message stringlengths 2 28.7k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 21 109k | cluster float64 7 7 | __index_level_0__ int64 42 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
n, q = map(int, input().split())
dl = [0]*(n+1)
cl = []
for i in range(q):
a, b = map(int, input().split())
for j in range(a, b+1):
dl[j]+=1
cl.append((a, b))
sm1 = 0
sm2 = 0
m1 = [0]
m2 = [0]
for i in range(1, n+1):
if dl[i]<=2:
sm2+=1
if dl[i]<=1:
sm1+=1
m1.append(sm1)
m2.append(sm2)
def integ(a1, b1, a2, b2):
if b1<a2 or b2<a1:
x = m1[b1]-m1[a1-1]+m1[b2]-m1[a2-1]
return x
else:
if a1<a2:
if b2>=b1:
x = m2[b1]-m2[a2-1]+m1[a2]-m1[a1-1]+m1[b2]-m1[b1-1]
else:
x = m1[a2] - m1[a1-1] + m2[b2] - m2[a2-1] + m1[b1] - m1[b2-1]
return x
else:
if b2<=b1:
x = m1[a1]-m1[a2-1]+m2[b2]-m2[a1-1]+m1[b1]-m1[b2-1]
else:
x = m1[a1] - m1[a2-1] + m2[b1] - m2[a1-1] + m1[b2] - m1[b1-1]
return x
mn = 10e5
for i in range(q):
for j in range(i+1, q):
m = integ(cl[i][0], cl[i][1], cl[j][0], cl[j][1])
if m<mn:
mn = m
print(n-mn)
``` | instruction | 0 | 55,662 | 7 | 111,324 |
No | output | 1 | 55,662 | 7 | 111,325 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
import sys
from functools import cmp_to_key
def cmp(a,b):
if (a[1]-a[0])>(b[1]-b[0]):
return 1
elif (a[1]-a[0])<(b[1]-b[0]):
return -1
else:
if a[1]>b[1]:
return 1
elif a[1]<b[1]:
return -1
return 0
n,t = map(int,sys.stdin.readline().split())
arr = []
for i in range(t):
arr.append(list(map(int,sys.stdin.readline().split())))
arr.sort(key=cmp_to_key(cmp),reverse=True)
res = []
res1 = [1 for _ in range(5001)]
for i in arr[:len(arr)-2]:
res1[i[0]:i[1]+1] = [0]*(i[1]+1-i[0])
print(res1.count(0))
``` | instruction | 0 | 55,663 | 7 | 111,326 |
No | output | 1 | 55,663 | 7 | 111,327 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
from itertools import accumulate
n,m = map(int,input().split())
p = [list(map(int,input().split())) for i in range(m)]
ans = 0
for i in range(m):
imos = [0]*(n+2)
anstmp = n
for j in range(i+1,m):
l,r = p[j]
imos[l] += 1
imos[r+1] -= 1
acc = list(accumulate(imos))
anstmp -= acc.count(0)-2
acc1 = [1 if acc[i] == 1 else 0 for i in range(n+2)]
aacc = list(accumulate(acc1))
for j in range(i+1,m):
l,r = p[j]
ans = max(ans,anstmp-aacc[r]+aacc[l-1])
print(ans)
``` | instruction | 0 | 55,664 | 7 | 111,328 |
No | output | 1 | 55,664 | 7 | 111,329 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You have a long fence which consists of n sections. Unfortunately, it is not painted, so you decided to hire q painters to paint it. i-th painter will paint all sections x such that l_i ≤ x ≤ r_i.
Unfortunately, you are on a tight budget, so you may hire only q - 2 painters. Obviously, only painters you hire will do their work.
You want to maximize the number of painted sections if you choose q - 2 painters optimally. A section is considered painted if at least one painter paints it.
Input
The first line contains two integers n and q (3 ≤ n, q ≤ 5000) — the number of sections and the number of painters availible for hire, respectively.
Then q lines follow, each describing one of the painters: i-th line contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n).
Output
Print one integer — maximum number of painted sections if you hire q - 2 painters.
Examples
Input
7 5
1 4
4 5
5 6
6 7
3 5
Output
7
Input
4 3
1 1
2 2
3 4
Output
2
Input
4 4
1 1
2 2
2 3
3 4
Output
3
Submitted Solution:
```
# from sys import stdin
# def rl():
# return [int(w) for w in stdin.readline().split()]
from collections import defaultdict
# from math import sqrt,factorial,gcd,log2,inf,ceil
# map(int,input().split())
# l = list(map(int,input().split()))
n,q=map(int,input().split())
a=[list(map(int,input().split())) for _ in range(q)]
c=[0]*5005
for i in range(q):
for j in range(a[i][0],a[i][1]+1):
c[j]+=1
ans=0
for i in range(q):
temp = 0
d = c[:]
for j in range(a[i][0],a[i][1] - 1):
d[j]-=1
for j in range(5005):
if d[j]>0:
temp+=1
b = [0]*(5005)
for j in range(1,n+1):
b[j] = b[j-1]
if d[j] == 1:
b[j]+=1
for j in range(i+1,q):
ans = max(ans,temp -b[a[j][1]]+b[a[j][0]-1])
print(ans)
``` | instruction | 0 | 55,665 | 7 | 111,330 |
No | output | 1 | 55,665 | 7 | 111,331 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,087 | 7 | 112,174 |
Tags: brute force, implementation, number theory
Correct Solution:
```
s=input()+' '
d,ans,DD={},0,{"R":0,"B":0,"Y":0,"G":0}
for i in range(len(s)):
if s[i] in "RGBY": d[i%4]=s[i]
for i in range(len(s)):
if s[i]=="!": DD[d[i%4]]+=1
print(DD["R"],DD["B"],DD["Y"],DD["G"])
``` | output | 1 | 56,087 | 7 | 112,175 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,088 | 7 | 112,176 |
Tags: brute force, implementation, number theory
Correct Solution:
```
inp=str(input())
res=[0,0,0,0]
rbyg=[-1,-1,-1,-1]
for i in range(len(inp)):
try:
rbyg["RBYG".index(inp[i])]=i%4
except:
res[i%4]+=1
print(res[rbyg[0]], res[rbyg[1]], res[rbyg[2]], res[rbyg[3]], sep=' ')
``` | output | 1 | 56,088 | 7 | 112,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,089 | 7 | 112,178 |
Tags: brute force, implementation, number theory
Correct Solution:
```
import sys
import os
def changeStdioToFile():
path = os.path.dirname(os.path.abspath(__file__))
sys.stdin = open(f'{path}/input.txt', 'r')
sys.stdout = open(f'{path}/output.txt', 'w')
#changeStdioToFile()
t = 1
#t = int(input())
for _ in range(t):
bulbList = input()
n = len(bulbList)
kaputBulb = {"R":0, "G":0, "B":0, "Y":0}
colors = ("R", "G", "B", "Y")
for color in colors:
pos = bulbList.find(color)
for i in range(4, max((pos - 0), (n - pos)) + 1, 4):
if pos - i >= 0 and bulbList[pos-i] == '!':
bulbList = bulbList[:pos-i] + color + bulbList[pos-i+1:]
kaputBulb[color] += 1
if pos + i < n and bulbList[pos+i] == '!':
bulbList = bulbList[:pos+i] + color + bulbList[pos+i+1:]
kaputBulb[color] += 1
print(kaputBulb["R"], kaputBulb["B"], kaputBulb["Y"], kaputBulb["G"])
``` | output | 1 | 56,089 | 7 | 112,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,090 | 7 | 112,180 |
Tags: brute force, implementation, number theory
Correct Solution:
```
arr = input()
pos = [0,1,2,3]
myd = {}
for c in pos:
cnt = 0
for l in range(c,len(arr), 4):
if(arr[l]!='!'):
k=arr[l]
else:
cnt = cnt + 1
myd[k] = cnt
print('%d %d %d %d' % (myd['R'], myd['B'], myd['Y'], myd['G']))
``` | output | 1 | 56,090 | 7 | 112,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,091 | 7 | 112,182 |
Tags: brute force, implementation, number theory
Correct Solution:
```
a=input()
b=a.index('B')%4
r=a.index('R')%4
g=a.index('G')%4
y=a.index('Y')%4
tb=0
tr=0
tg=0
ty=0
for i in range(len(a)):
if a[i]=='!':
z=i%4
if z==b:
tb+=1
if z==r:
tr+=1
