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.
There are N Reversi pieces arranged in a row. (A Reversi piece is a disc with a black side and a white side.) The state of each piece is represented by a string S of length N. If S_i=`B`, the i-th piece from the left is showing black; If S_i=`W`, the i-th piece from the left is showing white.
Consider performing the following operation:
* Choose i (1 \leq i < N) such that the i-th piece from the left is showing black and the (i+1)-th piece from the left is showing white, then flip both of those pieces. That is, the i-th piece from the left is now showing white and the (i+1)-th piece from the left is now showing black.
Find the maximum possible number of times this operation can be performed.
Constraints
* 1 \leq |S| \leq 2\times 10^5
* S_i=`B` or `W`
Input
Input is given from Standard Input in the following format:
S
Output
Print the maximum possible number of times the operation can be performed.
Examples
Input
BBW
Output
2
Input
BWBWBW
Output
6
Submitted Solution:
```
S = list(input())
count = 0
for i in range(len(S)):
for j in range(len(S)-1, i, -1):
if S[j] == "W" and S[j-1] == "B":
S[j], S[j-1] = S[j-1], S[j]
count += 1
print(count)
``` | instruction | 0 | 84,861 | 7 | 169,722 |
No | output | 1 | 84,861 | 7 | 169,723 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,950 | 7 | 169,900 |
"Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 998244353
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
rr = []
while True:
n,w = LI()
if n == 0:
break
a = [I() for _ in range(n)]
ma = max(a)
l = (ma // w + 1)
b = [0] * l
for c in a:
b[c//w] += 1
h = max(b)
r = 0.01
for i in range(l):
bi = b[i]
d = (l-i-1) / (l-1)
r += d * bi / h
rr.append(r)
return '\n'.join(map(str, rr))
print(main())
``` | output | 1 | 84,950 | 7 | 169,901 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,951 | 7 | 169,902 |
"Correct Solution:
```
while True:
n, w = map(int, input().split())
if n is 0 and w is 0:
break
l = [int(input()) for i in range(n)]
max_num = 0
for i in range(n):
max_num = max(max_num, l[i])
hist_size = int((max_num/w) + 1)
hist = [0.0] * int(hist_size)
for i in range(n):
hist[int(l[i]/w)] += 1
max_height= 0
for i in range(hist_size):
max_height = max(max_height,hist[i])
ans = 0.0
for i in range(hist_size-1):
ans += hist[i]/max_height*(hist_size - 1 - i)/(hist_size-1)
ans += 0.01
print(ans)
``` | output | 1 | 84,951 | 7 | 169,903 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,952 | 7 | 169,904 |
"Correct Solution:
```
# coding: utf-8
while 1:
n,w=map(int,input().split())
if n==0:
break
dic={}
mx=-1
max_section=-1
for i in range(1+100//w):
dic[i]=0
for i in range(n):
x=int(input())
dic[x//w]+=1
mx=max(mx,dic[x//w])
max_section=max(max_section,x//w)
ans=0
for i in range(1+100//w):
ans+=(dic[i]/mx)*(1-i*(1/(max_section)))
print(ans+0.01)
``` | output | 1 | 84,952 | 7 | 169,905 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,953 | 7 | 169,906 |
"Correct Solution:
```
while True:
n, w = list(map(int, input().split()))
if n==0:
break
v = [int(input()) for i in range(n)]
h = [0 for i in range(100)]
ma = 0
mi = 1e9
for i in range(n):
h[v[i]//w] += 1
ma = max(ma, v[i]//w)
mi = min(mi, v[i]//w)
sm = ma-mi
h = [h[i]/max(h) for i in range(100)]
ans = 0.01
for i in range(mi, ma):
ans += (sm-i+mi)/(sm+mi)*h[i]
print(ans)
``` | output | 1 | 84,953 | 7 | 169,907 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,954 | 7 | 169,908 |
"Correct Solution:
```
while True:
N,W = map(int,input().split())
if N == 0: break
src = [int(input()) for i in range(N)]
highest = max(src)
bands = highest//W + 1
hist = [0 for i in range(bands)]
for i in range(N):
hist[src[i] // W] += 1
ans = maxn = 0
for i,n in enumerate(hist):
ans += n * (bands-1-i)
maxn = max(n, maxn)
print(ans / maxn / (bands-1) + 0.01)
``` | output | 1 | 84,954 | 7 | 169,909 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,955 | 7 | 169,910 |
"Correct Solution:
```
while True:
N, W = map(int, input().split())
if not N:
break
s = list(sorted([int(input()) for i in range(N)]))
m = s[-1] // W + 1
ss = [0] * m
for c in s:
ss[c // W] += 1
unit = max(ss)
a = 0
for i, c in enumerate(ss):
a += (1 - i / (m - 1)) * (c / unit)
