print("Finite")
print(cnt)
```
| 12,507 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
n = int(input())
a = [int(i) for i in input().split()]
total = 0
suma = dict()
suma[1, 2] = 3
suma[1, 3] = 4
suma[2, 1] = 3
suma[2, 3] = 9999999999999999999999999999999999999999999999999
suma[3, 1] = 4
suma[3, 2] = 9999999999999999999999999999999999999999999999999
for i, j in zip(a[:-1], a[1:]):
total += suma[i, j]
for i in range(len(a)-2):
if a[i] == 3 and a[i+1] == 1 and a[i+2] == 2:
total -= 1
if total < 99999999999999999999999999999:
print("Finite")
print(total)
else:
print("Infinite")
```
| 12,508 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
input()
a = ''.join(input().split())
res = 0
for aij in zip(a, a[1:]):
if '1' in aij:
res += sum(map(int, aij))
else:
print('Infinite')
break
else:
print('Finite')
print(res - a.count('312'))
```
| 12,509 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
def solve(n,A):
cnt=0
for i in range(n-1):
a0,a1,a2=A[i],A[i+1],A[i+2]
if a2==1:
if a1==2:
cnt+=3
else:
cnt+=4
elif a2==2:
if a1==1:
if a0==3:
cnt+=2
else:
cnt+=3
else:
print('Infinite')
return
else:
if a1==1:
cnt+=4
else:
print('Infinite')
return
print('Finite')
print(cnt)
return
n=int(input())
A=[0]+list(map(int,input().split()))
solve(n,A)
```
| 12,510 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
n = int(input())
a = list(map(int,input().split()))
flag = 0
count = 0
for i in range(1,n):
if a[i] == 1:
if a[i-1] == 1:
flag = 1
break
else:
if a[i-1] == 2:
count += 3
else:
count += 4
elif a[i] == 2:
if a[i-1] == 3 or a[i-1] == 2:
flag = 1
break
else:
count += 3
else:
if a[i-1] == 2 or a[i-1] == 3:
flag = 1
break
else:
count += 4
for i in range(n-2):
if(a[i] == 3 and a[i+1] == 1 and a[i+2] == 2):
count -= 1
if flag == 1:
print("Infinite")
else:
print("Finite")
print(count)
```
| 12,511 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Tags: geometry
Correct Solution:
```
n = int(input())
vals = input().replace(" ", "")
if "23" in vals or "32" in vals:
print("Infinite")
exit(0)
else:
result = 0
for i in range(len(vals)-1):
temp = vals[i] + vals[i+1]
if temp == "12" or temp == "21":
result += 3
if temp == "31" or temp == "13":
result += 4
result -= vals.count("312")
print("Finite")
print(result)
```
| 12,512 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
import sys
n = int(input())
nums = [int(x) for x in input().split()]
fig = []
lst = 1
ans = 0
for i in range(0, len(nums)):
if(lst != 1 and nums[i] != 1):
print("Infinite")
sys.exit()
if(nums[i] != 1):
fig.append(nums[i])
ans += nums[i]+1
if(i != 0 and i != n-1):
ans += nums[i]+1;
lst = nums[i]
for i in range(0, len(fig)-1):
if(fig[i] == 3 and fig[i+1] == 2):
ans -= 1
print("Finite\n",ans,sep='')
```
Yes
| 12,513 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
import io, sys, atexit, os
import math as ma
from sys import exit
from decimal import Decimal as dec
from itertools import permutations
from random import randint as rand
def li ():
return list (map (int, input ().split ()))
def num ():
return map (int, input ().split ())
def nu ():
return int (input ())
def find_gcd ( x, y ):
while (y):
x, y = y, x % y
return x
def lcm ( x, y ):
gg = find_gcd (x, y)
return (x * y // gg)
mm = 1000000007
yp = 0
def solve ():
t = 1
for _ in range (t):
n = nu ()
a = li ()
cc = 0
fl = True
for i in range (1, n):
if ((a [ i ] == 2 and a [ i - 1 ] == 3) or (a [ i ] == 3 and a [ i - 1 ] == 2)):
fl = False
break
if ((a [ i ] == 1 and a [ i - 1 ] == 2) or ((a [ i ] == 2 and a [ i - 1 ] == 1))):
cc += 3
else:
if((a [ i ] == 1 and a [ i - 1 ] == 3)):
if((i+1)<n):
if(a[i+1]==2):
cc+=3
else:
cc+=4
else:
cc+=4
else:
cc+=4
if (fl):
print ("Finite")
print (cc)
else:
print ("Infinite")
if __name__ == "__main__":
solve ()
```
Yes
| 12,514 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
def main():
n = int(input())
arr = list(map(int,input().split()))
for i in range(n-1):
if arr[i] == 2 and arr[i+1] == 3:
print('Infinite')
return
if arr[i] == 3 and arr[i+1] == 2:
print('Infinite')
return
if arr[i] == arr[i+1]:
print('Infinite')
return
ans = 0
for i in range(n-1):
if arr[i] == 1 and arr[i+1] == 2:
ans += 3
if i > 0 and arr[i-1] == 3:
ans -= 1
elif arr[i] == 1 and arr[i+1] == 3:
ans += 4
elif arr[i] == 2 and arr[i+1] == 1:
ans += 3
elif arr[i] == 3 and arr[i+1] == 1:
ans += 4
print('Finite')
print(ans)
main()
```
Yes
| 12,515 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
n= int(input())
list= [int (x) for x in input().split()]
a= list[0]
cnt = 0
prev = a
preprev = -1;
inf = False;
for i in range(1,n):
if(list[i] == 1):
if(prev == 3) :cnt += 4;
else: cnt += 3;
elif(list[i]== 2):
if(prev == 3): inf = True;
else: cnt += 3;
else:
if(prev == 2): inf = True;
else: cnt += 4;
if(preprev == 3 and prev == 1 and list[i] == 2) :cnt-=1
if(preprev == 1 and prev == 3 and list[i]== 2) :cnt -= 2;
preprev = prev;
prev = list[i]
if(inf): print("Infinite")
else:
print("Finite")
print(cnt)
```
Yes
| 12,516 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
def solve(a):
res = 0
for i in range(1, len(a)):
u, v = min(a[i-1], a[i]), max(a[i-1], a[i])
if u == 2 and v == 3:
print("Infinite")
return
if v == 3:
res += 4
else:
res += 3
print("Finite")
print(res)
input()
a = list(map(int, input().split()))
solve(a)
```
No
| 12,517 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
@Project : CodeForces
@File : 1.py
@Time : 2019/5/1 21:13
@Author : Koushiro
"""
if __name__ == "__main__":
n = int(input())
nums = list(map(int, input().split()))
last=nums[0]
ret=0
Infinite=False
for i in range(1,len(nums)):
if last==1:
if nums[i]==1:
Infinite=True
break
elif nums[i]==2:
ret+=3
else:
ret+=4
last=nums[i]
elif last==2:
if nums[i]==1:
ret+=3
elif nums[i]==2:
ret+=3
else:
Infinite=True
break
last=nums[i]
else:
if nums[i]==1:
ret+=4
elif nums[i]==2:
Infinite=True
break
else:
Infinite=True
break
last=nums[i]
if Infinite:
print('Infinite')
else:
print('Finite')
print(ret)
```
No
| 12,518 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
n = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(n-1):
if A[i] == 1 and A[i+1] == 2:
ans += 3
elif A[i] == 1 and A[i+1] == 3:
ans += 4
elif A[i] == 2 and A[i+1] == 1:
ans += 3
elif A[i] == 2 and A[i+1] == 3:
print('Infinite')
exit()
elif A[i] == 3 and A[i+1] == 1:
ans += 4
else:
print('Infinite')
exit()
print('Finite')
print(ans)
```
No
| 12,519 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The math faculty of Berland State University has suffered the sudden drop in the math skills of enrolling students. This year the highest grade on the entrance math test was 8. Out of 100! Thus, the decision was made to make the test easier.
Future students will be asked just a single question. They are given a sequence of integer numbers a_1, a_2, ..., a_n, each number is from 1 to 3 and a_i ≠ a_{i + 1} for each valid i. The i-th number represents a type of the i-th figure:
1. circle;
2. isosceles triangle with the length of height equal to the length of base;
3. square.
The figures of the given sequence are placed somewhere on a Cartesian plane in such a way that:
* (i + 1)-th figure is inscribed into the i-th one;
* each triangle base is parallel to OX;
* the triangle is oriented in such a way that the vertex opposite to its base is at the top;
* each square sides are parallel to the axes;
* for each i from 2 to n figure i has the maximum possible length of side for triangle and square and maximum radius for circle.
Note that the construction is unique for some fixed position and size of just the first figure.
The task is to calculate the number of distinct points (not necessarily with integer coordinates) where figures touch. The trick is, however, that the number is sometimes infinite. But that won't make the task difficult for you, will it?
So can you pass the math test and enroll into Berland State University?
Input
The first line contains a single integer n (2 ≤ n ≤ 100) — the number of figures.
The second line contains n integer numbers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 3, a_i ≠ a_{i + 1}) — types of the figures.
Output
The first line should contain either the word "Infinite" if the number of distinct points where figures touch is infinite or "Finite" otherwise.
If the number is finite than print it in the second line. It's guaranteed that the number fits into 32-bit integer type.
Examples
Input
3
2 1 3
Output
Finite
7
Input
3
1 2 3
Output
Infinite
Note
Here are the glorious pictures for the examples. Note that the triangle is not equilateral but just isosceles with the length of height equal to the length of base. Thus it fits into a square in a unique way.
The distinct points where figures touch are marked red.
In the second example the triangle and the square touch each other for the whole segment, it contains infinite number of points.