if z==g:
tg+=1
if z==y:
ty+=1
print(tr,tb,ty,tg)
``` | output | 1 | 56,091 | 7 | 112,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,092 | 7 | 112,184 |
Tags: brute force, implementation, number theory
Correct Solution:
```
s=input()
n=len(s)
arr={'R':0,'B':0,'Y':0,'G':0}
for i in range(4):
j=i
count=0
while j<n:
if s[j]=='!':
count+=1
else:
chr=s[j]
j+=4
arr[chr]=count
for j in list(arr.values()):
print(j,end=" ")
``` | output | 1 | 56,092 | 7 | 112,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,093 | 7 | 112,186 |
Tags: brute force, implementation, number theory
Correct Solution:
```
s = list(input().strip())
dead = {'R':0,'B':0,'Y':0,'G':0}
for i in range(len(s)):
if s[i] == '!':
for j in range(i%4, len(s), 4):
if s[j] != '!':
s[i] = s[j]
dead[s[j]] += 1
break
print(*dead.values())
``` | output | 1 | 56,093 | 7 | 112,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements. | instruction | 0 | 56,094 | 7 | 112,188 |
Tags: brute force, implementation, number theory
Correct Solution:
```
s=list(input())
d = {'R':-1,'Y':-1,'G':-1,'B':-1}
j=0
for i in s:
if(i!='!'):
d[i] = j%4
j+=1
_d = {}
for key in d:
_d[d[key]] = key
j=0
l = [0,0,0,0]
for i in range(len(s)):
if(s[i]=='!'):
try:
key = _d[j%4]
except:
key = _d[-1]
if(key=='R'):
l[0]+=1
elif(key=='B'):
l[1]+=1
elif(key=='Y'):
l[2]+=1
elif(key=='G'):
l[3]+=1
j+=1
for i in l:
print(i,end=" ")
``` | output | 1 | 56,094 | 7 | 112,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
s = list(input())
result = {'R': 0, 'B': 0, 'Y': 0, 'G': 0}
for mod in [0, 1, 2, 3]:
w = filter(lambda x: s[x] == '!', range(mod, len(s), 4))
letter = next(filter(lambda x: s[x] != '!', range(mod, len(s), 4)))
result[s[letter]] = len(list(w))
print('{0} {1} {2} {3}'.format(result['R'], result['B'], result['Y'], result['G']))
``` | instruction | 0 | 56,095 | 7 | 112,190 |
Yes | output | 1 | 56,095 | 7 | 112,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
s=input()
l=len(s)
x={"R":0 , "B":0 , "Y":0 ,"G":0}
for i in range(4):
a=0
col=""
for j in range(i,l,4):
if(s[j]=="!"):
a+=1
else:
col=s[j]
x[col]=a
print(*x.values())
``` | instruction | 0 | 56,096 | 7 | 112,192 |
Yes | output | 1 | 56,096 | 7 | 112,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
import sys
def main():
s = sys.stdin.readline()
s = [c for c in s]
a = {}
for letter in ['R', 'B', 'Y', 'G']:
a[letter] = 0
ind = s.index(letter)
if ind >= 4:
for i in range(ind, -1, -4):
if s[i] == '!':
s[i] = letter
a[letter] += 1
if len(s) - ind > 4:
for i in range(ind, len(s), 4):
if s[i] == '!':
s[i] = letter
a[letter] += 1
print(a['R'], a['B'], a['Y'], a['G'])
if __name__ == "__main__":
main()
``` | instruction | 0 | 56,097 | 7 | 112,194 |
Yes | output | 1 | 56,097 | 7 | 112,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
s=input()
q,w,e,t=0,0,0,0
s=s+'aaaaaaaaaa'
r=(s.find('R'))%4
b=(s.find('B'))%4
y=(s.find('Y'))%4
g=(s.find('G'))%4
while(r<=len(s)-5):
if(s[r]=='!'):
q=q+1
r=r+4
while(b<=len(s)-5):
if(s[b]=='!'):
w=w+1
b=b+4
while(y<=len(s)-5):
if(s[y]=='!'):
e=e+1
y=y+4
while(g<=len(s)-5):
if(s[g]=='!'):
t=t+1
g=g+4
print(q,w,e,t)
``` | instruction | 0 | 56,098 | 7 | 112,196 |
Yes | output | 1 | 56,098 | 7 | 112,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
ch=input()
t=[]
m=[]
n=[]
k=0
ch1=''
i=0
a=(len(ch)//4)*4
while i<(a):
ch1=ch[i:i+4]
t.append(ch1)
i=i+4
ch1=''
ch2=ch[a:len(ch)]
i=0
for i in range(4-len(ch2)):
ch2=ch2+'*'
t.append(ch2)
i=0
for j in range(4):
i=0
while (t[i][j] in ['!','*']) and (i<(len(t)-1)):
i=i+1
if (not(t[i][j] in ['!','*'])) :
m.append(t[i][j])
else:
m.append('!')
s=0
ch2=0
for k in range(len(t)):
if t[k][j]=='!':
s=s+1
ch2=m[j]+str(s)
n.append(ch2)
n.sort()
b=n[0]
c=n[1]
d=n[2]
e=n[3]
n[0]=d
n[1]=b
n[2]=e
n[3]=c
ch3=''
for i in range(4):
ch3=ch3+n[i][1]+' '
print(ch3[:(len(ch3)-1)])
``` | instruction | 0 | 56,099 | 7 | 112,198 |
No | output | 1 | 56,099 | 7 | 112,199 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
#!/bin/python3
def rotate(l,n):
return l[n:] + l[:n]
s = input()
L = [ ch for ch in s]
F = [ '.' for i in range(0, len(L)) ]
sz = len(L)
seq = []
for i in range(0,sz-3):
x = i % 4
l = L[i:i+4]
#print(i, l)
l =rotate(l,4-x)
#print(l)
if( l.count('!') == 0):
seq = list(l)
break
if( l.count('!') == 1):
ix = None
for e in 'RGYB':
try:
ix = l.index(e)
except ValueError:
ix = -1
if( ix == -1):
ix = l.index('!')
l[ix] = e
break
seq = list(l)
#print('from_here:',seq)
break
#print('seq2', seq)
p = 0
for q in range(0,sz):
F[q] = seq[p]
p = (p+1)%4
#print(F)
d = dict( [ ('R', 0) , ('B' , 0) , ('G' , 0) , ('Y', 0) ] )
for i in range(0,sz):
if( L[i] == '!'):
d[ F[i] ] += 1
print( *list(d.values()) )
``` | instruction | 0 | 56,100 | 7 | 112,200 |
No | output | 1 | 56,100 | 7 | 112,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
input_string = input()
my_array= []
r = 0
b = 0
y = 0
g = 0
flag = dict(r=False, b=False, y=False, g=False)
counter = dict(r=0, b=0, y=0, g=0)
for i in range(len(input_string)):
my_array.append(input_string[i])
k = 0
for i in range(len(input_string)):
if (my_array[i] == 'R') & (flag['r']==False):
while i >= 4:
i -=4
r = i
k+=1
flag['r'] = True
if (my_array[i] == 'B') & (flag['b']==False):
while i >= 4:
i -=4
b = i
k+=1
flag['b'] = True
if (my_array[i] == 'Y') & (flag['y']==False):
while i >= 4:
i -=4
y = i
k+=1
flag['y'] = True
if (my_array[i] == 'G') & (flag['g']==False):
while i >= 4:
i -=4
g = i
k+=1
flag['g'] = True
if k==4:
break
print (r)
for i in range(len(my_array)):
if my_array[i]=="!":
if i % 4 == r:
counter['r']+=1
if i % 4 == b:
counter['b']+=1
if i % 4 == y:
counter['y']+=1
if i % 4 == g:
counter['g']+=1
print (counter['r'], ' ', counter['b'], ' ', counter['y'],' ',counter['g'])
``` | instruction | 0 | 56,101 | 7 | 112,202 |
No | output | 1 | 56,101 | 7 | 112,203 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nothing is eternal in the world, Kostya understood it on the 7-th of January when he saw partially dead four-color garland.
Now he has a goal to replace dead light bulbs, however he doesn't know how many light bulbs for each color are required. It is guaranteed that for each of four colors at least one light is working.
It is known that the garland contains light bulbs of four colors: red, blue, yellow and green. The garland is made as follows: if you take any four consecutive light bulbs then there will not be light bulbs with the same color among them. For example, the garland can look like "RYBGRYBGRY", "YBGRYBGRYBG", "BGRYB", but can not look like "BGRYG", "YBGRYBYGR" or "BGYBGY". Letters denote colors: 'R' — red, 'B' — blue, 'Y' — yellow, 'G' — green.