print(a + 0.01)
``` | output | 1 | 84,955 | 7 | 169,911 |
Provide a correct Python 3 solution for this coding contest problem.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667 | instruction | 0 | 84,956 | 7 | 169,912 |
"Correct Solution:
```
while 1:
n,w=map(int,input().split())
if n==w==0:break
a=[0]*100
s=h=int(0)
while n:
b=int(input())
a[b//w]+=1
s=max(s,b//w)
h=max(h,a[b//w])
n-=1
ans=0.01
for i in range(int(s)+1):
ans+=a[i]/h*(s-i)/s
print(ans)
``` | output | 1 | 84,956 | 7 | 169,913 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
def calcInk(wdt,val):
maxval=val[-1]
histnum=maxval//wdt
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(val)
for i in range(vl):
v=heappop(val)
hist[v//wdt].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
return ink+0.01
if __name__=='__main__':
while True:
n,w=map(int,input().split())
if n==w==0:
break
V=[]
for _ in range(n):
heappush(V,int(input()))
``` | instruction | 0 | 84,957 | 7 | 169,914 |
No | output | 1 | 84,957 | 7 | 169,915 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
``` | instruction | 0 | 84,958 | 7 | 169,916 |
No | output | 1 | 84,958 | 7 | 169,917 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
from heapq import *
def calcInk(wdt,val):
maxval=val[-1]
histnum=maxval//wdt
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(val)
for i in range(vl):
v=heappop(val)
hist[v//wdt].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
return ink+0.01
``` | instruction | 0 | 84,959 | 7 | 169,918 |
No | output | 1 | 84,959 | 7 | 169,919 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Dr. Grey is a data analyst, who visualizes various aspects of data received from all over the world everyday. He is extremely good at sophisticated visualization tools, but yet his favorite is a simple self-made histogram generator.
Figure 1 is an example of histogram automatically produced by his histogram.
<image>
A histogram is a visual display of frequencies of value occurrences as bars. In this example, values in the interval 0-9 occur five times, those in the interval 10-19 occur three times, and 20-29 and 30-39 once each.
Dr. Grey’s histogram generator is a simple tool. First, the height of the histogram is fixed, that is, the height of the highest bar is always the same and those of the others are automatically adjusted proportionately. Second, the widths of bars are also fixed. It can only produce a histogram of uniform intervals, that is, each interval of a histogram should have the same width (10 in the above example). Finally, the bar for each interval is painted in a grey color, where the colors of the leftmost and the rightmost intervals are black and white, respectively, and the darkness of bars monotonically decreases at the same rate from left to right. For instance, in Figure 1, the darkness levels of the four bars are 1, 2/3, 1/3, and 0, respectively.
In this problem, you are requested to estimate ink consumption when printing a histogram on paper. The amount of ink necessary to draw a bar is proportional to both its area and darkness.
Input
The input consists of multiple datasets, each of which contains integers and specifies a value table and intervals for the histogram generator, in the following format.
n w
v1
v2
.
.
vn
n is the total number of value occurrences for the histogram, and each of the n lines following the first line contains a single value. Note that the same value may possibly occur multiple times.
w is the interval width. A value v is in the first (i.e. leftmost) interval if 0 ≤ v < w, the second one if w ≤ v < 2w, and so on. Note that the interval from 0 (inclusive) to w (exclusive) should be regarded as the leftmost even if no values occur in this interval. The last (i.e. rightmost) interval is the one that includes the largest value in the dataset.
You may assume the following.
1 ≤ n ≤ 100
10 ≤ w ≤ 50
0 ≤ vi ≤ 100 for 1 ≤ i ≤ n
You can also assume that the maximum value is no less than w. This means that the histogram has more than one interval. The end of the input is indicated by a line containing two zeros.
Output
For each dataset, output a line containing the amount of ink consumed in printing the histogram.