<image>
Submitted Solution:
```
# import sys
# sys.stdin = open("#input.txt", "r")
def solve(n,ls):
ans = 0
for i in range(n-1):
s = ls[i] + ls[i+1]
if s in [3,4]:
ans += s
else:
return False,0
return True,ans
n = int(input())
ls = list(map(int, input().split()))
fin, an = solve(n,ls)
if fin:
print("Finite")
print(an)
else:
print("Infinite")
```
No
| 12,520 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
n = int(input())
odp = []
for i in range(n - 1):
odp.append([i + 1, i + 2])
odp.append([n, 1])
k = n
def prime(x):
i = 2
while i ** 2 <= x:
if x % i == 0:
return False
i += 1
return True
primes = [0] * 50000
for i in range(50000):
primes[i] = (1 if prime(i) else 0)
w = (n + 2) // 2
m = 1
while w <= n:
if primes[k] == 1:
break
else:
odp.append([m, w])
m += 1
w += 1
k += 1
print(len(odp))
for i in range(len(odp)):
print(odp[i][0], odp[i][1])
```
| 12,521 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
import sys
read = lambda: sys.stdin.readline().strip()
readi = lambda: map(int, read().split())
from collections import *
n = int(read())
nearestPrime = None
for i in range(n, 1010):
for j in range(2, int(i**0.5)+1):
if i % j == 0:
break
else:
nearestPrime = i
break
print(nearestPrime)
for i in range(1, n):
print(i, i+1)
print(n, 1)
for i in range(1, nearestPrime - n + 1):
print(i, n//2 + i)
```
| 12,522 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
from bisect import *
l = [3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097, 1103, 1109]
z = [1, 2, 5, 6, 9, 10, 13, 14, 17, 18, 21, 22, 25, 26, 29, 30, 33, 34, 37, 38]
x = [3, 4, 7, 8, 11, 12, 15, 16, 19, 20, 23, 24, 27, 28, 31, 32, 35, 36, 39, 40]
n = int(input())
idx = bisect_left(l,n)
rem = l[idx]-n
ans = []
for i in range(1,n):
ans.append([i,i+1])
ans.append([1,n])
for i in range(1,rem+1):
ans.append([z[i-1],x[i-1]])
print(len(ans))
for i in ans:
print(*i)
```
| 12,523 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
prost = [3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009]
n = int(input())
for i in prost:
if i>=n:
m=i
break
raz = m-n
if n<3:
print(-1)
elif n==3:
print(3)
print(1,2)
print(2,3)
print(1,3)
elif n==4:
print(5)
print(1,2)
print(2,3)
print(3,4)
print(1,4)
print(1,3)
elif n==5:
print(5)
print(1,2)
print(2,3)
print(3,4)
print(4,5)
print(1,5)
elif n==6:
print(7)
print(1,2)
print(2,3)
print(3,4)
print(4,5)
print(5,6)
print(6,1)
print(2,5)
else:
print(m)
arr = []
for i in range(1,n+1):
arr.append(i)
if n%4==3:
now = arr[:7]
arr = arr[7:]
for i in range(7):
print(now[i],now[(i+1)%7])
if raz>0:
print(now[0],now[2])
raz-=1
if raz>0:
print(now[3],now[5])
raz-=1
if raz>0:
print(now[4],now[6])
raz-=1
n-=7
elif n%4==2:
now=arr[:6]
arr=arr[6:]
for i in range(6):
print(now[i],now[(i+1)%6])
if raz>0:
print(now[0],now[2])
raz-=1
if raz>0:
print(now[3],now[5])
raz-=1
if raz>0:
print(now[1],now[4])
raz-=1
n-=6
elif n%4==1:
now = arr[:5]
arr=arr[5:]
for i in range(5):
print(now[i],now[(i+1)%5])
if raz>0:
print(now[1],now[3])
raz-=1
if raz>0:
print(now[2],now[4])
raz-=1
n-=5
while raz>=2:
now=arr[:4]
arr=arr[4:]
for i in range(4):
print(now[i],now[(i+1)%4])
print(now[0],now[2])
print(now[1],now[3])
raz-=2
if len(arr)>0:
now=arr[:4]
arr=arr[4:]
for i in range(4):
print(now[i],now[(i+1)%4])
if raz>0:
print(now[0],now[2])
raz-=1
if raz>0:
print(now[1],now[3])
raz-=1
while len(arr)>0:
now=arr[:4]
arr=arr[4:]
for i in range(4):
print(now[i],now[(i+1)%4])
```
| 12,524 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
import bisect
import decimal
from decimal import Decimal
import os
from collections import Counter
import bisect
from collections import defaultdict
import math
import random
import heapq
from math import sqrt
import sys
from functools import reduce, cmp_to_key
from collections import deque
import threading
from itertools import combinations
from io import BytesIO, IOBase
from itertools import accumulate
from random import randint
# sys.setrecursionlimit(200000)
# mod = 10**9+7
# mod = 998244353
decimal.getcontext().prec = 46
def primeFactors(n):
prime = set()
while n % 2 == 0:
prime.add(2)
n = n//2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
prime.add(i)
n = n//i
if n > 2:
prime.add(n)
return list(prime)
def getFactors(n) :
factors = []
i = 1
while i <= math.sqrt(n):
if (n % i == 0) :
if (n // i == i) :
factors.append(i)
else :
factors.append(i)
factors.append(n//i)
i = i + 1
return factors
def modefiedSieve():
mx=10**7+1
sieve=[-1]*mx
for i in range(2,mx):
if sieve[i]==-1:
sieve[i]=i
for j in range(i*i,mx,i):
if sieve[j]==-1:
sieve[j]=i
return sieve
def SieveOfEratosthenes(n):
prime = [True for i in range(n+1)]
p = 2
while (p * p <= n):
if (prime[p] == True):
for i in range(p * p, n+1, p):
prime[i] = False
p += 1
num = []
for p in range(2, n+1):
if prime[p]:
num.append(p)
return num
def lcm(a,b):
return (a*b)//math.gcd(a,b)
def sort_dict(key_value):
return sorted(key_value.items(), key = lambda kv:(kv[1], kv[0]))
def list_input():
return list(map(int,input().split()))
def num_input():
return map(int,input().split())
def string_list():
return list(input())
def decimalToBinary(n):
return bin(n).replace("0b", "")
def binaryToDecimal(n):
return int(n,2)
def DFS(n,s,adj):
visited = [False for i in range(n+1)]
stack = []
stack.append(s)
while (len(stack)):
s = stack[-1]
stack.pop()
if (not visited[s]):
visited[s] = True
for node in adj[s]:
if (not visited[node]):
stack.append(node)
def maxSubArraySum(a,size):
maxint = 10**10
max_so_far = -maxint - 1
max_ending_here = 0
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < max_ending_here):
max_so_far = max_ending_here
if max_ending_here < 0:
max_ending_here = 0
return max_so_far
def isPrime(n) :
if (n <= 1) :
return False
if (n <= 3) :
return True
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True
def solve():
n = int(input())
N = n
if isPrime(n):
print(n)
for i in range(n-1):
print(i+1,i+2)
print(n,1)
else:
for i in range(n+1,2000):
if isPrime(i):
n = i
break
print(n)
for i in range(N-1):
print(i+1,i+2)
print(N,1)
u,v = 1,N-1
for i in range(n-N):
print(u,v)
u += 1
v -= 1
t = 1
#t = int(input())
for _ in range(t):
solve()
```
| 12,525 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
prime=[True]*1501
prime[0]=False
prime[1]=False
for i in range(2,1501):
if prime[i]:
for j in range(i*i,1501,i):
prime[j]=False
n=int(input())
degree=[2]*n
ans=[]
edges=n
for i in range(n-1):
ans.append([i+1,i+2])
ans.append([1,n])
c=0
while(not prime[edges]):
ans.append([c+1,c+n//2+1])
#print(edges,c,c+n//2)
degree[c]+=1
degree[c+n//2]+=1
c+=1
edges+=1
print(len(ans))
for i in ans:
print(i[0],i[1])
```
| 12,526 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
import math
n = int(input())
def isPrime(x):
if x in [2, 3, 5, 7]:
return True
if x == 1:
return False
for i in range(2, int(math.sqrt(x)) + 1):
if x % i == 0:
return False
return True
ans = []
for i in range(1, n):
ans.append([i, i + 1])
ans.append([n, 1])
k = 1
if isPrime(n) == False:
while True:
ans.append([k, k + n // 2])
if isPrime(n + k):
break
k += 1
print(len(ans))
for x in ans:
print(*x)
```
| 12,527 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Tags: constructive algorithms, greedy, math, number theory
Correct Solution:
```
from sys import stdin,stdout,setrecursionlimit
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush,nlargest
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm , accumulate
from bisect import bisect_left as bl, bisect_right as br, bisect
from time import perf_counter
from fractions import Fraction
import copy
import time
setrecursionlimit(10**9)
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353
# from sys import stdin
# input = stdin.readline
#def data(): return sys.stdin.readline().strip()
def data(): return input()
def num():return int(input())
def L(): return list(sp())
def LF(): return list(spf())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def spf(): return map(int, input.readline().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
def pmat(A):
for ele in A:
print(*ele,end="\n")
def pmat2(A):
for ele in A:
for j in ele:
print(j,end='')
print()
def iseven(n):
return n%2==0
def seive(r):
prime=[1 for i in range(r+1)]
prime[0]=0
prime[1]=0
for i in range(r+1):
if(prime[i]):
for j in range(2*i,r+1,i):
prime[j]=0
return prime
#solution
#ACPC
#remeber cut ribbon problem
# set data structures faster than binary search sometimes
#bipartite match dfs
#think in problems with recursive manner.