Using the information that for each color at least one light bulb still works count the number of dead light bulbs of each four colors.
Input
The first and the only line contains the string s (4 ≤ |s| ≤ 100), which describes the garland, the i-th symbol of which describes the color of the i-th light bulb in the order from the beginning of garland:
* 'R' — the light bulb is red,
* 'B' — the light bulb is blue,
* 'Y' — the light bulb is yellow,
* 'G' — the light bulb is green,
* '!' — the light bulb is dead.
The string s can not contain other symbols except those five which were described.
It is guaranteed that in the given string at least once there is each of four letters 'R', 'B', 'Y' and 'G'.
It is guaranteed that the string s is correct garland with some blown light bulbs, it means that for example the line "GRBY!!!B" can not be in the input data.
Output
In the only line print four integers kr, kb, ky, kg — the number of dead light bulbs of red, blue, yellow and green colors accordingly.
Examples
Input
RYBGRYBGR
Output
0 0 0 0
Input
!RGYB
Output
0 1 0 0
Input
!!!!YGRB
Output
1 1 1 1
Input
!GB!RG!Y!
Output
2 1 1 0
Note
In the first example there are no dead light bulbs.
In the second example it is obvious that one blue bulb is blown, because it could not be light bulbs of other colors on its place according to the statements.
Submitted Solution:
```
from math import ceil
string=input()
l=int(4)
r=int(0)
b=int(0)
y=int(0)
g=int(0)
for x in range(ceil(len(string)/4)):
k=string[(l-4):l]
if('!' in k):
if(len(k)!=4):
for x in k:
if(x=='!'):
k=string[l-8]
if 'R' in k:
r+=1
if 'B' in k:
b+=1
if 'Y' in k:
y+=1
if 'G' in k:
g+=1
break
else:
if 'R' not in k:
r+=1
if 'B' not in k:
b+=1
if 'Y' not in k:
y+=1
if 'G' not in k:
g+=1
l+=4
print(r,b,y,g)
``` | instruction | 0 | 56,102 | 7 | 112,204 |
No | output | 1 | 56,102 | 7 | 112,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,540 | 7 | 113,080 |
Tags: math
Correct Solution:
```
import math
from collections import deque
import sys
sys.setrecursionlimit(10**4)
def Divisors(n) :
l=[]
i = 2
while i <= math.sqrt(n):
if (n % i == 0) :
if (n // i == i) :
l.append(i)
else :
l.append(i)
l.append(n//i)
i = i + 1
return l
def SieveOfEratosthenes(n):
l=[]
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
for p in range(2, n+1):
if prime[p]:
l.append(p)
return l
def primeFactors(n):
l=[]
while n % 2 == 0:
l.append(2)
n = n / 2
for i in range(3,int(math.sqrt(n))+1,2):
while n % i== 0:
l.append(i)
n = n / i
if n > 2:
l.append(n)
return(l)
def Factors(n) :
result = []
for i in range(2,(int)(math.sqrt(n))+1) :
if (n % i == 0) :
if (i == (n/i)) :
result.append(i)
else :
result.append(i)
result.append(n//i)
result.append(1)
result.append(n)
return result
def maxSubArraySum(a):
max_so_far = 0
max_ending_here = 0
size=len(a)
for i in range(0, size):
max_ending_here = max_ending_here + a[i]
if (max_so_far < abs(max_ending_here)):
max_so_far = max_ending_here
return max_so_far
def longestsubarray(arr, n, k):
current_count = 0
# this will contain length of
# longest subarray found
max_count = 0
for i in range(0, n, 1):
if (arr[i] % k != 0):
current_count += 1
else:
current_count = 0
max_count = max(current_count,
max_count)
return max_count
#print(SieveOfEratosthenes(100))
#print(Divisors(100))
#print(primeFactors(100))
#print(Factors(100))
#print(maxSubArraySum(a))
def main():
#n=int(input())
n,k=list(map(int,input().split()))
if n>k:
print((k*(k+1))//2)
else:
print(n*(n-1)//2 + 1)
t=int(input())
for i in range(0,t):
main()
``` | output | 1 | 56,540 | 7 | 113,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,541 | 7 | 113,082 |
Tags: math
Correct Solution:
```
import collections as cc
import math as mt
I=lambda:list(map(int,input().split()))
for tc in range(int(input())):
n,k=I()
d=0
if k>=n:
d=abs(n-k)
n-=1
print(n*(n+1)//2+1)
else:
print(k*(k+1)//2)
``` | output | 1 | 56,541 | 7 | 113,083 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,542 | 7 | 113,084 |
Tags: math
Correct Solution:
```
for i in range(int(input())):
n,r = map(int, input().split())
ct = 0
if r >= n:
ct = ((n-1)*(n)//2) +1
print(int(ct))
elif r < n:
ct = ((r)*(r+1)//2)
print(int(ct))
``` | output | 1 | 56,542 | 7 | 113,085 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,543 | 7 | 113,086 |
Tags: math
Correct Solution:
```
t = int(input())
while t!=0:
t-=1
n ,r= map(int, input().split())
if n > r:
print((r*(r+1))//2)
else:
print((n*(n-1))//2 + 1)
``` | output | 1 | 56,543 | 7 | 113,087 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,544 | 7 | 113,088 |
Tags: math
Correct Solution:
```
t=int(input())
for i in range(t):
n,r=map(int,input().split())
v=min(r,n-1)
s=(v*(v+1))//2
if r>=n:
s+=1
print(s)
``` | output | 1 | 56,544 | 7 | 113,089 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,545 | 7 | 113,090 |
Tags: math
Correct Solution:
```
for _ in range(int(input())):
n, r = map(int, input().split())
if n > r:
print((1+r)*r//2)
else:
ans = (1+n-1)*(n-1)//2 + 1
print(ans)
``` | output | 1 | 56,545 | 7 | 113,091 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,546 | 7 | 113,092 |
Tags: math
Correct Solution:
```
# cook your dish here
from math import *
t=int(input())
while t:
t=t-1
n,r=map(int,input().split())
out=0
if r<n:
out+=(r*(r+1))//2
else:
n=n-1
out+=(n*(n+1))//2
out+=1
print(out)
``` | output | 1 | 56,546 | 7 | 113,093 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format. | instruction | 0 | 56,547 | 7 | 113,094 |
Tags: math
Correct Solution:
```
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")
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10**9+7
Ri = lambda : [int(x) for x in sys.stdin.readline().split()]
ri = lambda : sys.stdin.readline().strip()
for _ in range(int(ri())):
n,r = Ri()
ans = 0
if r < n:
# r-=1
print(r*(r+1)//2)
elif n < r:
n-=1
print((n*(n+1))//2 +1)
else:
n-=1
print((n*(n+1))//2 + 1)
``` | output | 1 | 56,547 | 7 | 113,095 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
for _ in range(int(input())):
n,r = map(int,input().split())
if n>r:
print((r*(r+1))//2)
else:
n-=1
ans = (n*(n+1))//2
print(ans+1)
``` | instruction | 0 | 56,548 | 7 | 113,096 |
Yes | output | 1 | 56,548 | 7 | 113,097 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
a=int(input())
for i in range(a):
n,r=map(int,input().split())
if(r<n):
print((r*(r+1)//2))
else:
m=r-n+1
x=n-1
print((x*(x+1)//2)+1)
``` | instruction | 0 | 56,549 | 7 | 113,098 |
Yes | output | 1 | 56,549 | 7 | 113,099 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
t=int(input())
for i in range(t):
n,r=map(int,input().split())
if n<=r:
print(((n-1)*n)//2+1)
else:
print(((r+1)*r)//2)
``` | instruction | 0 | 56,550 | 7 | 113,100 |
Yes | output | 1 | 56,550 | 7 | 113,101 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
t=int(input())
for i in range(t):
n,r=map(int,input().split())
if(n>r):
re=(r*(r+1))//2
print(re)
else:
re=((n*(n+1))//2)-n+1
print(re)
``` | instruction | 0 | 56,551 | 7 | 113,102 |
Yes | output | 1 | 56,551 | 7 | 113,103 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
tests = int(input())
for _ in range(tests):
a, b = map(int, input().split())
if b == 1:
print(1)
continue
t = min(a, b)
print(t * (t - 1) // 2 + 1 + max(0, a - b))
``` | instruction | 0 | 56,552 | 7 | 113,104 |
No | output | 1 | 56,552 | 7 | 113,105 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
def solve():
[n, r] = [int(x) for x in input().split()]
ans = 1
small = min(n, r)
# for i in range(2, small+1):
# print(ans)
# x = n // i
# y = (x+1)
# z = (y*i) % n
# if z == 0:
# ans += 1
# else:
# ans += z + 2
# print(ans)
if r % 2 == 0:
if r > n:
print( (((small * (small+1) )// 2)) - n +1)
else:
print(((small * (small+1) )// 2))
else:
if r > n:
print( (((small * (small+1) )// 2 )-n) + 1)
else:
print((((small * (small+1) )// 2 )) )
'''
for i in range(2, small + 1):
# print(ans)
x = n // i
y = (x+1)
if (y*i)%n == 0:
ans += 1
else:
ans += (y*i)%n + 1
print(ans)
if (r*i) == 0:
ans += 1
break
else:
ans += n%i
print(ans)
'''
def main():
n = int(input())
while n > 0:
solve()
n -= 1
main()
``` | instruction | 0 | 56,553 | 7 | 113,106 |
No | output | 1 | 56,553 | 7 | 113,107 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
for _ in range(int(input())):
n,r=map(int,input().split())
if(r>=n):
s=int((n*(n-1))/2)
print(s+1)
else:
s=int((r*(r+1))/2)
print(s)
``` | instruction | 0 | 56,554 | 7 | 113,108 |
No | output | 1 | 56,554 | 7 | 113,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A competitive eater, Alice is scheduling some practices for an eating contest on a magical calendar. The calendar is unusual because a week contains not necessarily 7 days!