One unit of ink is necessary to paint one highest bar black. Assume that 0.01 units of ink per histogram is consumed for various purposes except for painting bars such as drawing lines and characters (see Figure 1). For instance, the amount of ink consumed in printing the histogram in Figure 1 is:
<image>
Each output value should be in a decimal fraction and may have an error less than 10-5 .
Example
Input
3 50
100
0
100
3 50
100
100
50
10 10
1
2
3
4
5
16
17
18
29
30
0 0
Output
0.51
0.26
1.4766666666666667
Submitted Solution:
```
if __name__=='__main__':
while True:
n,w=map(int,input().split())
if n==w==0:
break
V=[]
for _ in range(n):
heappush(V,int(input()))
maxV=V[-1]
histnum=maxV//w
hist=[]
for i in range(histnum+1):
hist.append([])
vl=len(V)
for i in range(vl):
v=heappop(V)
hist[v//w].append(v)
highest=0
for i in range(len(hist)):
highest=max(highest,len(hist[i]))
ink=0
for i in range(len(hist)):
ink+=(1-i/histnum)*(len(hist[i])/highest)
print(ink)
``` | instruction | 0 | 84,960 | 7 | 169,920 |
No | output | 1 | 84,960 | 7 | 169,921 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111. | instruction | 0 | 85,812 | 7 | 171,624 |
Tags: dfs and similar, dsu, graphs
Correct Solution:
```
from sys import stdin
input = stdin.readline
n , k = [int(i) for i in input().split()]
pairs = [i + k for i in range(k)] + [i for i in range(k)]
initial_condition = list(map(lambda x: x == '1',input().strip()))
data = [i for i in range(2*k)]
constrain = [-1] * (2*k)
h = [0] * (2*k)
L = [1] * k + [0] * k
dp1 = [-1 for i in range(n)]
dp2 = [-1 for i in range(n)]
for i in range(k):
input()
inp = [int(j) for j in input().split()]
for s in inp:
if dp1[s-1] == -1:dp1[s-1] = i
else:dp2[s-1] = i
pfsums = 0
ans = []
def remove_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums -= L[s1]
elif constrain[pairs[s1]] == 1:
pfsums -= L[pairs[s1]]
else:
pfsums -= min(L[s1],L[pairs[s1]])
def sh(i):
while i != data[i]:
i = data[i]
return i
def upd_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums += L[s1]
elif constrain[pairs[s1]] == 1:
pfsums += L[pairs[s1]]
else:
pfsums += min(L[s1],L[pairs[s1]])
def ms(i,j):
i = sh(i) ; j = sh(j)
cons = max(constrain[i],constrain[j])
if h[i] < h[j]:
data[i] = j
L[j] += L[i]
constrain[j] = cons
return j
else:
data[j] = i
if h[i] == h[j]:
h[i] += 1
L[i] += L[j]
constrain[i] = cons
return i
for i in range(n):
if dp1[i] == -1 and dp2[i] == -1:
pass
elif dp2[i] == -1:
s1 = sh(dp1[i])
remove_pfsum(s1)
constrain[s1] = 0 if initial_condition[i] else 1
constrain[pairs[s1]] = 1 if initial_condition[i] else 0
upd_pfsum(s1)
else:
s1 = sh(dp1[i]) ; s2 = sh(dp2[i])
if s1 == s2 or pairs[s1] == s2:
pass
else:
remove_pfsum(s1)
remove_pfsum(s2)
if initial_condition[i]:
new_s1 = ms(s1,s2)
new_s2 = ms(pairs[s1],pairs[s2])
else:
new_s1 = ms(s1,pairs[s2])
new_s2 = ms(pairs[s1],s2)
pairs[new_s1] = new_s2
pairs[new_s2] = new_s1
upd_pfsum(new_s1)
ans.append(pfsums)
for i in ans:
print(i)
``` | output | 1 | 85,812 | 7 | 171,625 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n lamps on a line, numbered from 1 to n. Each one has an initial state off (0) or on (1).
You're given k subsets A_1, …, A_k of \{1, 2, ..., n\}, such that the intersection of any three subsets is empty. In other words, for all 1 ≤ i_1 < i_2 < i_3 ≤ k, A_{i_1} ∩ A_{i_2} ∩ A_{i_3} = ∅.