def isprime(x):
if x==2:return True
elif x%2==0:return False
sq=int(sqrt(x))+1
for i in range(3,sq,2):
if x%i==0:
return False
return True
n=int(input());m=n
while not isprime(m):m+=1
print(m)
print(1,n)
for i in range(n-1):
print(i+1,i+2)
for i in range(m-n):
print(i+1,i+1+n//2)
endtime = time.time()
#print(f"Runtime of the program is {endtime - starttime}")
```
| 12,528 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
n = int(input())
def pr(x):
for i in range(2, x):
if x % i == 0:
return False
return True
e = []
for i in range(n):
e += [[i, (i+1) % n]]
x = n
u = 0
v = n // 2
while not pr(x):
e += [[u, v]]
x += 1
u += 1
v += 1
print(x)
for g in e:
print(g[0]+1, g[1]+1)
```
Yes
| 12,529 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
from sys import stdin, stdout, exit
huge_p = 998244353
N = 10**6
ps = [True]*N
for i in range(2,N):
if ps[i]:
for k in range(2*i, N, i):
ps[k] = False
def opp(i):
return (i + (n // 2)) % n
n = int(input())
for i in range(n//2):
if ps[n+i]:
stdout.write(str(n+i) + "\n")
for j in range(n):
stdout.write(str(j+1) + " " + str((j+1)%n + 1) + "\n")
for j in range(i):
stdout.write(str(j + 1) + " " + str(opp(j) + 1) + "\n")
exit()
print(-1)
```
Yes
| 12,530 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
line = input()
tmp = int(line)
def is_prime(x):
for i in range(2, x):
if x % i == 0:
return False
return True
def find_next_prime(x):
x = x + 3
while not is_prime(x):
x = x + 1
return x
def find_last_prime(x):
while not is_prime(x):
x = x - 1
return x
if tmp == 4:
print(5)
print('1 2')
print('1 3')
print('2 3')
print('2 4')
print('3 4')
elif tmp == 6:
print(7)
print('1 2')
print('1 3')
print('2 3')
print('4 5')
print('5 6')
print('4 6')
print('1 4')
elif is_prime(tmp):
print(tmp)
print(1, tmp)
for i in range(1, tmp):
print(i, i + 1)
else:
np = find_next_prime(tmp)
m = np-tmp
print(np)
print(1, m)
for i in range(1, m):
print(i, i + 1)
c = m + 1
print(c, tmp)
for i in range(c, tmp):
print(i, i + 1)
for i in range(m):
print(1 + i, 1 + i + m)
```
Yes
| 12,531 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
data = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, 1009, 1013, 1019, 1021, 1031, 1033, 1039, 1049, 1051, 1061, 1063, 1069, 1087, 1091, 1093, 1097]
key = int(input())
if key in data:
print(key)
for each in range(key):
print(each % key + 1,(each + 1) % key + 1)
else:
ans = 0
for each in data:
if(each > key):
ans = each
break
print(ans)
x = (ans - key) * 2
y = key - x
index_start = 1
index_mid = 2
index_end = key
flag = []
for each in range(key):
flag.append(0)
for each_1 in range(key):
if(each_1 < x):
sub = flag[index_start - 1]
for each_2 in range(3 - sub):
if(index_start >= index_mid):
index_mid = index_start + 1
print(index_start,index_mid)
flag[index_start - 1] = flag[index_start - 1] + 1
flag[index_mid - 1] = flag[index_mid - 1] + 1
index_mid = index_mid + 1
if(index_mid > index_end):
index_mid = index_start + 1
index_start = index_start + 1
else:
sub = flag[index_start - 1]
for each_2 in range(2 - sub):
if(index_start >= index_mid):
index_mid = index_start + 1
print(index_start,index_mid)
flag[index_start - 1] = flag[index_start - 1] + 1
flag[index_mid - 1] = flag[index_mid - 1] + 1
index_mid = index_mid + 1
if(index_mid > index_end):
index_mid = index_start + 1
index_start = index_start + 1
```
Yes
| 12,532 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
n = int(input())
arr = [i for i in range(1,n+1)]
ans = []
while len(arr)-5>0:
a,b,c,d = arr[0],arr[1],arr[2],arr[3]
arr = arr[4:]
ans.append((a,b))
ans.append((b,c))
ans.append((c,d))
ans.append((d,a))
ans.append((a,c))
if len(arr)==1:
print(-1)
elif len(arr)<=5:
if len(arr)==5:
a,b,c,d,e = arr[0],arr[1],arr[2],arr[3],arr[4]
ans.append((a,b))
ans.append((b,c))
ans.append((a,c))
ans.append((d,e))
elif len(arr)==4:
a,b,c,d = arr[0],arr[1],arr[2],arr[3]
ans.append((a,b))
ans.append((b,c))
ans.append((c,d))
ans.append((d,a))
ans.append((a,c))
elif len(arr)==3:
a,b,c, = arr[0],arr[1],arr[2]
ans.append((a,b))
ans.append((b,c))
ans.append((a,c))
elif len(arr)==2:
a,b = arr[0],arr[1]
ans.append((a,b))
print(len(ans))
for k in ans:
print(" ".join([str(k[0]),str(k[1])]))
```
No
| 12,533 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
def isPrime(n):
# a prime(except 2 or 3) is of the form 6k-1 or 6k+1
if n == 2 or n == 3:
return True
if n % 2 == 0 or n % 3 == 0:
return False
i = 5
w = 2
sqN = int(pow(n, .5))
while i <= sqN:
if n % i == 0:
return False
i += w
w = 6 - w
return True
n = int(input())
edges = []
for i in range(1, n):
edges.append((i, i+1))
edges.append((n, 1))
ind = 1
while not isPrime(n):
edges.append((ind, ind+2))
ind = (ind+2)%n + 1
n += 1
print(len(edges))
for edge in edges:
print(*edge)
```
No
| 12,534 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
n = int(input())
for i in range(n - 1):
print(i + 1, i + 2)
print(n, 1)
k = n
def prime(x):
i = 2
while i ** 2 <= x:
if x % i == 0:
return False
i += 1
return True
primes = [0] * 50000
for i in range(50000):
primes[i] = (1 if prime(i) else 0)
w = (n + 1) // 2
m = 1
while w <= n:
if primes[k] == 1:
break
else:
print(m, w)
m += 1
w += 1
k += 1
```
No
| 12,535 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every person likes prime numbers. Alice is a person, thus she also shares the love for them. Bob wanted to give her an affectionate gift but couldn't think of anything inventive. Hence, he will be giving her a graph. How original, Bob! Alice will surely be thrilled!
When building the graph, he needs four conditions to be satisfied:
* It must be a simple undirected graph, i.e. without multiple (parallel) edges and self-loops.
* The number of vertices must be exactly n — a number he selected. This number is not necessarily prime.
* The total number of edges must be prime.
* The degree (i.e. the number of edges connected to the vertex) of each vertex must be prime.
Below is an example for n = 4. The first graph (left one) is invalid as the degree of vertex 2 (and 4) equals to 1, which is not prime. The second graph (middle one) is invalid as the total number of edges is 4, which is not a prime number. The third graph (right one) is a valid answer for n = 4.
<image>
Note that the graph can be disconnected.
Please help Bob to find any such graph!
Input
The input consists of a single integer n (3 ≤ n ≤ 1 000) — the number of vertices.
Output
If there is no graph satisfying the conditions, print a single line containing the integer -1.
Otherwise, first print a line containing a prime number m (2 ≤ m ≤ (n(n-1))/(2)) — the number of edges in the graph. Then, print m lines, the i-th of which containing two integers u_i, v_i (1 ≤ u_i, v_i ≤ n) — meaning that there is an edge between vertices u_i and v_i. The degree of each vertex must be prime. There must be no multiple (parallel) edges or self-loops.
If there are multiple solutions, you may print any of them.
Note that the graph can be disconnected.
Examples
Input
4
Output
5
1 2
1 3
2 3
2 4
3 4
Input
8
Output
13
1 2
1 3
2 3
1 4
2 4
1 5
2 5
1 6
2 6
1 7
1 8
5 8
7 8
Note
The first example was described in the statement.
In the second example, the degrees of vertices are [7, 5, 2, 2, 3, 2, 2, 3]. Each of these numbers is prime. Additionally, the number of edges, 13, is also a prime number, hence both conditions are satisfied.
<image>
Submitted Solution:
```
n=int(input())
ans=[]
i=1
if n==3:
print(3)
print(1,2)
print(2,3)
print(3,1)
exit()
while n-4>=4:
ans.append([i,i+1])
ans.append([i+1,i+2])
ans.append([i+2,i+3])
ans.append([i+3,i])
ans.append([i+1,i+3])
ans.append([i+2,i+4])
n-=4
i+=4
if n==4:
ans.append([i, i + 1])
ans.append([i + 1, i + 2])
ans.append([i + 2, i + 3])
ans.append([i + 3, i])
ans.append([i+1,i+3])
elif n==5:
ans.append([i, i + 1])
ans.append([i + 1, i + 2])
ans.append([i + 2, i + 3])
ans.append([i + 3, i])
ans.append([i + 1, i + 3])
ans.append([i,i+4])
ans.append([i+4,i+2])
elif n==6:
ans.append([i, i + 1])
ans.append([i + 2, i + 3])
ans.append([i + 3, i])
ans.append([i+1,i+4])
ans.append([i+2,i+4])
ans.append([i,i+5])
ans.append([i+3,i+5])
else:
ans.append([i, i + 1])
ans.append([i + 1, i + 2])
ans.append([i + 2, i + 3])
ans.append([i + 3, i])
i+=4
ans.append([i,i+1])
ans.append([i+1,i+2])
ans.append([i+2,i])
print(len(ans))
for i in range(len(ans)):
print(*ans[i])
```
No
| 12,536 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
import sys
def countR(ip):
c=0
for i in ip:
if(i=='R'):
c+=1
return c
def countB(ip):
c=0
for i in ip:
if(i=='B'):
c+=1
return c
def countG(ip):
c=0
for i in ip:
if(i=='G'):
c+=1
return c
# sys.stdin.readline()
t=int(sys.stdin.readline())
x='RGB'*680
y='GBR'*680
z='BRG'*680
for i in range(t):
n,k=list(map(int,sys.stdin.readline().strip().split()))
a=sys.stdin.readline().strip()
xk=x[:k]
yk=y[:k]
zk=z[:k]
# print(k,xk,zk)
# xc=[]
# yc=[]
# zc=[]
# xc.append(countR(xk))
# xc.append(countG(xk))
# xc.append(countB(xk))
# yc.append(countR(yk))
# yc.append(countG(yk))
# yc.append(countB(yk))
# zc.append(countR(zk))
# zc.append(countG(zk))
# zc.append(countB(zk))
op=2001
for j in range(n-k+1):
b=a[j:j+k]
# print(len(b),xc,zc)
# bc=[]
# bc.append(countR(b))
# bc.append(countG(b))
# bc.append(countB(b))
xd=0
yd=0
zd=0
# print(a,b,xc,yc,zc,bc)
for jj in range(len(b)):
if(b[jj]!=xk[jj]):
xd+=1
if(b[jj]!=yk[jj]):
yd+=1
if(b[jj]!=zk[jj]):
zd+=1
# print(a,b,xd,yd,zd,z)
op=min(op,xd,yd,zd)
print(op)
```
| 12,537 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
import sys
for _ in range(int(sys.stdin.readline())):