In detail, she can choose any integer k which satisfies 1 ≤ k ≤ r, and set k days as the number of days in a week.
Alice is going to paint some n consecutive days on this calendar. On this calendar, dates are written from the left cell to the right cell in a week. If a date reaches the last day of a week, the next day's cell is the leftmost cell in the next (under) row.
She wants to make all of the painted cells to be connected by side. It means, that for any two painted cells there should exist at least one sequence of painted cells, started in one of these cells, and ended in another, such that any two consecutive cells in this sequence are connected by side.
Alice is considering the shape of the painted cells. Two shapes are the same if there exists a way to make them exactly overlapped using only parallel moves, parallel to the calendar's sides.
For example, in the picture, a week has 4 days and Alice paints 5 consecutive days. [1] and [2] are different shapes, but [1] and [3] are equal shapes.
<image>
Alice wants to know how many possible shapes exists if she set how many days a week has and choose consecutive n days and paints them in calendar started in one of the days of the week. As was said before, she considers only shapes, there all cells are connected by side.
Input
The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Next t lines contain descriptions of test cases.
For each test case, the only line contains two integers n, r (1 ≤ n ≤ 10^9, 1 ≤ r ≤ 10^9).
Output
For each test case, print a single integer — the answer to the problem.
Please note, that the answer for some test cases won't fit into 32-bit integer type, so you should use at least 64-bit integer type in your programming language.
Example
Input
5
3 4
3 2
3 1
13 7
1010000 9999999
Output
4
3
1
28
510049495001
Note
In the first test case, Alice can set 1,2,3 or 4 days as the number of days in a week.
There are 6 possible paintings shown in the picture, but there are only 4 different shapes. So, the answer is 4. Notice that the last example in the picture is an invalid painting because all cells are not connected by sides.
<image>
In the last test case, be careful with the overflow issue, described in the output format.
Submitted Solution:
```
T = int(input())
def solve(n,r):
if n<=r:
if (n-1)%2 == 0:
a = int((n-1)/2)
ret = int(a*n)+1
else:
a = int((n)/2)
ret = int(a*(n-1))+1
elif n>r:
if (r+1)%2 == 0:
a = int((r+1)/2)
ret = int(a*r)
else:
a = int((r)/2)
print(a)
ret = int(a*(r+1))
return ret
def solve2(n,r):
ret = 1
for k in range(2,r+1):
if n>k:
ret+=(k)
else:
ret+=1
break
return ret
while T>0:
n,r = [int(el) for el in input().split(' ')]
print(solve(n,r))
T-=1
``` | instruction | 0 | 56,555 | 7 | 113,110 |
No | output | 1 | 56,555 | 7 | 113,111 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,056 | 7 | 116,112 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n=int(input())
a=list(map(int,input().split()))
from collections import defaultdict as dc
d=dc(int)
for i in a:
d[i]+=1
ch=0
for i in d:
if d[i]%2!=0:ch+=1
if ch>1:print('NO');exit()
l=len(d)
ch=0
a1=(n//2)**2
if n%2==0:
if l>a1:ch=1
else:
if l>a1+(n**2-a1*4-1)//2+1:ch=1
if ch==1:
print('NO');exit()
ans=[[0 for j in range(n)] for i in range(n)]
sa=[[i,d[i]] for i in d]
sa.sort(reverse=True,key=lambda x:x[1])
ch1=0
try:
for i in range(n//2):
for j in range(n//2):
while sa[ch1][1]<4:ch1+=1
ans[i][j]=sa[ch1][0]
ans[n-i-1][j]=sa[ch1][0]
ans[i][n-j-1]=sa[ch1][0]
ans[n-i-1][n-j-1]=sa[ch1][0]
sa[ch1][1]-=4
if n%2!=0:
ss=[]
for i,j in sa:
if j!=0:
ss.append([i,j])
ss.sort(reverse=True,key=lambda x:x[1])
ch1=0
for i in range(n//2):
while ss[ch1][1]<2:ch1+=1
ans[n//2][i]=ss[ch1][0]
ans[n//2][n-i-1]=ss[ch1][0]
ss[ch1][1]-=2
for i in range(n//2):
while ss[ch1][1]<2:ch1+=1
ans[i][n//2]=ss[ch1][0]
ans[n-i-1][n//2]=ss[ch1][0]
ss[ch1][1]-=2
for i,j in ss:
if j!=0:ans[n//2][n//2]=i;break
except:
print('NO');exit()
print('YES')
for i in ans:print(*i)
``` | output | 1 | 58,056 | 7 | 116,113 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,057 | 7 | 116,114 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def print_mat(m):
n = len(m)
for i in range(n):
print(' '.join([str(j) for j in m[i]]))
def mat_unity4(b):
global n
b1, b2, b3, b4 = b, tr_mat(b, 2), tr_mat(b, 3), tr_mat(b, 4)
r = [['' for i in range(n)] for j in range(n)]
for i in range(n // 2):
for j in range(n):
if j < n // 2:
r[i][j] = b1[i][j]
else:
r[i][j] = b2[i][j % (n // 2)]
for i in range(n // 2):
for j in range(n):
if j < n // 2:
r[i + n // 2][j] = b4[i][j]
else:
r[i + n // 2][j] = b3[i][j % (n // 2)]
return r
def mat_unity9(b, m, c):
global n
b1, b2, b3, b4 = b, tr_mat(b, 2), tr_mat(b, 3), tr_mat(b, 4)
b5, b6, b7, b8 = m[0], m[1], m[0][::-1], m[1][::-1]
b9 = c
r = [['' for i in range(n)] for j in range(n)]
for i in range(n // 2):
for j in range(n):
if j < n // 2:
r[i][j] = b1[i][j]
elif j == n // 2:
r[i][j] = b5[i]
else:
r[i][j] = b2[i][j % (n // 2 + 1)]
for j in range(n):
if j < n // 2:
r[n // 2][j] = b8[j]
elif j == n // 2:
r[n // 2][j] = b9
else:
r[n // 2][j] = b6[j % (n // 2 + 1)]
for i in range(n // 2):
for j in range(n):
if j < n // 2:
r[i + n // 2 + 1][j] = b4[i][j]
elif j == n // 2:
r[i + n // 2 + 1][j] = b7[i]
else:
r[i + n // 2 + 1][j] = b3[i][j % (n // 2 + 1)]
return r
def tr_mat(m, t):
n = len(m)
if n == 1:
return m
r = [['' for i in range(n)] for j in range(n)]
if t == 2:
for i in range(n):
for j in range(n):
r[i][j] = m[i][n - j - 1]
elif t == 4:
for j in range(n):
for i in range(n):
r[i][j] = m[n - i - 1][j]
else:
r = tr_mat(tr_mat(m, 2), 4)
return r
def test(m):
for i in range(2, 5):
if m != tr_mat(m, i):
return False
return True
from collections import Counter
n = int(input())
a = [int(i) for i in input().split()]
d = Counter(a)
if n % 2 == 0:
i = 0
c = list(d.keys())
while i < len(d) and d[c[i]] % 4 == 0:
i += 1
else:
if i < len(d):
print('NO')
else:
print('YES')
b = [['' for i in range(n // 2)] for j in range(n // 2)]
sc = 0
for i in c:
d[i] //= 4
for j in range(d[i]):
b[sc // (n // 2)][sc % (n // 2)] = i
sc += 1
print_mat(mat_unity4(b))
#print(test(mat_unity4(b)))
else:
c = list(d.keys())
k1, k2, k4 = [], [], []
b = 0
for i in range(len(c)):
it = d[c[i]]
if it % 4 == 1:
k1.append(c[i])
elif it % 4 == 2:
k2.append(c[i])
elif it % 4 == 3:
k2.append(c[i])
k1.append(c[i])
for j in range(it // 4):
k4.append(c[i])
s = len(k4) - (n // 2) ** 2
for i in range(s):
k2.append(k4[0])
k2.append(k4[0])
del k4[0]
if len(k1) != 1 or len(k2) != 2 * (n // 2) or len(k4) != (n // 2) ** 2:
print('NO')
else:
print('YES')
b1 = [['' for i in range(n // 2)] for j in range(n // 2)]
b2 = [['' for i in range(n // 2)] for j in range(2)]
b3 = k1[0]
sc = 0
for i in k4:
b1[sc // (n // 2)][sc % (n // 2)] = i
sc += 1
sc = 0
for i in k2:
b2[sc // (n // 2)][sc % (n // 2)] = i
sc += 1
print_mat(mat_unity9(b1, b2, b3))
#print(test(mat_unity9(b1, b2, b3)))
``` | output | 1 | 58,057 | 7 | 116,115 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,058 | 7 | 116,116 |
Tags: constructive algorithms, implementation
Correct Solution:
```
def fastio():
import sys
from io import StringIO
from atexit import register
global input
sys.stdin = StringIO(sys.stdin.read())
input = lambda : sys.stdin.readline().rstrip('\r\n')
sys.stdout = StringIO()
register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
from collections import Counter as C
n, = I()
l = I()
l.sort()
c = C(l)
if n == 1:
print('YES')
print(l[0])
exit()
if n%2 == 0:
for i in c:
if c[i]%4:
print('NO')
exit()
print('YES')
ans = []
i = 0
while i < n**2:
ans.append(l[i])
i += 4
res = []
# print
for i in range(n//2):
res.append(ans[i*n//2:i*n//2 + n//2])
for i in res:
print(*i, *reversed(i))
for i in reversed(res):
print(*i, *reversed(i))
if n%2:
temp = l[:]
l = []
k = []
s = []
n -= 1
for i in c:
while c[i] >= 4 and len(l) < n**2:
for x in range(4):
l.append(i)
c[i] -= 4
while c[i] >= 2:
for x in range(2):
s.append(i)
c[i] -= 2
while c[i] >= 1:
k.append(i)
c[i] = 0
if len((k)) != 1 or len(s) > 2*n or len(l) < n**2:
print('NO')
exit()
print('YES')
l.sort()
ans = []
i = 0
while i < n**2:
ans.append(l[i])
i += 4
res = []
aiyo = []
i = 0
while i < len(s):
aiyo.append(s[i])
i += 2
for i in range(n//2):
res.append(ans[i*n//2:i*n//2 + n//2])
j = 0
zz = []
el = aiyo[j]
for i in res:
print(*i, el, *reversed(i))
zz.append(el)
j += 1
el = aiyo[j]
kkk = aiyo[j:]
print(*kkk, k[0], *reversed(kkk))
j = 0
zz = list(reversed(zz))
el = zz[j]
for i in reversed(res):
print(*i, el, *reversed(i))
j += 1
if j < len(zz):
el = zz[j]
``` | output | 1 | 58,058 | 7 | 116,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,059 | 7 | 116,118 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from collections import defaultdict, deque
import sys
def input():
return sys.stdin.readline().rstrip()
def getcnt(LIST):
d = defaultdict(int)
for v in LIST:
d[v] += 1
return d
def odd_only_1(dic):
cnt = 0
res_value = 0
for key, value in dic.items():
if value % 2 != 0:
cnt += 1
res_value = key
if cnt == 1:
return res_value
return 0
def slv():
n = int(input())
a = list(map(int, input().split()))
d = getcnt(a)
if n % 2 == 0:
#even case
if all(value % 4 == 0 for value in d.values()):
stack = deque([])
for key, value in d.items():
for _ in range(value):
stack.append(key)
print("YES")
ans = [[-1]*(n) for i in range(n)]
for i in range(n//2):
for j in range(n//2):
nums = stack[0]
ans[i][j] = ans[i][-(j + 1)] = ans[-(i + 1)
][j] = ans[-(i + 1)][-(j + 1)] = nums
for _ in range(4):
stack.popleft()
for i in range(n):
print(*ans[i])
return
else:
print("NO")
return
else:
if not odd_only_1(d):
print("NO")
return
val = odd_only_1(d)
d[val] -= 1
ans = [[0]*(n) for i in range(n)]
ans[n//2][n//2] = val
if sum(value//4 for value in d.values()) < (n//2) ** 2:
print("NO")
return
print("YES")
stack = deque([])
for key, value in d.items():
u, v = divmod(value, 4)
for _ in range(4 * u):
stack.appendleft(key)
for _ in range(v):
stack.append(key)
for i in range(n//2):
for j in range(n//2):
nums = stack[0]
ans[i][j] = ans[i][-(j + 1)] = ans[-(i + 1)
][j] = ans[-(i + 1)][-(j + 1)] = nums
for _ in range(4):
stack.popleft()
for i in range(n//2):
num = stack[0]
ans[n//2][i] = ans[n//2][-(i + 1)] = num
for _ in range(2):
stack.popleft()
for i in range(n//2):
num = stack[0]
ans[i][n//2] = ans[-(i + 1)][n//2] = num
for _ in range(2):
stack.popleft()
for i in range(n):
print(*ans[i])
return
def main():
t = 1
for i in range(t):
slv()
return
if __name__ == "__main__":
main()
``` | output | 1 | 58,059 | 7 | 116,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,060 | 7 | 116,120 |
Tags: constructive algorithms, implementation
Correct Solution:
```
n=int(input())
l=list(map(int,input().split()))
d=dict()
for j in range(n*n):
if l[j] not in d:
d.update({l[j]:1})
else:
d[l[j]]+=1
four=0
t=0
f=0
#print(d)
for j in d:
if d[j]%2==1:
if n%2==1 and t==0:
t=1
else:
f=1
break
if d[j]>=4:
four+=d[j]//4
if f==1 or four<(n//2)**2:
print("NO")
else:
print("YES")
a=[[0 for j in range(n)]for i in range(n)]
for i in range(n//2):
for j in range(n//2):
for k in d:
if d[k]>=4:
a[i][j]=k
a[n-1-i][j]=k
a[i][n-1-j]=k
a[n-1-i][n-1-j]=k
d[k]-=4
if d[k]==0:
del d[k]
break
#print(d)
if n%2==1:
for i in range(n//2):
for k in d:
if d[k]>=2:
#print(k)
a[i][n//2]=k
a[n-1-i][n//2]=k
d[k]-=2
if d[k]==0:
del d[k]
break
for i in range(n//2):
for k in d:
if d[k]>=2:
a[n//2][i]=k
a[n//2][n-1-i]=k
d[k]-=2
if d[k]==0:
del d[k]
break
for k in d:
if d[k]==1:
a[n//2][n//2]=k
for i in range(n):
print(*a[i],sep=" ")
``` | output | 1 | 58,060 | 7 | 116,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,061 | 7 | 116,122 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt, sin, cos, ceil, floor
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(1000000000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]
def inside(y, x, H, W):
return 0 <= y < H and 0 <= x < W
def ceil(a, b):
return (a + b - 1) // b
# aとbの最大公約数
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
# aとbの最小公倍数
def lcm(a, b):
g = gcd(a, b)
return a / g * b
def euclidean_distance(y1, x1, y2, x2):
return sqrt((y1 - y2) ** 2 + (x1 - x2) ** 2)
def manhattan_distance(y1, x1, y2, x2):
return abs(y1 - y2) + abs(x1 - x2)
def main():
N = int(input())
D = list(map(int, input().split()))
f, t, o = [], [], []
C = Counter(D)
for k, v in C.items():
while v >= 4:
v -= 4
f.append(k)
while v >= 2:
v -= 2
t.append(k)
if v:
o.append(k)
ans = [[0] * N for _ in range(N)]
if N % 2 == 0:
if len(t) > 0 or len(o) > 0:
print(NO)
return
i = 0
for y in range(N // 2):
for x in range(N // 2):
ans[y][x] = ans[-y - 1][x] = ans[y][-x - 1] = ans[-y - 1][-x - 1] = f[i]
i += 1
else:
if len(f) < ((N // 2) * (N // 2)) or len(o) != 1:
print(NO)
return
i = 0
for y in range(N // 2):
for x in range(N // 2):
ans[y][x] = ans[-y - 1][x] = ans[y][-x - 1] = ans[-y - 1][-x - 1] = f[i]
i += 1
while i < len(f):
t.append(f[i])
t.append(f[i])
i += 1
i = 0
for k in range(N // 2):
ans[N // 2][k] = ans[N // 2][-k - 1] = t[i]
i += 1
for k in range(N // 2):