In one operation, you can choose one of these k subsets and switch the state of all lamps in it. It is guaranteed that, with the given subsets, it's possible to make all lamps be simultaneously on using this type of operation.
Let m_i be the minimum number of operations you have to do in order to make the i first lamps be simultaneously on. Note that there is no condition upon the state of other lamps (between i+1 and n), they can be either off or on.
You have to compute m_i for all 1 ≤ i ≤ n.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 3 ⋅ 10^5).
The second line contains a binary string of length n, representing the initial state of each lamp (the lamp i is off if s_i = 0, on if s_i = 1).
The description of each one of the k subsets follows, in the following format:
The first line of the description contains a single integer c (1 ≤ c ≤ n) — the number of elements in the subset.
The second line of the description contains c distinct integers x_1, …, x_c (1 ≤ x_i ≤ n) — the elements of the subset.
It is guaranteed that:
* The intersection of any three subsets is empty;
* It's possible to make all lamps be simultaneously on using some operations.
Output
You must output n lines. The i-th line should contain a single integer m_i — the minimum number of operations required to make the lamps 1 to i be simultaneously on.
Examples
Input
7 3
0011100
3
1 4 6
3
3 4 7
2
2 3
Output
1
2
3
3
3
3
3
Input
8 6
00110011
3
1 3 8
5
1 2 5 6 7
2
6 8
2
3 5
2
4 7
1
2
Output
1
1
1
1
1
1
4
4
Input
5 3
00011
3
1 2 3
1
4
3
3 4 5
Output
1
1
1
1
1
Input
19 5
1001001001100000110
2
2 3
2
5 6
2
8 9
5
12 13 14 15 16
1
19
Output
0
1
1
1
2
2
2
3
3
3
3
4
4
4
4
4
4
4
5
Note
In the first example:
* For i = 1, we can just apply one operation on A_1, the final states will be 1010110;
* For i = 2, we can apply operations on A_1 and A_3, the final states will be 1100110;
* For i ≥ 3, we can apply operations on A_1, A_2 and A_3, the final states will be 1111111.
In the second example:
* For i ≤ 6, we can just apply one operation on A_2, the final states will be 11111101;
* For i ≥ 7, we can apply operations on A_1, A_3, A_4, A_6, the final states will be 11111111.
Submitted Solution:
```
from sys import stdin
input = stdin.readline
n , k = [int(i) for i in input().split()]
pairs = [i + n for i in range(n)] + [i for i in range(n)]
initial_condition = list(map(lambda x: x == '1',input().strip()))
data = [i for i in range(2*n)]
constrain = [-1] * (2*n)
h = [0] * (2*n)
L = [1] * n + [0] * n
dp1 = [-1 for i in range(n)]
dp2 = [-1 for i in range(n)]
for i in range(k):
input()
inp = [int(j) for j in input().split()]
for s in inp:
if dp1[s-1] == -1:dp1[s-1] = i
else:dp2[s-1] = i
pfsums = 0
ans = []
def remove_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums -= L[s1]
elif constrain[pairs[s1]] == 1:
pfsums -= L[pairs[s1]]
else:
pfsums -= min(L[s1],L[pairs[s1]])
def sh(i):
while i != data[i]:
i = data[i]
return i
def upd_pfsum(s1):
global pfsums
if constrain[s1] == 1:
pfsums += L[s1]
elif constrain[pairs[s1]] == 1:
pfsums += L[pairs[s1]]
else:
pfsums += min(L[s1],L[pairs[s1]])
def ms(i,j):
i = sh(i) ; j = sh(j)
cons = max(constrain[i],constrain[j])
if h[i] < h[j]:
data[i] = j
h[i] += 1
L[j] += L[i]
constrain[j] = cons
return j
else:
data[j] = i
h[j] += 1
L[i] += L[j]
constrain[i] = cons
return i
for i in range(n):
if dp1[i] == -1 and dp2[i] == -1:
pass
elif dp2[i] == -1:
s1 = sh(dp1[i])
remove_pfsum(s1)
constrain[s1] = 0 if initial_condition[i] else 1
constrain[pairs[s1]] = 1 if initial_condition[i] else 0
upd_pfsum(s1)
else:
s1 = sh(dp1[i]) ; s2 = sh(dp2[i])
if s1 == s2 or pairs[s1] == s2:
pass
else:
remove_pfsum(s1)
remove_pfsum(s2)
if initial_condition[i]:
new_s1 = ms(s1,s2)