n, k = [int(i) for i in sys.stdin.readline().split()]
word = sys.stdin.readline().strip()
ans = 0
for col in ["RGB", "GBR", "BRG"]:
cnt = 0
for i in range(k):
# print(word[i],"VS",col[i%3])
if word[i] == col[i%3]:
cnt += 1
mx = cnt
for i in range(n-k):
if word[i+k] == col[(i+k)%3]:
cnt += 1
if word[i] == col[i%3]:
cnt -= 1
if cnt > mx:
mx = cnt
ans = max(ans, mx)
sys.stdout.write(str(k - ans) + "\n")
```
| 12,538 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
s=input()
s1="RGB"
mini=10**9
for i in range(n-k+1):
for j in range(3):
curr=0
for x in range(k):
curr+=(s[i+x]!=s1[(x+j)%3])
mini=min(mini,curr)
print(mini)
```
| 12,539 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
import sys
input = lambda: sys.stdin.readline().strip()
from math import ceil
def mismatch(s1, s2):
cnt = 0
for i in range(len(s1)):
if s1[i]!=s2[i]: cnt+=1
return cnt
T = int(input())
for _ in range(T):
n, k = map(int, input().split())
check = ''
for i in range(ceil((k+2)/3)):
check+='RGB'
ls = []
for i in range(3):
ls.append(check[i:i+k])
s = input()
m = n
for i in range(n-k+1):
for j in ls:
m = min(m, mismatch(s[i:i+k], j))
print(m)
```
| 12,540 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
import sys
t=sys.stdin.buffer.readline()
for g in range(int(t)):
n,k=map(int,input().split())
s=input()
t = 2000
for i in range(n-k+1):
s1 = 'RGB'
'''if s[i]=="R":
var=0
elif s[i]=="G":
var=1
elif s[i]=='B':
var=2'''
for h in range(3):
var=h
count = 0
for j in range(k):
if s[i + j] == s1[var]:
var += 1
var = var % 3
else:
count += 1
var += 1
var = var % 3
t=min(count,t)
print(t)
```
| 12,541 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
a = int(input())
ot = ''
for i in range(a):
b, c = map(int,input().split())
s = input()
pref = [0] * (b + 1)
pref1 = [0] * (b + 1)
pref2 = [0] * (b + 1)
n = 'RGB'
n1 = 'GBR'
n2 = 'BRG'
for j in range(b):
pref[j + 1] = pref[j]
if s[j] != n[j % 3]:
pref[j + 1] += 1
pref1[j + 1] = pref1[j]
if s[j] != n1[j % 3]:
pref1[j + 1] += 1
pref2[j + 1] = pref2[j]
if s[j] != n2[j % 3]:
pref2[j + 1] += 1
mi = 1000000000
#print(*pref)
#print(*pref1)
#print(*pref2)
for j in range(c, b + 1):
mi = min(pref[j] - pref[j - c], pref1[j] - pref1[j - c], pref2[j] - pref2[j - c], mi)
#print(mi, j)
mi = min(pref[-1], pref1[-1], pref2[-1], mi)
ot += str(mi) + '\n'
print(ot)
```
| 12,542 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
for _ in range(int(input())):
n, k = [int(i) for i in input().split()]
word = input()
ans = 0
for col in ["RGB", "GBR", "BRG"]:
cnt = 0
for i in range(k):
if word[i] == col[i%3]:
cnt += 1
mx = cnt
for i in range(n-k):
if word[i+k] == col[(i+k)%3]:
cnt += 1
if word[i] == col[i%3]:
cnt -= 1
if cnt > mx:
mx = cnt
ans = max(ans, mx)
print(k - ans)
```
| 12,543 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Tags: implementation
Correct Solution:
```
q = int(input())
for _ in range(q):
n, k = list(map(int, input().split()))
s = input()
rgb = 0
gbr = 0
brg = 0
RGB = "RGB"
for i in range(k):
if i % 3 == 0:
rgb += int(s[i] != 'R')
gbr += int(s[i] != 'G')
brg += int(s[i] != 'B')
if i % 3 == 1:
rgb += int(s[i] != 'G')
gbr += int(s[i] != 'B')
brg += int(s[i] != 'R')
if i % 3 == 2:
rgb += int(s[i] != 'B')
gbr += int(s[i] != 'R')
brg += int(s[i] != 'G')
minimum = min(rgb, brg, gbr)
for i in range(k, n):
if (i - k) % 3 == 0:
rgb -= int(s[i - k] != 'R')
gbr -= int(s[i - k] != 'G')
brg -= int(s[i - k] != 'B')
if (i - k) % 3 == 1:
rgb -= int(s[i - k] != 'G')
gbr -= int(s[i - k] != 'B')
brg -= int(s[i - k] != 'R')
if (i - k) % 3 == 2:
rgb -= int(s[i - k] != 'B')
gbr -= int(s[i - k] != 'R')
brg -= int(s[i - k] != 'G')
if i % 3 == 0:
rgb += int(s[i] != 'R')
gbr += int(s[i] != 'G')
brg += int(s[i] != 'B')
if i % 3 == 1:
rgb += int(s[i] != 'G')
gbr += int(s[i] != 'B')
brg += int(s[i] != 'R')
if i % 3 == 2:
rgb += int(s[i] != 'B')
gbr += int(s[i] != 'R')
brg += int(s[i] != 'G')
minimum = min(minimum, rgb, brg, gbr)
print(minimum)
```
| 12,544 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
from collections import deque
#just calculate it for the first string length k for the 3 different states and from then on its just DP
def conv(let):
if let=="R":
return 0
elif let=="G":
return 1
else:
return 2
for j in range(int(input())):
n,k=map(int,input().split())
vals=[conv(i) for i in input()]
store=[[0,(k-1)%3],[1,k%3],[2,(k+1)%3]]
cost=[0,0,0];mincost=10**9
for s in range(3):
for i in range(k):
if vals[i]!=(s+i)%3:
cost[s]+=1
mincost=min(cost)
for i in range(1,n+1-k):
for s in range(3):
if vals[i-1]!=store[s][0]:
cost[s]-=1
store[s][0]+=1;store[s][1]+=1
store[s][0]=store[s][0]%3;store[s][1]=store[s][1]%3
if vals[i+k-1]!=store[s][1]:
cost[s]+=1
mincost=min(mincost,min(cost))
print(mincost)
```
Yes
| 12,545 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
from sys import stdin
input=stdin.readline
R = lambda:map(int,input().split())
q, = R()
a = 'RGB'
an =[0]*3
for q1 in range(q):
n,k = R()
s = input()
for i2 in range(3):
ans = [0]*n
for i in range(n):
ans[i]=s[i]==a[(i+i2)%3]
m = ans[0:k].count(0)
m1 = m
for i in range(k,n):
m1 += -ans[i]+ans[i-k]
m = min(m,m1)
an[i2]=m
print(min(an),flush=False)
```
Yes
| 12,546 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
a=input()
s=["RGB"*(k//3)+"RGB"[:k%3],"GBR"*(k//3)+"GBR"[:k%3],"BRG"*(k//3)+"BRG"[:k%3]]
minn=n
for i in range(n-k+1):
for j in s:
total=0
for l in range(k):
total+=1 if a[i+l]!=j[l] else 0
minn=min(total,minn)
if minn==0:
break
if minn==0:
break
print(minn)
```
Yes
| 12,547 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
t="RGB"*100000
q=int(input())
for _ in range(q):
ans=999999999999999999999999999999999999999999999999999999999999999999999999999
n,k=map(int,input().split())
s=input()
for i in range(3):
to=t[i:i+n+1]
dp=[0 for i in range(n)]
for j in range(n):
if s[j] !=to[j]:
dp[j]+=1
for j in range(1,n):
dp[j] += dp[j-1]
ii=0
jj=k-1
while jj<n:
if ii==0:
cnt=dp[jj]
else:
cnt=dp[jj]-dp[ii-1]
ans=min(ans,cnt)
jj += 1
ii += 1
print(ans)
```
Yes
| 12,548 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
for _ in range(int(input())):
n,k=map(int, input().split())
s=input()
if(k==1):
print(0)
elif(k==2):
if('RG' in s or 'GB' in s or 'BR' in s):
print(0)
else:
print(1)
elif(k==3):
if('RGB' in s or 'GBR' in s or 'BRG' in s):
print(0)
elif('RG' in s or 'GB' in s or 'BR' in s):
print(1)
else:
print(2)
else:
s1='RGB'*(k//3)
s2='GBR'*(k//3)
s3='BRG'*(k//3)
if(k%3==1):
s1+='R'
s2+='G'
s3+='B'
elif(k%3==2):
s1+='RG'
s2+='GB'
s3+='BR'
m=2005
for i in range(n):
if(i+k>n):
break
st=s[i:i+k]
c=0
if(st[0]=='R'):
for j in range(k):
if(st[j]!=s1[j]):
c+=1
elif(st[0]=='G'):
for j in range(k):
if(st[j]!=s2[j]):
c+=1
elif(st[0]=='B'):
for j in range(k):
if(st[j]!=s3[j]):
c+=1
if(c<m):
m=c
print(m)
```
No
| 12,549 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
#575_D1
import sys
q = int(input())
r = ["R", "G", "B"]
for i in range(0, q):
ns = [int(j) for j in input().split(" ")]
n = ns[0]
k = ns[1]
st = list(sys.stdin.readline().rstrip())
mx = 10000000000
for j in range(0, len(st)):
if len(st) - j < k:
break
rinc = r.index(st[j])
strinc = rinc
c = 0
for l in range(j + 1, len(st)):
rinc += 1
if rinc - strinc >= k:
break
if r[rinc % 3] != st[l]:
c += 1
mx = min(mx, c)
print(mx)
```
No
| 12,550 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
q=int(input())
while(q>0):
l=list(map(int,input().split()))
n=l[0]
k=l[1]
s=input()
min=1000
for j in range(0,n-k):
x=list(s)
w=1
c=0
i=j
while(w<k):
if x[i]=="R":
if x[i+1]=="G":
w+=1
else:
w+=1
x[i+1]="G"
c+=1
elif x[i]=="G":
if x[i+1]=="B":
w+=1
else:
w+=1
x[i+1]="B"
c+=1
elif x[i]=="B":
if x[i+1]=="R":
w+=1
else:
w+=1
x[i+1]="R"
c+=1
i+=1
if min>c:
min=c
print(min)
q-=1
```
No
| 12,551 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The only difference between easy and hard versions is the size of the input.
You are given a string s consisting of n characters, each character is 'R', 'G' or 'B'.
You are also given an integer k. Your task is to change the minimum number of characters in the initial string s so that after the changes there will be a string of length k that is a substring of s, and is also a substring of the infinite string "RGBRGBRGB ...".
A string a is a substring of string b if there exists a positive integer i such that a_1 = b_i, a_2 = b_{i + 1}, a_3 = b_{i + 2}, ..., a_{|a|} = b_{i + |a| - 1}. For example, strings "GBRG", "B", "BR" are substrings of the infinite string "RGBRGBRGB ..." while "GR", "RGR" and "GGG" are not.
You have to answer q independent queries.
Input
The first line of the input contains one integer q (1 ≤ q ≤ 2000) — the number of queries. Then q queries follow.
The first line of the query contains two integers n and k (1 ≤ k ≤ n ≤ 2000) — the length of the string s and the length of the substring.
The second line of the query contains a string s consisting of n characters 'R', 'G' and 'B'.
It is guaranteed that the sum of n over all queries does not exceed 2000 (∑ n ≤ 2000).
Output
For each query print one integer — the minimum number of characters you need to change in the initial string s so that after changing there will be a substring of length k in s that is also a substring of the infinite string "RGBRGBRGB ...".
Example
Input
3
5 2
BGGGG
5 3
RBRGR
5 5
BBBRR
Output
1
0
3
Note
In the first example, you can change the first character to 'R' and obtain the substring "RG", or change the second character to 'R' and obtain "BR", or change the third, fourth or fifth character to 'B' and obtain "GB".
In the second example, the substring is "BRG".