ans[k][N // 2] = ans[-k - 1][N // 2] = t[i]
i += 1
ans[N // 2][N // 2] = o[0]
print(YES)
for line in ans:
print(*line, sep=" ")
if __name__ == '__main__':
main()
``` | output | 1 | 58,061 | 7 | 116,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,062 | 7 | 116,124 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from collections import Counter
n = int(input().strip())
entries = list(map(int, input().strip().split()))
# n = 5
# entries = []
# entries += [1]
# entries += [2]*2
# entries += [3]*2
# entries += [4]*2
# entries += [5]*2
# entries += [6]*(4*4)
if n == 1:
print("Yes")
print(entries[0])
exit()
size_4_entries = (n//2) * (n//2)
size_2_entries = 2 * (n//2) * (n%2)
size_1_entries = n%2
counts = Counter(entries)
size_1 = []
size_2 = []
size_4 = []
digits = list(counts.keys())
dcounts = list(counts.values())
# Early exists
has_one = size_1_entries
for c in dcounts:
if (c%2) == 1:
has_one -= 1
if has_one < 0:
print("No")
exit()
has_two = size_1_entries + size_2_entries
for c in dcounts:
if (c%4) != 0:
has_two -= 1
if has_two < 0:
print("No")
exit()
def get_assign(s4, s2, s1, dcounts):
if s4 == s2 == s1 == 0:
res = dict()
res["4"] = []
res["2"] = []
res["1"] = []
yield res
return
if s4 != 0:
for idx, count in enumerate(dcounts):
if count >= 4:
dcounts[idx] -= 4
for sol in get_assign(s4-1, s2, s1, dcounts):
if sol is not None:
sol["4"].append(digits[idx])
yield sol
dcounts[idx] += 4
if s2 != 0:
for idx, count in enumerate(dcounts):
if count >= 2:
dcounts[idx] -= 2
for sol in get_assign(s4, s2-1, s1, dcounts):
if sol is not None:
sol["2"].append(digits[idx])
yield sol
dcounts[idx] += 2
if s1 != 0:
for idx, count in enumerate(dcounts):
if count >= 1:
dcounts[idx] -= 1
for sol in get_assign(s4, s2, s1-1, dcounts):
if sol is not None:
sol["1"].append(digits[idx])
yield sol
dcounts[idx] += 1
yield None
for res in get_assign(size_4_entries, size_2_entries, size_1_entries, dcounts):
break
# from pprint import pprint
# pprint(memory)
if res is None:
print("No")
exit()
else:
size_4 = res["4"]
size_2 = res["2"]
size_1 = res["1"]
print("Yes")
low = n//2
if (n%2)==0:
high = n//2
else:
high = n//2 + 1
for i in range(n):
row = []
for j in range(n):
if i<low and j<low:
row.append(size_4[j+n//2*i])
elif i<low and j>=high:
row.append(size_4[n-1-j+n//2*i])
elif i>=high and j<low:
row.append(size_4[j+n//2*(n-1-i)])
elif i>=high and j>=high:
row.append(size_4[n-1-j+n//2*(n-1-i)])
elif (n%2)==1 and i==n//2:
if j<low:
row.append(size_2[j])
elif j == n//2:
row.append(size_1[0])
else:
row.append(size_2[n-1-j])
elif (n%2)==1 and j==n//2:
if i<low:
row.append(size_2[n//2+i])
else:
row.append(size_2[n-1-(n//2+i)])
print(" ".join(map(str, row)))
``` | output | 1 | 58,062 | 7 | 116,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples. | instruction | 0 | 58,063 | 7 | 116,126 |
Tags: constructive algorithms, implementation
Correct Solution:
```
from collections import Counter
n = int(input())
s = list(map(int, input().split()))
c = Counter(s)
m = {}
two_pos = []
four_pos = []
one_pos = []
if n % 2 == 1:
one_pos.append((n // 2, n // 2))
for i in range(n // 2):
two_pos.append(((i, n // 2), (n - i - 1, n // 2)))
two_pos.append(((n // 2, i), (n // 2, n - i - 1)))
for i in range(n // 2):
for j in range(n // 2):
four_pos.append(((i, j), (i, n - j - 1), (n - i - 1, j), (n - i - 1, n - j - 1)))
IS_POSSIBLE = True
for k, v in c.items():
if v % 2 == 1:
if len(one_pos) == 0:
IS_POSSIBLE = False
break
pos = one_pos.pop()
m[pos] = k
v -= 1
while v >= 4 and len(four_pos) > 0:
for e in four_pos.pop():
m[e] = k
v -= 4
while v >= 2 and len(two_pos) > 0:
for e in two_pos.pop():
m[e] = k
v -= 2
if v > 0:
IS_POSSIBLE = False
break
if not IS_POSSIBLE:
print('NO')
else:
print('YES')
for i in range(n):
for j in range(n):
print(m[(i, j)], end= ' ')
print()
``` | output | 1 | 58,063 | 7 | 116,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples.
Submitted Solution:
```
from collections import Counter
def is_possible(n, counter):
dup_4_arr = []
dup_2_arr = []
only_one_value = None
if n & 1 == 0:
for v in counter.values():
if v % 4 != 0:
return (False, dup_4_arr, dup_2_arr, only_one_value)
else:
#for 4
count_4 = 0
min_count_4 = (n // 2)**2
for k, v in counter.items():
while (v >= 4):
v = v - 4
count_4 += 1
counter[k] -= 4
dup_4_arr.append(k)
if count_4 == min_count_4:
break
if count_4 == min_count_4:
break
if count_4 < min_count_4:
return (False, dup_4_arr, dup_2_arr, only_one_value)
# for 2
count_2 = 0
min_count_2 = (n // 2) * 2
for k, v in counter.items():
while (v >= 2):
v = v - 2
count_2 += 1
counter[k] -= 2
dup_2_arr.append(k)
if count_2 == min_count_2:
break
if count_2 == min_count_2:
break
if count_2 < min_count_2:
return (False, dup_4_arr, dup_2_arr, only_one_value)
for k, v in counter.items():
if v == 1:
only_one_value = k
return True, dup_4_arr, dup_2_arr, only_one_value
def make_matrix(n, arr, counter, dup_4_arr, dup_2_arr, only_one_value):
ret = [["-1"] * n for _ in range(n)]
arr = sorted(arr)
if n & 1 == 0:
current = 0
for i in range(n//2):
for j in range(n//2):
ret[i][j] = str(arr[current])
current += 1
ret[i][n-1-j] = str(arr[current])
current += 1
ret[n-1-i][j] = str(arr[current])
current += 1
ret[n-1-i][n-1-j] = str(arr[current])
current += 1
else:
# for 4
current_4 = 0
for i in range(n//2):
for j in range(n//2):
ret[i][j] = str(dup_4_arr[current_4])
ret[i][n-1-j] = str(dup_4_arr[current_4])
ret[n-1-i][j] = str(dup_4_arr[current_4])
ret[n-1-i][n-1-j] = str(dup_4_arr[current_4])
current_4 += 1
current_2 = 0
center = n//2
for i in range(n//2):
ret[center][i] = str(dup_2_arr[current_2])
ret[center][n-1-i] = str(dup_2_arr[current_2])
current_2 += 1
for i in range(n//2):
ret[i][center] = str(dup_2_arr[current_2])
ret[n-1-i][center] = str(dup_2_arr[current_2])
current_2 += 1
ret[center][center] = str(only_one_value)
return ret
if __name__ == "__main__":
n = int(input())
arr = tuple(map(int, input().split(" ")))
counter = Counter(arr)
possible, dup_4_arr, dup_2_arr, only_one_value = is_possible(n, counter)
if possible:
print("YES")
mat = make_matrix(n, arr, counter, dup_4_arr, dup_2_arr, only_one_value)
for i in range(n):
print(" ".join(mat[i]))
else:
print("NO")
``` | instruction | 0 | 58,067 | 7 | 116,134 |
Yes | output | 1 | 58,067 | 7 | 116,135 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Let's call some square matrix with integer values in its cells palindromic if it doesn't change after the order of rows is reversed and it doesn't change after the order of columns is reversed.