new_s2 = ms(pairs[s1],pairs[s2])
else:
new_s1 = ms(s1,pairs[s2])
new_s2 = ms(pairs[s1],s2)
pairs[new_s1] = new_s2
pairs[new_s2] = new_s1
upd_pfsum(new_s1)
ans.append(pfsums)
for i in ans:
print(i)
``` | instruction | 0 | 85,813 | 7 | 171,626 |
No | output | 1 | 85,813 | 7 | 171,627 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,131 | 7 | 172,262 |
Tags: implementation
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m = LI()
a = [LI() for _ in range(n)]
s = set()
for c in a:
s |= set(c[1:])
if len(s) == m:
return 'YES'
return 'NO'
print(main())
``` | output | 1 | 86,131 | 7 | 172,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,132 | 7 | 172,264 |
Tags: implementation
Correct Solution:
```
n,m=map(int,input().split())
yy=[]
for i in range(n):
x=list(map(int,input().split()))
x=x[1:]
yy.append(x)
main=[]
for i in yy:
for j in i:
main.append(j)
main=set(main)
gg={0}
for i in range(1,m+1):
gg.add(i)
gg.remove(0)
if(gg==main):
print("YES")
else:
print("NO")
``` | output | 1 | 86,132 | 7 | 172,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,133 | 7 | 172,266 |
Tags: implementation
Correct Solution:
```
n,m=map(int,input().split(" "))
L=[0]*(m)
for k in range(n):
i=0
for j in map(int,input().split(" ")):
if(i!=0):
L[j-1]=1
i+=1
if(sum(L)==m):
print("YES")
else:
print("NO")
``` | output | 1 | 86,133 | 7 | 172,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,134 | 7 | 172,268 |
Tags: implementation
Correct Solution:
```
if __name__ == '__main__':
n, m = map(int, input().split())
s = set()
for _ in range(n):
s.update(map(int, input().split()[1:]))
print('YES' if len(s) == m else 'NO')
``` | output | 1 | 86,134 | 7 | 172,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,135 | 7 | 172,270 |
Tags: implementation
Correct Solution:
```
a,b=map(int,input().split())
y=[]
k=0
for i in range(0,a):
x=list(map(int,input().split()))
y.extend(x[1:])
for i in range(1,b+1):
if(y.count(i)<1):
k=-1
print("NO")
break
if(k==0):
print("YES")
``` | output | 1 | 86,135 | 7 | 172,271 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,136 | 7 | 172,272 |
Tags: implementation
Correct Solution:
```
n , m = list(map(int,input().split()))
p = []
for i in range(n):
l = list(map(int,input().split()))
for k in range(1,len(l)):
p.append(l[k])
setp = set(p)
o = []
for i in range(1,m+1):
o.append(i)
seto = set(o)
if len(seto)==len(setp):
print("YES")
else:
print("NO")
``` | output | 1 | 86,136 | 7 | 172,273 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,137 | 7 | 172,274 |
Tags: implementation
Correct Solution:
```
def main():
n, m = map(int, input().split())
a = set({})
for i in range(n):
b = list(map(int, input().split()))
a.update(b[1:])
print("YES" if(len(a) == m) else "NO")
if __name__ == "__main__":
main()
``` | output | 1 | 86,137 | 7 | 172,275 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp. | instruction | 0 | 86,138 | 7 | 172,276 |
Tags: implementation
Correct Solution:
```
(n, m) = map(int, input().split())
a = []
for i in range(n):
a += list(map(int, input().split()))[1:]
print( 'YES' if len(set(a)) == m else 'NO')
``` | output | 1 | 86,138 | 7 | 172,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n, m = [int(s) for s in input().split(' ')]
bulbs = set()
for i in range(n):
for bulb in input().split(' ')[1:]:
bulbs.add(int(bulb))
if len(bulbs) < m:
print('NO')
else:
print('YES')
``` | instruction | 0 | 86,139 | 7 | 172,278 |
Yes | output | 1 | 86,139 | 7 | 172,279 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n,m=input().split();
n=int(n);
m=int(m);
a=[1]*(m+1);
for i in range(n):
k=input().split();
w=int(k[0]);
for j in range(w):
x=int(k[j+1]);
m-=a[x];
a[x]=0;
if m==0 :print('YES');
else :print('NO');
``` | instruction | 0 | 86,140 | 7 | 172,280 |