Submitted Solution:
```
import sys
n = int(input())
I = lambda : map(int,sys.stdin.readline().split())
for _ in range (n) :
n , m = I()
minn = float("inf")
st = input()
a1 = (m//3+1)*("RGB")
a1 = a1[:m]
a2 = (m//3+1)*("BRG")
a2 = a2[:m]
a3 = (m//3+1)*("GBR")
a3 = a3[:m]
print(a2)
for i in range (n-m+1) :
c1 = 0 ; c2 = 0 ; c3 =0 ;
for j in range (m) :
if st[i+j] == a1[j] :
pass
else :
c1+=1
if st[i+j] == a2[j] :
pass
else :
c2+=1
if st[i+j] == a3[j] :
pass
else :
c3+=1
minn = min(minn , c1)
minn = min(minn , c2)
minn = min(minn , c3)
print(minn)
```
No
| 12,552 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
def main():
n = int(sys.stdin.readline().strip())
s1 = sys.stdin.readline().strip()
s2 = sys.stdin.readline().strip()
for c1 in 'abc':
for c2 in 'abc':
for c3 in 'abc':
if len(set(c1 + c2 + c3)) != 3:
continue
ss = (c1 + c2 + c3) * n
if s1 in ss or s2 in ss:
continue
print('YES')
print(ss)
return
for c1 in 'abc':
for c2 in 'abc':
for c3 in 'abc':
if len(set(c1 + c2 + c3)) != 3:
continue
ss = c1 * n + c2 * n + c3 * n
if s1 in ss or s2 in ss:
continue
print('YES')
print(ss)
return
print('NO')
main()
```
| 12,553 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
def mp():
return map(int, input().split())
n = int(input())
s = input()
t = input()
if s[1] == t[1] and s[0] != t[0] and s[1] != s[0] and t[1] != t[0]:
print('YES')
print(s[1] * n + s[0] * n + t[0] * n)
elif s[0] == t[0] and s[1] != t[1] and s[0] != s[1] and t[0] != t[1]:
print('YES')
print(s[1] * n + t[1] * n + s[0] * n)
else:
for w in ['a', 'b', 'c']:
ans = [w]
cnt = [0, 0, 0]
cnt[ord(w) - 97] += 1
for i in range(3 * n - 1):
p = ans[-1]
may = [1] * 3
if s[0] == p:
may[ord(s[1]) - 97] = 0
if t[0] == p:
may[ord(t[1]) - 97] = 0
jj = -1
for j in range(3):
if may[j] and cnt[j] < n and (jj == -1 or cnt[j] <= cnt[jj]):
jj = j
if jj == -1:
break
ans.append(chr(jj + 97))
cnt[jj] += 1
if len(ans) == 3 * n:
print('YES')
print(''.join(ans))
break
else:
for w in ['a', 'b', 'c']:
ans = [w]
cnt = [0, 0, 0]
cnt[ord(w) - 97] += 1
for i in range(3 * n - 1):
p = ans[-1]
may = [1] * 3
if s[0] == p:
may[ord(s[1]) - 97] = 0
if t[0] == p:
may[ord(t[1]) - 97] = 0
jj = -1
for j in range(3):
if may[j] and cnt[j] < n and (jj == -1 or cnt[j] < cnt[jj]):
jj = j
if jj == -1:
break
ans.append(chr(jj + 97))
cnt[jj] += 1
if len(ans) == 3 * n:
print('YES')
print(''.join(ans))
break
else:
print('NO')
```
| 12,554 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
l = ["abc" * n, "acb" * n, "bac" * n, "bca" * n, "cba" * n, "cab" * n,
'b' * n + 'c' * n + 'a' * n, 'c' * n + 'a' * n + 'b' * n,
'a' * n + 'b' * n + 'c' * n]
s1 = input().strip()
s2 = input().strip()
print("YES")
for x in l:
if s1 not in x and s2 not in x:
print(x)
break
```
| 12,555 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
n = int(input())
s = input()
t = input()
a = [[0,0,0],[0,0,0],[0,0,0]]
b = ['a','b','c']
a[b.index(s[0])][b.index(s[1])] = 1
a[b.index(t[0])][b.index(t[1])] = 1
tr = a[0][0] + a[1][1] + a[2][2]
print("YES")
if (s == t):
if (s[0] == s[1]):
print("abc"*n)
else:
i = 0
while b[i] == s[0] or b[i] == s[1]:
i += 1
print(s[0]*n + b[i]*n + s[1]*n)
else:
max_row_sum = 0
for i in range(3):
if a[i][0] + a[i][1] + a[i][2] > max_row_sum:
max_row_sum = a[i][0] + a[i][1] + a[i][2]
index = i
max_col_sum = 0
for i in range(3):
if a[0][i] + a[1][i] + a[2][i] > max_col_sum:
max_col_sum = a[0][i] + a[1][i] + a[2][i]
index_col = i
if tr == 2:
print('abc'*n)
elif tr == 0 and max_row_sum == 2:
i = 0
while i == index:
i += 1
j = 0
while i == j or index == j:
j += 1
print(b[i]*n + b[j]*n + b[index]*n)
elif tr == 1:
for i in range(3):
for j in range(3):
if a[i][j] and i != j:
for k in range(3):
if k != i and k != j:
print((b[j] + b[i] + b[k])*n)
else:
if max_col_sum == 2:
i = 0
while i == index_col:
i += 1
j = 0
while i == j or index_col == j:
j += 1
print(b[index_col]*n + b[j]*n + b[i]*n)
elif (s[0] == t[1] and s[1] == t[0]):
i = 0
while b[i] == s[0] or b[i] == s[1]:
i += 1
print(s[0]*n + b[i]*n + s[1]*n)
else:
i = 0
while a[index][i] != 0 or i == index:
i += 1
for j in range(3):
if j != index and j != i:
print((b[index] + b[i] + b[j])*n)
```
| 12,556 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
n=int(input())
a=['abc'*n,'acb'*n,'bac'*n,'bca'*n,'cab'*n,'cba'*n,'a'*n+'b'*n+'c'*n,'a'*n+'c'*n+'b'*n,'b'*n+'a'*n+'c'*n,'b'*n+'c'*n+'a'*n,'c'*n+'b'*n+'a'*n,'c'*n+'a'*n+'b'*n]
b=input()
c=input()
for i in a:
if b not in i and c not in i:
print("YES")
print(i)
exit()
print("NO")
```
| 12,557 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
n = int(input())
a = {'a', 'b', 'c'}
s, t = input(), input()
if t[0] == t[1]:
s, t = t, s
print('YES')
if s[0] == s[1]:
if t[0] == t[1]:
print('abc'*n)
elif t[0] == s[0]:
print((s[0]+next(iter(a-set(t)))+t[1])*n)
elif t[1] == s[0]:
print((s[0]+t[0]+next(iter(a-set(t))))*n)
else:
print((s[0]+t[::-1])*n)
else:
if t[0] == s[0]:
print(''.join(a-{t[0]})*n+t[0]*n)
elif t[1] == s[1]:
print(t[1]*n+''.join(a-{t[1]})*n)
elif s == t[::-1]:
print(s[0]*n+next(iter(a-set(s)))*n+s[1]*n)
elif s[1] == t[0]:
print(t[1]*n+t[0]*n+s[0]*n)
else:
print(s[1]*n+s[0]*n+t[0]*n)
```
| 12,558 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
'''input
1
cb
ac
'''
import sys
from collections import defaultdict as dd
from itertools import permutations as pp
from itertools import combinations as cc
from collections import Counter as ccd
from random import randint as rd
from bisect import bisect_left as bl
from heapq import heappush as hpush
from heapq import heappop as hpop
mod=10**9+7
def ri(flag=0):
if flag==0:
return [int(i) for i in sys.stdin.readline().split()]
else:
return int(sys.stdin.readline())
n =ri(1)
s = input()
t = input()
a="abc"
for i in pp(a):
new = "".join(i)
one= new*n
two= new[0]*n + new[1]*n + new[2]*n
if s not in one and t not in one:
print("YES")
print(one)
sys.exit(0)
if s not in two and t not in two:
print("YES")
print(two)
sys.exit(0)
print("NO")
```
| 12,559 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Tags: brute force, constructive algorithms
Correct Solution:
```
n = int(input())
s = input()
t = input()
ans = ''
def gen_singular(c):
global ans
ans += (c * n)
def gen_pairs(p, q):
global ans
ans += (p+q) * n
def d(s):
return s[0] != s[1]
def e(p):
return p[0] == p[1]
def rest(s, t):
return [el for el in ['a', 'b', 'c'] if el not in set(s+t)][0]
def attempt(s, t):
global ans
#1
if s[0] == t[0] and s[1] != t[1] and d(s) and d(t):
gen_singular(s[1])
gen_singular(t[1])
gen_singular(s[0])
#2
if s[1] == t[1] and s[0] != t[0] and d(s) and d(t):
gen_singular(s[1])
gen_singular(t[0])
gen_singular(s[0])
#3
if e(s) and t[0] == s[0] and d(t):
gen_singular(t[1])
gen_pairs(s[0], rest(s, t))
#4
if e(s) and t[1] == s[0] and d(t):
gen_pairs(s[0], rest(s, t))
gen_singular(t[0])
#5
if d(s) and d(t) and s[1] == t[0] and s[0] != t[1]:
gen_singular(t[1])
gen_singular(s[1])
gen_singular(s[0])
#6
if s[0] == s[1] == t[0] == t[1]:
if s[0] == 'a':
gen_pairs(s[0], 'c')
gen_singular('b')
else:
gen_pairs(s[0], 'a')
gen_singular(rest('a', s[0]))
#7
if d(s) and d(t) and t == s[::-1]:
gen_singular(s[0])
gen_singular(rest(s[0], s[1]))
gen_singular(s[1])
#8
if e(s) and d(t) and s[0] != t[0] and s[0] != t[1]:
gen_pairs(t[0], s[0])
gen_singular(t[1])
#9
if d(s) and s == t:
gen_singular(s[0])
gen_singular(rest(s[0], s[1]))
gen_singular(s[1])
#10
if e(s) and e(t) and s[0] != t[0]:
gen_pairs(s[0], t[0])
gen_singular(rest(s[0], t[0]))
attempt(s, t)
if len(ans) == 0:
attempt(t, s)
print('YES')
print(ans)
```
| 12,560 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
n = int(input())
s = input()
t = input()
print('YES')
if len(set(s + t)) == 1:
print('abc' * n)
elif len(set(s + t)) == 2:
if t[0] == t[1]:
s, t = t, s
if s[0] == s[1]:
a = s[0]
b = list(set(s + t) - {a})[0]
c = list(set('abc') - {a, b})[0]
if s[0] == t[1]:
b, c = c, b
print((b + a + c) * n)
else:
a, b = s
c = list(set('abc') - {a, b})[0]
print(a * n + c * n + b * n)
else:
if t[0] == t[1]:
s, t = t, s
if s[0] == s[1]:
a = s[0]
b, c = t
print((b + a + c) * n)
else:
if s[0] == t[0]:
c, a = s
b = t[1]
elif s[1] == t[1]:
b, a = s
c = t[0]
elif s[0] == t[1]:
c, b = s
a = t[0]
else:
b, a = s
c = t[1]
print(a * n + b * n + c * n)
```
Yes
| 12,561 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
from itertools import permutations
n = int(input())
s1, s2 = input()
t1, t2 = input()
print("YES")
if s1==t2 and s2==t1 and s1 != s2:
x3 = list(set("abc")-set([s1,s2]))[0]
print(s1*n+x3*n+s2*n)
else:
for x1,x2,x3 in permutations("abc"):
possible = True
for a1, a2 in [(x1, x2), (x2, x3), (x3, x1)]:
if (a1==s1 and a2==s2) or (a1==t1 and a2==t2):
possible = False
break
if possible:
print((x1+x2+x3)*n)
exit()
if s1==t1:
print((s2+t2)*n+s1*n)
exit()
if s2==t2:
print(s2*n+(s1+t1)*n)
exit()
```
Yes
| 12,562 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
n = int(input())
s = input()
t = input()
characters = ['a', 'b', 'c']
print('YES')
if s[0] == s[1] and t[0] == t[1]:
x = 'abc'
print(x*n)
elif s[0] == s[1]:
if 'a' not in t:
x = t[0] + 'a' + t[1]
elif 'b' not in t:
x = t[0] + 'b' + t[1]
else:
x = t[0] + 'c' + t[1]
print(x*n)
elif t[0] == t[1]:
if 'a' not in s:
x = s[0] + 'a' + s[1]
elif 'b' not in s:
x = s[0] + 'b' + s[1]
else:
x = s[0] + 'c' + s[1]
print(x*n)
elif (s[1] == t[1]) or (s[0] == t[0]):
if s[1] == t[1]:
characters.remove(s[1])
print(s[1]*n + ''.join(characters)*n)
else:
characters.remove(s[0])
print(''.join(characters)*n + s[0]*n)
elif s[0] == t[1] and s[1] == t[0]:
characters.remove(s[0])
characters.remove(s[1])
print(s[0]*n + characters[0]*n + s[1]*n)
else:
if 'a' not in s:
x = s[0] + 'a' + s[1]
elif 'b' not in s:
x = s[0] + 'b' + s[1]
else:
x = s[0] + 'c' + s[1]
print(x*n)
```
Yes
| 12,563 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
n=int(input())
a=n
s1=input()
s2=input()
print('YES')
def solver(x,n,r):
s1=''
s2=''
for i in range(n):
s1+=x
s2+=r[0]
s2+=r[1]
print(s1+s2,end="")
exit(0)
def rolver(x,s1,s2,n):
if(x=='a'):
r='bc'
if(x=='b'):
r='ca'
if(x=='c'):
r='ab'
if(r==s1 or r==s2):
r=r[::-1]
x1=''
g1=''
for i in range(n):
x1+=x
g1+=r
if(str(r[1]+x)!=s1 and str(r[1]+x)!=s2):
print(g1+x1)
exit(0)
print(x1+g1)
exit(0)
g1=['abc','acb','bac','bca','cab','cba']
if(s1[0]!=s1[1] and s2[0]!=s2[1]):
for i in range(6):
t=g1[i]
if(s1 not in t and s2 not in t):
for x in range(3):
for j in range(n):
print(g1[i][x],end="")
exit(0)
for i in range(6):
t=g1[i]+g1[i]
if(s1 not in t and s2 not in t):
for j in range(n):
print(g1[i],end="")
exit(0)
```
Yes
| 12,564 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