For example, the following matrices are palindromic:
<image>
The following matrices are not palindromic because they change after the order of rows is reversed:
<image>
The following matrices are not palindromic because they change after the order of columns is reversed:
<image>
You are given n^2 integers. Put them into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic. If there are multiple answers, print any. If there is no solution, print "NO".
Input
The first line contains one integer n (1 ≤ n ≤ 20).
The second line contains n^2 integers a_1, a_2, ..., a_{n^2} (1 ≤ a_i ≤ 1000) — the numbers to put into a matrix of n rows and n columns.
Output
If it is possible to put all of the n^2 numbers into a matrix of n rows and n columns so that each number is used exactly once, each cell contains exactly one number and the resulting matrix is palindromic, then print "YES". Then print n lines with n space-separated numbers — the resulting matrix.
If it's impossible to construct any matrix, then print "NO".
You can print each letter in any case (upper or lower). For example, "YeS", "no" and "yES" are all acceptable.
Examples
Input
4
1 8 8 1 2 2 2 2 2 2 2 2 1 8 8 1
Output
YES
1 2 2 1
8 2 2 8
8 2 2 8
1 2 2 1
Input
3
1 1 1 1 1 3 3 3 3
Output
YES
1 3 1
3 1 3
1 3 1
Input
4
1 2 1 9 8 4 3 8 8 3 4 8 9 2 1 1
Output
NO
Input
1
10
Output
YES
10
Note
Note that there exist multiple answers for the first two examples.
Submitted Solution:
```
from collections import defaultdict
n = int(input())
arr = list(map(int,input().split()))
counts = defaultdict(int)
for a in arr: counts[a]+=1
if n==1:
print('YES')
print(arr[0])
exit(0)
matrix = [ [None]*n for _ in range(n)]
if n%2==0:
vals= []
for val,cnt in counts.items():
if cnt%4!=0:
print('NO')
exit(0)
vals.append( (val,cnt) )
for i in range(n//2):
for j in range(n//2):
val,cnt = vals.pop(0)
matrix[i][j] = val
matrix[i][n-j-1] = val
matrix[n-i-1][j] = val
matrix[n-i-1][n-j-1] = val
if cnt-4 != 0:
vals.append( (val,cnt-4) )
print('YES')
for row in matrix: print(*row)
else:
vals = []
avail=True
c = 0
for val,cnt in counts.items():
if cnt%4==1 and avail:
avail=False
matrix[(n//2)][(n//2)] = val
cnt-=1
if cnt!=0: vals.append( (val,cnt) )
elif cnt%4==3 and avail:
avail = False
matrix[(n//2)][(n//2)] = val
if c < n//2:
matrix[ n//2 ][c] = val
matrix[ n//2 ][ n-c-1 ] = val
c+=1
elif c < n-1:
i = c%(n//2)
matrix[ i ][n//2] = val
matrix[n-i-1][n//2] = val
c+=1
else:
print('NO');exit(0)
if (cnt-3) > 0 :
vals.append( (val,cnt-2) )
elif cnt%4==0:
vals.append( (val,cnt) )
elif cnt%2==0:
if c < n//2:
matrix[ n//2 ][c] = val
matrix[ n//2 ][ n-c-1 ] = val
c+=1
elif c < n-1:
i = c%(n//2)
matrix[ i ][n//2] = val
matrix[n-i-1][n//2] = val
c+=1
else:
print('NO');exit(0)
if (cnt-2) > 0 :
vals.append( (val,cnt-2) )
else:
print('NO');exit(0)
while c<n-1:
val,cnt = vals.pop(0)
if c < n//2:
matrix[ n//2 ][c] = val
matrix[ n//2][ n-c-1 ] = val
c+=1
else:
i = c%(n//2)
matrix[ i ][n//2] = val
matrix[n-i-1][n//2] = val
c+=1
if cnt-2 > 0:
vals.insert(0,(val,cnt-2))
if vals[0][1] % 4 !=0: print('NO');exit(0)
if avail: print('NO'); exit(0)
for i in range( n//2):
for j in range( n//2):
val,cnt = vals.pop(0)
matrix[i][j] = val
matrix[i][n-j-1] = val
matrix[n-i-1][j] = val
matrix[n-i-1][n-j-1] = val
if cnt-4 != 0:
vals.append( (val,cnt-4) )
print('YES')
for row in matrix: print(*row)
``` | instruction | 0 | 58,068 | 7 | 116,136 |
No | output | 1 | 58,068 | 7 | 116,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of n light bulbs in a single row. Each bulb has a number from 1 to n (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
<image>
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 2). For example, the complexity of 1 4 2 3 5 is 2 and the complexity of 1 3 5 7 6 4 2 is 1.
No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of light bulbs on the garland.
The second line contains n integers p_1,\ p_2,\ …,\ p_n (0 ≤ p_i ≤ n) — the number on the i-th bulb, or 0 if it was removed.
Output
Output a single number — the minimum complexity of the garland.
Examples
Input
5
0 5 0 2 3
Output
2
Input
7
1 0 0 5 0 0 2
Output
1
Note
In the first example, one should place light bulbs as 1 5 4 2 3. In that case, the complexity would be equal to 2, because only (5, 4) and (2, 3) are the pairs of adjacent bulbs that have different parity.
In the second case, one of the correct answers is 1 7 3 5 6 4 2. | instruction | 0 | 58,146 | 7 | 116,292 |
Tags: dp, greedy, sortings
Correct Solution:
```
n = int(input())
girl = list(map(int, input().split()))
odd = 0
even = 0
for g in girl:
if g != 0:
if g % 2 != 0:
odd += 1
else:
even += 1
left_odd = (n + 1) // 2 - odd
left_even = n // 2 - even
series = []
left = 0
right = n - 1
while girl[left] == 0:
left += 1
if left == n:
break
while girl[right] == 0:
right -= 1
if right == -1:
break
if left > right:
print(1 if n > 1 else 0)
exit()
cur = None
ans = 0
for i in range(left, right + 1):
if girl[i] != 0:
if i < n - 1 and girl[i + 1] != 0 and girl[i + 1] % 2 != girl[i] % 2:
ans += 1
if cur is not None:
cur[1] = girl[i] % 2
series.append(cur)
cur = None
else:
if cur is None:
cur = [0, 0, 1]
if i > 0:
cur[0] = girl[i - 1] % 2
else:
cur[2] += 1
series.sort(key=lambda tup: tup[2])
#print(series, left_even, left_odd)
#print(left, right)
for i in range(len(series)):
s = series[i]
if s[0] != s[1]:
ans += 1
elif s[0] % 2 != 0:
if left_odd >= s[2]:
left_odd -= s[2]
else:
ans += 2
else:
if left_even >= s[2]:
left_even -= s[2]
else:
ans += 2
if girl[left] % 2 == 0:
if left_even < left:
ans += 1
else:
left_even -= left
else:
if left_odd < left:
ans += 1
else:
left_odd -= left
if girl[right] % 2 == 0:
if left_even < n - 1 - right:
ans += 1
else:
if left_odd < n - 1 - right:
ans += 1
print(ans)
``` | output | 1 | 58,146 | 7 | 116,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of n light bulbs in a single row. Each bulb has a number from 1 to n (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
<image>
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 2). For example, the complexity of 1 4 2 3 5 is 2 and the complexity of 1 3 5 7 6 4 2 is 1.
No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of light bulbs on the garland.
The second line contains n integers p_1,\ p_2,\ …,\ p_n (0 ≤ p_i ≤ n) — the number on the i-th bulb, or 0 if it was removed.
Output
Output a single number — the minimum complexity of the garland.