Yes | output | 1 | 86,140 | 7 | 172,281 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
'''
Amirhossein Alimirzaei
Telegram : @HajLorenzo
Instagram : amirhossein_alimirzaei
University of Bojnourd
'''
n=list(map(int,input().split()))
blb=[]
for _ in range(n[0]):
tmp=(list(map(int,input().split())))
for __ in tmp[1:]:
if __ not in blb:
blb.append(__)
print("YES" if len(blb)==n[1] else "NO")
``` | instruction | 0 | 86,141 | 7 | 172,282 |
Yes | output | 1 | 86,141 | 7 | 172,283 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n,m = map(int,input().split())
ls=[]
x = []
no=[]
B=[]
for i in range(1,m+1):
B.append(i)
while(n):
n-=1
ls = list(map(int,input().split()))
x.append(ls[0])
for i in ls[1:]:
no.append(i)
no = set(no)
no=list(no)
if(no==B):
print('YES')
else:
print('NO')
``` | instruction | 0 | 86,142 | 7 | 172,284 |
Yes | output | 1 | 86,142 | 7 | 172,285 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 3 13:36:06 2020
@author: zuoxichen
"""
s1=input()
l1=s1.split()
n=int(l1[0])
m=int(l1[1])
i=0
l4=list()
while i<n:
s2=input()
l2=s2.split()
l3=list(int(l) for l in l2)
l3.pop(0)
for k in l3:
l4.append(k)
i+=1
else:
for i in range(1,m):
if i not in l4:
print('NO')
break
else:
print('YES')
``` | instruction | 0 | 86,143 | 7 | 172,286 |
No | output | 1 | 86,143 | 7 | 172,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
my_input = input()
buttons,bulbs = my_input.split()
bulbs = int(bulbs)
temp = []
for x in range(int(buttons)):
inp = input()
my_list = inp.split()
for item in my_list:
if item not in temp:
temp.append(item)
if len(temp) == bulbs:
print("YES")
else:
print("NO")
``` | instruction | 0 | 86,144 | 7 | 172,288 |
No | output | 1 | 86,144 | 7 | 172,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
import sys
si = True
i=0
j=0
n=[]
no=[]
m=None
for line in sys.stdin:
l = line.split()
if(len(l)!=0):
#print(l)
if(i==0):
m=int(l[1])
for x in range(m):
n.append(x+1)
#print("n=", n)
else:
for x in range(int(l[0])):
#print("'", l[x+1], "'")
if(int(l[x+1]) in n and not int(l[x+1]) in n):
#print("true")
no.append(int(l[x+1]))
j+=1
else: si = False
i+=1
#print(i)
else: break
#print("j= ", j)
if(si and j==len(n)): print("YES")
else: print("NO")
``` | instruction | 0 | 86,145 | 7 | 172,290 |
No | output | 1 | 86,145 | 7 | 172,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 ≤ n, m ≤ 100) — the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 ≤ xi ≤ m) — the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 ≤ yij ≤ m) — the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
num_but, num_lmp = map(int,input().split(" "))
switched = list("-" * num_lmp)
for i in range(num_but):
print("inputting button {}".format(i))
inp = input()
inp = inp.split(" ")
inp = inp[1:]
for lamp in inp:
print("switching on lamp {}".format(lamp))
switched[int(lamp) - 1] = "+"
for lamp in range(len(switched)):
if switched[lamp] == "-":
print("NO")
break
else:
print("YES")
``` | instruction | 0 | 86,146 | 7 | 172,292 |
No | output | 1 | 86,146 | 7 | 172,293 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,110 | 7 | 174,220 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n,a,b,c,d = map(int, input().split())
num = 0
i = 0
while n > i:
i += 1
if i + b - c < 1 or i + b - c > n:
continue
if i + a - d < 1 or i + a - d > n:
continue
if i + a - d + b - c < 1 or i + a - d + b - c > n:
continue
num += 1
print(num * n)
``` | output | 1 | 87,110 | 7 | 174,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,111 | 7 | 174,222 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
import sys
sys.stderr = sys.stdout
def painting(n, a, b, c, d):
lo = min(a+b, c+d, a+c, b+d)
hi = max(a+b, c+d, a+c, b+d)
delta = hi - lo
return max(n - delta, 0) * n