n = int(input())
s = input()
s1 = input()
p = ['a', 'b', 'c']
if s[0] == s1[0] or s1[1] == s[1] :
if s == s1:
for i in range(3):
for j in range(3):
for k in range(3):
ss = p[i] + p[j] + p[k]
if p[i] != p[j] and p[i] != p[k] and p[j] != p[k]:
if s not in ss and s1 not in ss:
if n == 1:
print("YES")
print(ss)
exit()
if ss[-1] + ss[0] != s and ss[-1] + ss[0] != s1:
print("YES")
print(ss * n)
exit()
print("NO")
exit()
else:
if s[0] == s[1] or s1[0] == s1[1]:
if s[0] == s[1]:
k = ['a', 'b', 'c']
k.pop(k.index(s1[0]))
k.pop(k.index(s1[1]))
print("YES")
print((s1[0] + k[0] + s1[1]) * n)
else:
k = ['a', 'b', 'c']
k.pop(k.index(s[0]))
k.pop(k.index(s[1]))
print("YES")
print((s[0] + k[0] + s[1]) * n)
exit()
print('YES')
if s[0] == s1[0]:
print(s[1] * n + s1[1] * n + s[0] * n)
else:
print(s[1] * n + s[0] * n + s1[0] * n)
exit()
if s == s1[::-1]:
if n == 1:
print("YES")
p.pop(p.index(s[0]))
p.pop(p.index(s[1]))
print(s[0] + p[0] + s[1])
exit()
print("NO")
exit()
for i in range(3):
for j in range(3):
for k in range(3):
ss = p[i] + p[j] + p[k]
if p[i] != p[j] and p[i] != p[k] and p[j] != p[k]:
if s not in ss and s1 not in ss:
if n == 1:
print("YES")
print(ss)
exit()
if ss[-1] + ss[0] != s and ss[-1] + ss[0] != s1:
print("YES")
print(ss * n)
exit()
print("NO")
```
No
| 12,565 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
import math
import sys
input = sys.stdin.readline
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input().strip()
return(list(s[:len(s)]))
def invr():
return(map(int,input().split()))
n=inp()
s=input().strip()
t=input().strip()
l=['abc','acb','bac','bca','cab','cba']
templ=[]
for each in l:
if s in each :
continue
elif t in each:
continue
else:
templ.append(each)
res=''
flag=1
l=templ
for i in range(n):
count=0
temp=res+l[count]
while(s in temp or t in temp):
if count>=len(l):
print('NO')
flag=0
break
else:
temp=res+l[count]
count+=1
res=temp
if flag:
print('YES')
print(res)
```
No
| 12,566 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
n = int(input())
s1, s2 = input(), input()
print("YES")
if s1[0] != s1[1] and s1 == s2[::-1]:
print(s1[0]*n + (({'a','b','c'}-set(s1)).pop())*n + s1[1]*n)
exit()
if s1[0] != s1[1] and s2[0] != s2[1]:
if s1[1] == s2[0]:
print(s2[1]*n + s2[0]*n + (({'a','b','c'}-set(s2)).pop())*n)
else:
print(s1[1]*n + s1[0]*n + (({'a','b','c'}-set(s1)).pop())*n)
exit()
if s1[0] == s1[1] and s2[0] == s2[1]:
print('abc'*n)
exit()
if s1[0] == s1[1]:
s = s2[::-1]
else:
s = s1[::-1]
s += ({'a', 'b', 'c'} - set(s)).pop()
print(s*n)
```
No
| 12,567 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two strings s and t both of length 2 and both consisting only of characters 'a', 'b' and 'c'.
Possible examples of strings s and t: "ab", "ca", "bb".
You have to find a string res consisting of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
A substring of a string is a contiguous subsequence of that string. So, the strings "ab", "ac" and "cc" are substrings of the string "abacc", but the strings "bc", "aa" and "cb" are not substrings of the string "abacc".
If there are multiple answers, you can print any of them.
Input
The first line of the input contains one integer n (1 ≤ n ≤ 10^5) — the number of characters 'a', 'b' and 'c' in the resulting string.
The second line of the input contains one string s of length 2 consisting of characters 'a', 'b' and 'c'.
The third line of the input contains one string t of length 2 consisting of characters 'a', 'b' and 'c'.
Output
If it is impossible to find the suitable string, print "NO" on the first line.
Otherwise print "YES" on the first line and string res on the second line. res should consist of 3n characters, n characters should be 'a', n characters should be 'b' and n characters should be 'c' and s and t should not occur in res as substrings.
If there are multiple answers, you can print any of them.
Examples
Input
2
ab
bc
Output
YES
acbbac
Input
3
aa
bc
Output
YES
cacbacbab
Input
1
cb
ac
Output
YES
abc
Submitted Solution:
```
import math
import sys
from collections import defaultdict
# input = sys.stdin.readline
rt = lambda: map(int, input().split())
ri = lambda: int(input())
rl = lambda: list(map(int, input().split()))
def main():
n = ri()
s = input()
t = input()
trios = ['abc', 'acb', 'bac', 'bca', 'cab', 'cba']
if n == 1:
found = None
for trio in trios:
if s not in trio and t not in trio:
found = trio
break
if found:
print('YES')
print(found)
else:
print('NO')
else:
found = None
for t1 in trios:
x = t1 + t1
if s not in x and t not in x:
found = t1
break
if found:
print('YES')
print(found*n)
else:
print('NO')
if __name__ == '__main__':
main()
```
No
| 12,568 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
c,l=0,[]
for i in range(n):
l.append(int(input()))
for i in l:
if i%2!=0:
c+=1
cnt,p=int(c/2),0
for i in l:
if i%2==0:
print(int(i/2))
elif i%2!=0 and p<cnt and i<0:
print(int(i/2))
p+=1
elif i%2!=0 and p<cnt and i>0:
print(int(i/2)+1)
p+=1
elif i%2!=0 and p>=cnt and i<0:
print(int(i/2)-1)
p+=1
else:
print(int(i/2))
```
| 12,569 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
st=1
for i in range(n):
a=int(input())
if a%2==0:
print(a//2)
continue
if a==0:
print(0)
if a<0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
if a>0:
if st==1:
print(a//2)
st=0
else:
print(a//2+1)
st=1
```
| 12,570 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l=[]
no=0
po=0
for i in range(n):
x=int(input())
if(x<0 and x%2==1):
no+=1
elif(x>0 and x%2==1):
po+=1
l.append(x)
ans=[]
no//=2
po//=2
for i in l:
x=i
if(x%2==1):
if(x<0 and no>0):
# print("....",x)
x-=1
no-=1
elif(x>0 and po>0):
x+=1
po-=1
# print(x,x//2)
if(x<0):
ans.append(-(abs(x)//2))
else:
ans.append(x//2)
for i in ans:
print(i)
```
| 12,571 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
l1=[]
odd=0
for i in range (n):
l1.append(int(input()))
l=[]
sum1=0
for i in range (n):
if(l1[i]%2==0):
sum1+=int(l1[i]/2)
l.append(int(l1[i]/2))
elif(l1[i]%2!=0 and odd==0):
l.append(int(l1[i]/2))
sum1+=int(l1[i]/2)
if(sum1==0):
for i in range (n):
print(l[i])
elif(sum1<0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]>0):
sum1+=1
l[i]+=1
if(sum1==0):
break
for i in range (n):
print(l[i])
elif(sum1>0):
for i in range (n):
if(l1[i]%2!=0 and l1[i]<0):
sum1-=1
l[i]-=1
if(sum1==0):
break
for i in range (n):
print(l[i])
```
| 12,572 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
n=int(input())
mod=1
for _ in range(n):
a=int(input())
if(a%2)==0:
print(a//2)
else:
print(a//2 + mod)
mod=1-mod
```
| 12,573 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
l=[]
for i in range(int(input())):
l.append(int(input()))
ans=[]
f=1
for i in l:
if i%2==0:
ans.append(i//2)
else:
ans.append((i+f)//2)
f*=-1
for i in ans:
print(i)
```
| 12,574 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
from collections import Counter
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
##########################################################
#for _ in range(int(input())):
#import math
import sys
# from collections import deque
#from collections import Counter
# ls=list(map(int,input().split()))
# for i in range(m):
# for i in range(int(input())):
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=sys.stdin.readline()
#n=int(n)
#n,k= map(int, input().split())
#arr=list(map(int,input().split()))
#n=int(inaput())
#for _ in range(int(input())):
import math
n = int(input())
f=0
for i in range(n):
u= int(input())
if u%2==0:
print(u//2)
else:
if f==0:
print(math.floor(u/2))
f=1
else:
print(math.ceil(u/2))
f=0
#arr=list(map(int,input().split()))
#for _ in range(int(input())):
#n, k = map(int, input().split())
#arr=list(map(int,input().split()))
```
| 12,575 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Tags: implementation, math
Correct Solution:
```
sum=0
for i in range(int(input())):
x=int(input())
y=x//2
if y==x/2:
print(y)
else:
if sum==0:
sum=x-y*2
print(y)
else:
y=y+sum
sum=0
print(y)
```
| 12,576 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import sys
import math
def main():
#for _ in range(int(input())):
n = int(sys.stdin.readline())
#r, p, s= [int(x) for x in sys.stdin.readline().split()]
#l = list(set(int(x) for x in sys.stdin.readline().split()))
#str = sys.stdin.readline()
flag=0
for i in range(n):
x=int(sys.stdin.readline())
if x%2==0:
print(x//2)
else:
if flag==0:
print(x//2)
flag=1
else:
q=x//2
q+=1
print(q)
flag=0
if __name__ == "__main__":
main()
```
Yes
| 12,577 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
l=[]
low=0
even=0
for i in range(0,n):
t=int(input())
if t%2==0:
even=even+t//2
l.append((t//2,0))
else:
low=low+(t//2)
l.append((t//2))
if low!=-1*even:
count=(-1*even)-low
for i in range(0,len(l)):
if count==0:
break
if type(l[i])!=tuple:
l[i]=l[i]+1
count=count-1
for i in range(0,len(l)):
if type(l[i])==tuple:
print(l[i][0])
else:
print(l[i])
```
Yes
| 12,578 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
k = []
p = 0
for x in range(n):
a = int(input())
if a%2 == 0:
a = int(a/2)
k.append(a)
else:
if p%2 == 0:
a = floor(a/2)
k.append(a)
else:
a = ceil(a/2)
k.append(a)
p += 1
for i in range(len(k)):
print(k[i])
```
Yes
| 12,579 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
import math
n = int(input())
check1 = 0
while n:
n -= 1
a = int(input())
if a % 2 == 0: print(a//2)
elif a % 2 and check1 % 2 == 0:
print(int(math.floor(a / 2)))
check1 += 1
elif a % 2 and check1 % 2:
print(int(math.ceil(a / 2)))
check1 += 1
else: print(a)
```
Yes
| 12,580 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
o=0
for i in range(n):
l=int(input())
if(l>=0):
if(l%2==0):
print(l//2)
else:
if(o==0):
o=o+1
print(l//2)
else:
print(l//2+1)
o+=1
else:
l=abs(l)
if(l%2==0):
print(-(l//2))
else:
if(o==0):
o=o+1
print(-(l//2))
else:
print(-(l//2+1))
o=o+1
```
No
| 12,581 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import ceil,floor
n=int(input())
cou=0
while n:
a=int(input())
if(cou%2==0 and a%2==1):
print(ceil(a/2))
else:
print(floor(a/2))
cou+=1
n-=1
```
No
| 12,582 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
from math import *
n = int(input())
sum = 0
found = 1
for i in range(n):
x = int(input())
if x&1:
if found == 1:
print(floor(x / 2))
found = 0
else:
print(ceil(x / 2))
found = 1
else:
print(x / 2)
```
No
| 12,583 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Another Codeforces Round has just finished! It has gathered n participants, and according to the results, the expected rating change of participant i is a_i. These rating changes are perfectly balanced — their sum is equal to 0.