Examples
Input
5
0 5 0 2 3
Output
2
Input
7
1 0 0 5 0 0 2
Output
1
Note
In the first example, one should place light bulbs as 1 5 4 2 3. In that case, the complexity would be equal to 2, because only (5, 4) and (2, 3) are the pairs of adjacent bulbs that have different parity.
In the second case, one of the correct answers is 1 7 3 5 6 4 2. | instruction | 0 | 58,147 | 7 | 116,294 |
Tags: dp, greedy, sortings
Correct Solution:
```
def arr_inp():
return [int(x) for x in stdin.readline().split()]
def dp(i, odd, even, pre):
if i >= n:
return 0
ret = mem[i, odd, even, pre]
if ret != -1:
return ret
ret = float('inf')
if a[i] > 0:
if pre == 0:
ret = min(ret, dp(i + 1, odd, even, a[i]))
elif (pre & 1 == 0 and a[i] & 1) or (pre & 1 and a[i] & 1 == 0):
ret = min(ret, 1 + dp(i + 1, odd, even, a[i]))
else:
ret = min(ret, dp(i + 1, odd, even, a[i]))
else:
if pre == 0:
if odd:
ret = min(ret, dp(i + 1, odd - 1, even, 1))
if even:
ret = min(ret, dp(i + 1, odd, even - 1, 2))
elif pre & 1:
if odd:
ret = min(ret, dp(i + 1, odd - 1, even, 1))
if even:
ret = min(ret, 1 + dp(i + 1, odd, even - 1, 2))
else:
if even:
ret = min(ret, dp(i + 1, odd, even - 1, 2))
if odd:
ret = min(ret, 1 + dp(i + 1, odd - 1, even, 1))
mem[i, odd, even, pre] = ret
return ret
from sys import stdin
from collections import defaultdict, Counter
n, a, mem, o, e = int(input()), arr_inp(), defaultdict(lambda: -1), 0, 0
c = defaultdict(int, Counter(a))
for i in range(1, n + 1):
if c[i] == 0:
if i & 1:
o += 1
else:
e += 1
print(dp(0, o, e, 0))
# print(mem)
``` | output | 1 | 58,147 | 7 | 116,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of n light bulbs in a single row. Each bulb has a number from 1 to n (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
<image>
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 2). For example, the complexity of 1 4 2 3 5 is 2 and the complexity of 1 3 5 7 6 4 2 is 1.
No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of light bulbs on the garland.
The second line contains n integers p_1,\ p_2,\ …,\ p_n (0 ≤ p_i ≤ n) — the number on the i-th bulb, or 0 if it was removed.
Output
Output a single number — the minimum complexity of the garland.
Examples
Input
5
0 5 0 2 3
Output
2
Input
7
1 0 0 5 0 0 2
Output
1
Note
In the first example, one should place light bulbs as 1 5 4 2 3. In that case, the complexity would be equal to 2, because only (5, 4) and (2, 3) are the pairs of adjacent bulbs that have different parity.
In the second case, one of the correct answers is 1 7 3 5 6 4 2. | instruction | 0 | 58,148 | 7 | 116,296 |
Tags: dp, greedy, sortings
Correct Solution:
```
def calculate(ar, num):
s = 0
for x in range(len(ar)):
if(num >= ar[x][0]):
num -= ar[x][0]
else:
s += (1 if ar[x][1] else 2)
return s
def calculate1(ar, num):
s = calculate(ar, num)
ar1 = []
for x in range(len(ar)):
if(ar[x][1]):
ar1.append(x)
if(len(ar1) >= 1):
tmp = ar[ar1[0]]
ar = ar[:ar1[0]] + ar[ar1[0] + 1:]
s = min(s, calculate(ar, num) + 1)
ar = ar[:ar1[0]] + [tmp] + ar[ar1[0]:]
if(len(ar1) == 2):
ar = ar[:ar1[1]] + ar[ar1[1] + 1:]
s = min(s, calculate(ar, num) + 1)
ar = ar[:ar1[0]] + ar[ar1[0] + 1:]
s = min(s, calculate(ar, num) + 2)
return s
n = int(input())
p = list(map(int, input().split(" ")))
ar = [False] * (n + 1)
for x in p:
if(x != 0):
ar[x] = True
if(ar == [False] * (n + 1)):
if(n == 1):
print(0)
else:
print(1)
else:
ones = 0
zeros = 0
for x in range(1, n + 1):
if(not ar[x]):
if(x % 2 == 0):
zeros+=1
else:
ones += 1
c = 0
for x in range(1, n):
if(p[x] != 0 and p[x - 1] != 0 and p[x] % 2 != p[x - 1] % 2):
c += 1
s = 0
for x in range(n):
if(p[x] != 0):
s = x
break
c1 = 0
s1 = -1
zerosp = []
onesp = []
c1 = 0
for x in range(s + 1, n):
if(p[x] == 0):
if(s1 == -1):
s1 = x
c1 += 1
elif(c1 != 0):
if(p[s1 - 1] % 2 != p[x] % 2):
c += 1
else:
if(p[s1 - 1] % 2 == 0):
zerosp.append((c1, False))
else:
onesp.append((c1, False))
s1 = -1
c1 = 0
if(n >= 2):
if(p[0] == 0):
if(p[s] % 2 == 0):
zerosp.append((s, True))
else:
onesp.append((s, True))
if(p[-1] == 0):
c1 = 0
s2 = 0
for x in range(n - 2, -1, -1):
c1 += 1
if(p[x] != 0):
s2 = x
break
if(p[s2] % 2 == 0):
zerosp.append((c1, True))
else:
onesp.append((c1, True))
c1 = 0
zerosp = sorted(zerosp, key=lambda x: x[0])
onesp = sorted(onesp, key=lambda x: x[0])
print(c + calculate1(zerosp, zeros) + calculate1(onesp, ones))
``` | output | 1 | 58,148 | 7 | 116,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vadim loves decorating the Christmas tree, so he got a beautiful garland as a present. It consists of n light bulbs in a single row. Each bulb has a number from 1 to n (in arbitrary order), such that all the numbers are distinct. While Vadim was solving problems, his home Carp removed some light bulbs from the garland. Now Vadim wants to put them back on.
<image>
Vadim wants to put all bulb back on the garland. Vadim defines complexity of a garland to be the number of pairs of adjacent bulbs with numbers with different parity (remainder of the division by 2). For example, the complexity of 1 4 2 3 5 is 2 and the complexity of 1 3 5 7 6 4 2 is 1.
No one likes complexity, so Vadim wants to minimize the number of such pairs. Find the way to put all bulbs back on the garland, such that the complexity is as small as possible.
Input
The first line contains a single integer n (1 ≤ n ≤ 100) — the number of light bulbs on the garland.
The second line contains n integers p_1,\ p_2,\ …,\ p_n (0 ≤ p_i ≤ n) — the number on the i-th bulb, or 0 if it was removed.
Output
Output a single number — the minimum complexity of the garland.
Examples
Input
5
0 5 0 2 3
Output
2
Input
7
1 0 0 5 0 0 2
Output
1
Note
In the first example, one should place light bulbs as 1 5 4 2 3. In that case, the complexity would be equal to 2, because only (5, 4) and (2, 3) are the pairs of adjacent bulbs that have different parity.
In the second case, one of the correct answers is 1 7 3 5 6 4 2. | instruction | 0 | 58,149 | 7 | 116,298 |
Tags: dp, greedy, sortings
Correct Solution:
```
n=int(input())
z=float("INF")
a=[int(x) for x in input().split()]
dp=[[[[z]*2 for k in range(n+1)] for j in range(n+1)]for i in range(n+1)]
dp[0][0][0][0]=0
dp[0][0][0][1]=0
for i in range(1,n+1):
for j in range(n+1):
for k in range(n+1):
if a[i-1]&1:
dp[i][j][k][1]=min(dp[i-1][j][k-1][0]+1,dp[i-1][j][k-1][1])
elif a[i-1]==0:
dp[i][j][k][1]=min(dp[i-1][j][k-1][0]+1,dp[i-1][j][k-1][1])
dp[i][j][k][0]=min(dp[i-1][j-1][k][0],dp[i-1][j-1][k][1]+1)
else:
dp[i][j][k][0]=min(dp[i-1][j-1][k][0],dp[i-1][j-1][k][1]+1)
e=(n-(n&1))//2
o=(n+(n&1))//2
ans=[dp[n][e][o][0],dp[n][e][o][1]]
if a[n-1]==0:
print(min(ans))
else:
print(ans[a[n-1]&1])
``` | output | 1 | 58,149 | 7 | 116,299 |
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