def main():
n, a, b, c, d = readinti()
print(painting(n, a, b, c, d))
##########
def readint():
return int(input())
def readinti():
return map(int, input().split())
def readintt():
return tuple(readinti())
def readintl():
return list(readinti())
def readinttl(k):
return [readintt() for _ in range(k)]
def readintll(k):
return [readintl() for _ in range(k)]
def log(*args, **kwargs):
print(*args, **kwargs, file=sys.__stderr__)
if __name__ == '__main__':
main()
``` | output | 1 | 87,111 | 7 | 174,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,112 | 7 | 174,224 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
k = 0
for u in range(1, n + 1):
v = u + b - c
if (v >= 1) and (v <= n):
z = a + v - d
if (z >= 1) and (z <= n):
y = c + z - b
if (y >= 1) and (y <= n):
u = d + y - a
if (u >= 1) and (u <= n):
k += 1
print(k * n)
``` | output | 1 | 87,112 | 7 | 174,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,113 | 7 | 174,226 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
lo = min(a+b, a+c, c+d, b+d)
hi = max(a+b, a+c, c+d, b+d)
ans = 0
for x in range(1, n+1):
ans += max(0, n-(hi-lo))
print(ans)
``` | output | 1 | 87,113 | 7 | 174,227 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,114 | 7 | 174,228 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
from re import *
from sys import stderr
def readint():
return int(input())
def readfloat():
return float(input())
def readarray(N, foo=input):
return [foo() for i in range(N)]
def readlinearray(foo=int):
return map(foo, input().split())
def NOD(a, b):
while b:
a,b = b, a%b
return a
def gen_primes(max):
primes = [1]*(max+1)
for i in range(2, max+1):
if primes[i]:
for j in range(i+i, max+1, i):
primes[j] = 0
primes[0] = 0
return [x for x in range(max+1) if primes[x]]
def is_prime(N):
i = 3
if not(N % 2):
return 0
while i*i < N:
if not(N % i):
return 0
i += 3
return 1
n, a, b, c, d = readlinearray()
maxx = min(n, n - b + c, n - a + d, n - a + c - b + d)
minx = max(1, 1 - b + c, 1 - a + d, 1 - a + c - b + d)
if maxx < minx:
print(0)
else:
print((maxx - minx + 1) * n)
``` | output | 1 | 87,114 | 7 | 174,229 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,115 | 7 | 174,230 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
import math as m
n,a,b,c,d=(map(int,input().split()))
ans=0
for i in range(n):
x=i+1
t=x+b-c
curr=1
if t>=1 and t<=n:
curr=1
else:
curr=0
t=x+a-d
if t>=1 and t<=n and curr!=0:
curr=1
else:
curr*=0
t=x+a+b-d-c
if t>=1 and t<=n and curr!=0:
curr=1
else:
curr*=0
ans+=curr*n
# print(x,curr)
print(ans)
``` | output | 1 | 87,115 | 7 | 174,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,116 | 7 | 174,232 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
from collections import *
n,a,b,c,d = list(map(int,input().split()))
print((n-min(n,max(abs(a+b-c-d),abs(b-c),abs(a-d),abs(a+c-b-d))))*n)
``` | output | 1 | 87,116 | 7 | 174,233 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image> | instruction | 0 | 87,117 | 7 | 174,234 |
Tags: brute force, constructive algorithms, math
Correct Solution:
```
n, a, b, c, d = map(int, input().split())
count = 0
for x in range(1, n + 1):
w = x + c - b
if w >= 1 and w <= n:
y = a + w - d
if y >= 1 and y <= n:
z = a + x - d
if z >= 1 and z <= n:
count = count + 1
print(count * n)
``` | output | 1 | 87,117 | 7 | 174,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
for centro in range(1, n+1):
q1 = a + b + centro
q2 = a + c + centro
q3 = b + d + centro
q4 = c + d + centro
v = [q1, q2, q3, q4]
v.sort()
q_min = v[0]
q_max = v[-1]
if q_min + n < q_max + 1:
continue
elif q_min + n == q_max + 1:
ans += 1
else:
ans += (q_min + n) - (q_max + 1) + 1
print(ans)
``` | instruction | 0 | 87,118 | 7 | 174,236 |