Unfortunately, due to minor technical glitches, the round is declared semi-rated. It means that all rating changes must be divided by two.
There are two conditions though:
* For each participant i, their modified rating change b_i must be integer, and as close to (a_i)/(2) as possible. It means that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. In particular, if a_i is even, b_i = (a_i)/(2). Here ⌊ x ⌋ denotes rounding down to the largest integer not greater than x, and ⌈ x ⌉ denotes rounding up to the smallest integer not smaller than x.
* The modified rating changes must be perfectly balanced — their sum must be equal to 0.
Can you help with that?
Input
The first line contains a single integer n (2 ≤ n ≤ 13 845), denoting the number of participants.
Each of the next n lines contains a single integer a_i (-336 ≤ a_i ≤ 1164), denoting the rating change of the i-th participant.
The sum of all a_i is equal to 0.
Output
Output n integers b_i, each denoting the modified rating change of the i-th participant in order of input.
For any i, it must be true that either b_i = ⌊ (a_i)/(2) ⌋ or b_i = ⌈ (a_i)/(2) ⌉. The sum of all b_i must be equal to 0.
If there are multiple solutions, print any. We can show that a solution exists for any valid input.
Examples
Input
3
10
-5
-5
Output
5
-2
-3
Input
7
-7
-29
0
3
24
-29
38
Output
-3
-15
0
2
12
-15
19
Note
In the first example, b_1 = 5, b_2 = -3 and b_3 = -2 is another correct solution.
In the second example there are 6 possible solutions, one of them is shown in the example output.
Submitted Solution:
```
n=int(input())
a=[]
b=[]
for i in range(n):
x=int(input())
a.append(x)
if x>=0:
b.append(x//2)
else:
b.append(0-(abs(x)//2))
if sum(b)==(-1):
for i in range(n):
if a[i]>0 and a[i]%2==1:
b[i]+=1
break
elif sum(b)==1:
for i in range(n):
if a[i]<0 and a[i]%2==1:
b[i]-=1
break
for i in b:
print(i)
```
No
| 12,584 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We had a really tough time generating tests for problem D. In order to prepare strong tests, we had to solve the following problem.
Given an undirected labeled tree consisting of n vertices, find a set of segments such that:
1. both endpoints of each segment are integers from 1 to 2n, and each integer from 1 to 2n should appear as an endpoint of exactly one segment;
2. all segments are non-degenerate;
3. for each pair (i, j) such that i ≠ j, i ∈ [1, n] and j ∈ [1, n], the vertices i and j are connected with an edge if and only if the segments i and j intersect, but neither segment i is fully contained in segment j, nor segment j is fully contained in segment i.
Can you solve this problem too?
Input
The first line contains one integer n (1 ≤ n ≤ 5 ⋅ 10^5) — the number of vertices in the tree.
Then n - 1 lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ n, x_i ≠ y_i) denoting the endpoints of the i-th edge.
It is guaranteed that the given graph is a tree.
Output
Print n pairs of integers, the i-th pair should contain two integers l_i and r_i (1 ≤ l_i < r_i ≤ 2n) — the endpoints of the i-th segment. All 2n integers you print should be unique.
It is guaranteed that the answer always exists.
Examples
Input
6
1 2
1 3
3 4
3 5
2 6
Output
9 12
7 10
3 11
1 5
2 4
6 8
Input
1
Output
1 2
Tags: constructive algorithms, dfs and similar, divide and conquer, trees
Correct Solution:
```
def main():
import sys
input = sys.stdin.buffer.readline
N = int(input())
adj = [[] for _ in range(N+1)]
for _ in range(N-1):
a, b = map(int, input().split())
adj[a].append(b)
adj[b].append(a)
st = [1]
seen = [0] * (N+1)
seen[1] = 1
LL = [0] * (2 * N + 3)
LL[-1] = 1
LL[1] = 0
LL_rev = [0] * (2 * N + 3)
LL_rev[1] = -1
LL_rev[-1] = 0
st_append = st.append
while st:
v = st.pop()
for u in adj[v]:
if not seen[u]:
seen[u] = 1
st_append(u)
v_next = LL[v]
v_prev = LL_rev[v]
LL[v] = u
LL[u] = v_next
LL_rev[v_next] = u
LL_rev[u] = v
u_ = -u
LL_rev[v] = u_
#LL_rev[u_] = v_prev
LL[v_prev] = u_
LL[u_] = v
s = -1
ans1 = [0] * (N + 1)
ans2 = [0] * (N + 1)
for i in range(1, 2 * N + 1):
if s < 0:
ans1[-s] = i
else:
ans2[s] = i
s = LL[s]
ans = '\n'.join([' '.join(map(str, [ans1[i], ans2[i]])) for i in range(1, N + 1)])
sys.stdout.write(ans)
if __name__ == '__main__':
main()
```
| 12,585 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
import math
n = int(input())
kords = list()
for _ in range(n):
kords.append(list(map(int, input().split())))
if n % 2 == 0:
first_part = kords[:len(kords) // 2 + 1]
second_part = kords[len(kords) // 2:] + [kords[0]]
for i in range(n // 2):
if abs( first_part[i][0] - second_part[i + 1][0] ) == abs( first_part[i + 1][0] - second_part[i][0] ) and abs( first_part[i + 1][1] - second_part[i][1] ) == abs( first_part[i][1] - second_part[i + 1][1] ):
continue
else:
print('NO')
break
else:
print('YES')
else:
print('NO')
```
| 12,586 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
import sys
n = int(sys.stdin.readline().strip())
X = []
Y = []
for i in range (0, n):
x, y = list(map(int, sys.stdin.readline().strip().split()))
X.append(x)
Y.append(y)
X.append(X[0])
Y.append(Y[0])
if n % 2 == 1:
print("NO")
else:
v = True
m = n // 2
for i in range (0, m):
if X[i+1]-X[i] != X[i+m]-X[i+m+1]:
v = False
if Y[i+1]-Y[i] != Y[i+m]-Y[i+m+1]:
v = False
if v == True:
print("YES")
else:
print("NO")
```
| 12,587 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
from sys import stdin
from collections import deque
mod = 10**9 + 7
import sys
# sys.setrecursionlimit(10**6)
from queue import PriorityQueue
# def rl():
# return [int(w) for w in stdin.readline().split()]
from bisect import bisect_right
from bisect import bisect_left
from collections import defaultdict
from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# # l = list(map(int,input().split()))
# from itertools import permutations
import heapq
# input = lambda: sys.stdin.readline().rstrip()
input = lambda : sys.stdin.readline().rstrip()
from sys import stdin, stdout
from heapq import heapify, heappush, heappop
from itertools import permutations
n = int(input())
ba = []
if n%2!=0:
print('NO')
exit()
for i in range(n):
a,b = map(int,input().split())
ba.append([a,b])
slope = []
seti = set()
for i in range(n):
a,b = ba[i]
c,d = ba[(i+n//2)%n]
x,y = (a+c)/2,(b+d)/2
seti.add((x,y))
# print(seti)
if len(seti) == 1:
print('YES')
else:
print('NO')
```
| 12,588 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
# https://codeforces.com/contest/1300/problem/D
n = int(input())
p = [list(map(int, input().split())) for _ in range(n)]
def is_parallel(x1, y1, x2, y2):
if x1 == -x2 and y1 == -y2:
return True
return False
flg = True
if n % 2 == 1:
flg=False
else:
for i in range(n//2):
p1, p2 = p[i], p[i+1]
p3, p4 = p[i+n//2], p[(i+n//2+1)%n]
flg = is_parallel(p2[0]-p1[0], p2[1]-p1[1], p4[0]-p3[0], p4[1]-p3[1])
if flg==False:
break
if flg==False:
print('no')
else:
print('yes')
```
| 12,589 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
def main():
N = int(input())
if N % 2:
print("NO")
return
pts = [complex(*map(int, input().strip().split())) for _ in range(N)]
for i in range(N // 2):
if pts[i] + pts[i + N // 2] != pts[0] + pts[N // 2]:
print("NO")
return
print("YES")
main()
```
| 12,590 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
def main():
from sys import stdin, stdout
n = int(stdin.readline())
inp = tuple((tuple(map(float, stdin.readline().split())) for _ in range(n)))
avg_xt2 = sum((inp_i[0] for inp_i in inp)) / n * 2
avg_yt2 = sum((inp_i[1] for inp_i in inp)) / n * 2
sym = tuple(((avg_xt2 - x, avg_yt2 - y) for (x, y) in inp))
nd2 = n // 2
stdout.write("YES" if inp == sym[nd2:] + sym[:nd2] else "NO")
main()
```
| 12,591 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
from __future__ import division
N = int(input())
points = []
for p in range(N):
x, y = [int(x) for x in input().split()]
points.append((x, y))
# T always has central symmetry
# To determine: if P has a point of central symmetry
# function to determine the bottom and top most points of our polygon
def bot_top(points):
y_val = []
for p in points:
y_val.append(p[1])
minn = y_val[0]
maxx = y_val[1]
for y in y_val:
if y < minn:
minn = y
if y > maxx:
maxx = y
return minn, maxx
def left_right(points):
y_val = []
for p in points:
y_val.append(p[0])
minn = y_val[0]
maxx = y_val[0]
for y in y_val:
if y < minn:
minn = y
if y > maxx:
maxx = y
return minn, maxx
# point of central symmetry
def cs():
x_1, x_2 = left_right(points)
y_1, y_2 = bot_top(points)
return (x_2 + x_1)/2, (y_1 + y_2)/2
def midpoint(p1, p2):
x_1, y_1 = p1
x_2, y_2 = p2
return (x_2 + x_1)/2, (y_2 + y_1)/2
CS = cs()
zz = True
if N % 2 == 0:
for pos in range(int(N/2)):
p1 = points[pos]
p2 = points[pos + int(N/2)]
mp = midpoint(p1, p2)
if mp != CS:
zz = False
break
if zz:
print ('YES')
else:
print ('NO')
else:
print ('NO')
```
| 12,592 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Tags: geometry
Correct Solution:
```
def Input():
tem = input().split()
ans = []
for it in tem:
ans.append(int(it))
return ans
n = Input()[0]
a = []
ma = {}
for i in range(n):
x, y = Input()
a.append((x,y))
a.append(a[0])
flag = True
for i in range(1,n+1):
if (a[i][0]-a[i-1][0])==0:
v = None
else:
v = (a[i][1]-a[i-1][1])/(a[i][0]-a[i-1][0])
dis = ((a[i][1]-a[i-1][1])**2+(a[i][0]-a[i-1][0])**2)**0.5
if v not in ma:
ma[v]=dis
else:
if abs(dis-ma[v])<0.00000001:
del ma[v]
else:
flag = False
break
if len(ma)>0:
flag = False
if flag:
print('YES')
else:
print('NO')
```
| 12,593 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
def mid(a,b):
return [(a[0]+b[0])/2,(a[1]+b[1])/2]
n=int(input())
l=[]
for i in range(n):
l.append(list(map(int,input().split())))
if n%2==0:
k=mid(l[0],l[n//2])
flag=True
for i in range(1,n//2):
if k!=mid(l[i],l[n//2+i]):
flag=False
break
if flag==True:
print("YES")
else:
print("NO")
else:
print("NO")
```
Yes
| 12,594 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
from sys import stdin
from os import getenv
if not getenv('PYCHARM'):
def input():
return next(stdin)[:-1]
def main():
n = int(input())
pp = []
for _ in range(n):
pp.append(list(map(int, input().split())))
if n%2 != 0:
print("NO")
return
for i in range(n//2):
x1 = pp[i+1][0] - pp[i][0]
y1 = pp[i+1][1] - pp[i][1]
x2 = pp[(i+1+n//2) % n][0] - pp[i+n//2][0]
y2 = pp[(i+1+n//2) % n][1] - pp[i+n//2][1]
if x1 != -x2 or y1 != -y2:
print("NO")
return
print("YES")
if __name__ == "__main__":
main()
```
Yes
| 12,595 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import sys
input=sys.stdin.buffer.readline #FOR READING PURE INTEGER INPUTS (space separation ok)
def isParallel(p1a,p1b,p2a,p2b):
dx1=p1a[0]-p1b[0]
dy1=p1a[1]-p1b[1]
dx2=p2a[0]-p2b[0]
dy2=p2a[1]-p2b[1]
return dx1==dx2 and dy1==dy2
n=int(input())
xy=[] #[[x,y]]
for _ in range(n):
a,b=[int(z) for z in input().split()]
xy.append([a,b])
#YES if each line is parallel to another line
xy.sort(key=lambda z:(z[0],z[1])) #sort by x then y.
#xy[0] will match with xy[n-1],xy[1] will match with xy[n-2] etc.
if n%2==1:
print('NO')
else:
ans='YES'
for i in range(n//2):
p1a=xy[i]
p1b=xy[i+1]
j=n-1-i
p2a=xy[j-1]
p2b=xy[j]
if not isParallel(p1a,p1b,p2a,p2b):
ans='NO'
break
print(ans)
```
Yes
| 12,596 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import os,io
import sys
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
n=int(input())
shape=[]
for _ in range(n):
x,y=map(int,input().split())
shape.append([x,y])
if n%2==1:
print('NO')
sys.exit()
for i in range(n):
if shape[i][0]-shape[i-1][0]!=shape[(n//2+i-1)%n][0]-shape[(n//2+i)%n][0]:
print('NO')
sys.exit()
if shape[i][1]-shape[i-1][1]!=shape[(n//2+i-1)%n][1]-shape[(n//2+i)%n][1]:
print('NO')
sys.exit()
print('YES')
```
Yes
| 12,597 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
import sys
input = sys.stdin.readline
import math
import copy
import collections
from collections import deque
n = int(input())
points = []
for i in range(n):
points.append(list(map(int,input().split())))
vec = []
for i in range(n):
a = points[i]
b = points[(i+1)%n]
vec.append([b[0]-a[0],b[1]-a[1]])
dirs = set()
for ele in vec:
x,y = ele
d = math.atan(y/x)
if d<0:
d+=math.pi
if d in dirs:
dirs.remove(d)
else:
dirs.add(d)
if dirs:
print("NO")
else:
print("YES")
```
No
| 12,598 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Guy-Manuel and Thomas are going to build a polygon spaceship.
You're given a strictly convex (i. e. no three points are collinear) polygon P which is defined by coordinates of its vertices. Define P(x,y) as a polygon obtained by translating P by vector \overrightarrow {(x,y)}. The picture below depicts an example of the translation:
<image>
Define T as a set of points which is the union of all P(x,y) such that the origin (0,0) lies in P(x,y) (both strictly inside and on the boundary). There is also an equivalent definition: a point (x,y) lies in T only if there are two points A,B in P such that \overrightarrow {AB} = \overrightarrow {(x,y)}. One can prove T is a polygon too. For example, if P is a regular triangle then T is a regular hexagon. At the picture below P is drawn in black and some P(x,y) which contain the origin are drawn in colored:
<image>
The spaceship has the best aerodynamic performance if P and T are similar. Your task is to check whether the polygons P and T are [similar](https://tinyurl.com/vp5m7vl).
Input
The first line of input will contain a single integer n (3 ≤ n ≤ 10^5) — the number of points.
The i-th of the next n lines contains two integers x_i, y_i (|x_i|, |y_i| ≤ 10^9), denoting the coordinates of the i-th vertex.
It is guaranteed that these points are listed in counterclockwise order and these points form a strictly convex polygon.
Output
Output "YES" in a separate line, if P and T are similar. Otherwise, output "NO" in a separate line. You can print each letter in any case (upper or lower).
Examples
Input
4
1 0
4 1
3 4
0 3
Output
YES
Input
3
100 86
50 0
150 0
Output
nO
Input
8
0 0
1 0
2 1
3 3
4 6
3 6
2 5
1 3
Output
YES
Note
The following image shows the first sample: both P and T are squares. The second sample was shown in the statements.
<image>
Submitted Solution:
```
# from math import factorial as fac
from collections import defaultdict
# from copy import deepcopy
import sys, math
f = None
try:
f = open('q1.input', 'r')
except IOError:
f = sys.stdin
if 'xrange' in dir(__builtins__):
range = xrange
# print(f.readline())
sys.setrecursionlimit(10**2)
def print_case_iterable(case_num, iterable):
print("Case #{}: {}".format(case_num," ".join(map(str,iterable))))
def print_case_number(case_num, iterable):
print("Case #{}: {}".format(case_num,iterable))
def print_iterable(A):
print (' '.join(A))
def read_int():
return int(f.readline().strip())
def read_int_array():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def read_string():
return list(f.readline().strip())
def ri():
return int(f.readline().strip())
def ria():
return [int(x) for x in f.readline().strip().split(" ")]
def rns():
a = [x for x in f.readline().split(" ")]
return int(a[0]), a[1].strip()
def rs():
return list(f.readline().strip())
def bi(x):
return bin(x)[2:]
from collections import deque
import math
NUMBER = 10**9 + 7
# NUMBER = 998244353
def factorial(n) :
M = NUMBER
f = 1
for i in range(1, n + 1):
f = (f * i) % M # Now f never can
# exceed 10^9+7
return f
def mult(a,b):
return (a * b) % NUMBER
def minus(a , b):
return (a - b) % NUMBER
def plus(a , b):
return (a + b) % NUMBER
def egcd(a, b):
if a == 0:
return (b, 0, 1)
else:
g, y, x = egcd(b % a, a)
return (g, x - (b // a) * y, y)
def modinv(a):
m = NUMBER
g, x, y = egcd(a, m)
if g != 1:
raise Exception('modular inverse does not exist')
else:
return x % m
def choose(n,k):
if n < k:
assert false
return mult(factorial(n), modinv(mult(factorial(k),factorial(n-k)))) % NUMBER
from collections import deque, defaultdict
import heapq
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
def dfs(g, timeIn, timeOut,depths,parents):
# assign In-time to node u
cnt = 0
# node, neig_i, parent, depth
stack = [[1,0,0,0]]
while stack:
v,neig_i,parent,depth = stack[-1]
parents[v] = parent
depths[v] = depth
# print (v)
if neig_i == 0:
timeIn[v] = cnt
cnt+=1
while neig_i < len(g[v]):
u = g[v][neig_i]
if u == parent:
neig_i+=1
continue
stack[-1][1] = neig_i + 1
stack.append([u,0,v,depth+1])
break
if neig_i == len(g[v]):
stack.pop()
timeOut[v] = cnt
cnt += 1
# def isAncestor(u: int, v: int, timeIn: list, timeOut: list) -> str:
# return timeIn[u] <= timeIn[v] and timeOut[v] <= timeOut[u]
cnt = 0
@bootstrap
def dfs(v,adj,timeIn, timeOut,depths,parents,parent=0,depth=0):
global cnt
parents[v] = parent
depths[v] = depth
timeIn[v] = cnt
cnt+=1
for u in adj[v]:
if u == parent:
continue
yield dfs(u,adj,timeIn,timeOut,depths,parents,v,depth+1)
timeOut[v] = cnt
cnt+=1
yield
def gcd(a,b):
if a == 0:
return b
return gcd(b % a, a)
# Function to return LCM of two numbers
def lcm(a,b):
return (a*b) / gcd(a,b)
def get_num_2_5(n):
twos = 0
fives = 0
while n>0 and n%2 == 0:
n//=2
twos+=1
while n>0 and n%5 == 0:
n//=5
fives+=1
return (twos,fives)
def shift(a,i,num):
for _ in range(num):
a[i],a[i+1],a[i+2] = a[i+2],a[i],a[i+1]
def equal(x,y):
return abs(x-y) <= 1e-9
# def leq(x,y):
# return x-y <= 1e-9
def solution(a,n):
if n > 3:
return 'yes'
return 'no'
def main():
T = 1
# T = ri()
for i in range(T):
n = ri()
# s = rs()
data = []
for j in range(n):
data.append(ria())
# n,m = ria()
# a = ria()
# b = ria()
x = solution(data,n)
# continue
if 'xrange' not in dir(__builtins__):
print(x)
else:
print >>output,str(x)# "Case #"+str(i+1)+':',
if 'xrange' in dir(__builtins__):
print(output.getvalue())
output.close()
if 'xrange' in dir(__builtins__):
import cStringIO
output = cStringIO.StringIO()
#example usage:
# for l in res:
# print >>output, str(len(l)) + ' ' + ' '.join(l)
if __name__ == '__main__':
main()
```
No
| 12,599 |
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