Yes | output | 1 | 87,118 | 7 | 174,237 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
import sys,os,io
import math,bisect,operator
inf,mod = float('inf'),10**9+7
# sys.setrecursionlimit(10 ** 6)
from itertools import groupby,accumulate
from heapq import heapify,heappop,heappush
from collections import deque,Counter,defaultdict
input = iter(sys.stdin.buffer.read().decode().splitlines()).__next__
Neo = lambda : list(map(int,input().split()))
# test, = Neo()
n,a,b,c,d = Neo()
Ans = 0
for i in range(1,n+1):
p = i+a+b
q = p-a-c
r = p-b-d
s = p-c-d
if 1<=q<=n and 1<=r<=n and 1<=s<=n:
Ans += 1
print(n*Ans)
``` | instruction | 0 | 87,119 | 7 | 174,238 |
Yes | output | 1 | 87,119 | 7 | 174,239 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = n * (n - abs(a - d) - abs(b - c))
print(max(0, ans))
``` | instruction | 0 | 87,120 | 7 | 174,240 |
Yes | output | 1 | 87,120 | 7 | 174,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n,a,b,c,d=tuple(map(int,input().split()))
ans=0
sum=0
def ch(c):
return (n>=c and c>0)
for i in range(1,n+1):
sum=i+a+b
x2=sum-a-c
x3=sum-c-d
x4=sum-b-d
if ch(i) and ch(x2) and ch(x3) and ch(x4):
ans+=1
print(ans*n)
``` | instruction | 0 | 87,121 | 7 | 174,242 |
Yes | output | 1 | 87,121 | 7 | 174,243 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n,a,b,c,d=map(int,input().split())
print((max(a+b+1,b+c+1,c+d+1,a+d+1)-min(a+b+n,b+c+n,c+d+n,d+a+n)+1)*n)
``` | instruction | 0 | 87,122 | 7 | 174,244 |
No | output | 1 | 87,122 | 7 | 174,245 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
nabcd = list(map(int, input().split()))
RANGE = range(1,nabcd[0])
a = nabcd[1]
b = nabcd[2]
c = nabcd[3]
d = nabcd[4]
dif1 = a-d
dif2 = b-c
dif = abs(dif1) + abs(dif2)
counter = nabcd[0]*(nabcd[0]-dif)
print(counter)
``` | instruction | 0 | 87,123 | 7 | 174,246 |
No | output | 1 | 87,123 | 7 | 174,247 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
l = [a+b, a+c, b+d, d+c]
sum = (min(l)+n)-(max(l)+1)+1
ans = sum*n
print(ans)
``` | instruction | 0 | 87,124 | 7 | 174,248 |
No | output | 1 | 87,124 | 7 | 174,249 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya works as a watchman in the gallery. Unfortunately, one of the most expensive paintings was stolen while he was on duty. He doesn't want to be fired, so he has to quickly restore the painting. He remembers some facts about it.
* The painting is a square 3 × 3, each cell contains a single integer from 1 to n, and different cells may contain either different or equal integers.
* The sum of integers in each of four squares 2 × 2 is equal to the sum of integers in the top left square 2 × 2.
* Four elements a, b, c and d are known and are located as shown on the picture below.
<image>
Help Vasya find out the number of distinct squares the satisfy all the conditions above. Note, that this number may be equal to 0, meaning Vasya remembers something wrong.
Two squares are considered to be different, if there exists a cell that contains two different integers in different squares.
Input
The first line of the input contains five integers n, a, b, c and d (1 ≤ n ≤ 100 000, 1 ≤ a, b, c, d ≤ n) — maximum possible value of an integer in the cell and four integers that Vasya remembers.
Output
Print one integer — the number of distinct valid squares.
Examples
Input
2 1 1 1 2
Output
2
Input
3 3 1 2 3
Output
6
Note
Below are all the possible paintings for the first sample. <image> <image>
In the second sample, only paintings displayed below satisfy all the rules. <image> <image> <image> <image> <image> <image>
Submitted Solution:
```
n, a, b, c, d = map(int, input().split())
ans = 0
if n >= 10 :
ans = 0
else :
S1 = set([a,b,c,d])
for i in range(1, n+1) :
s = a+b+i
j = s - a - c
k = s - b - d
l = s - c - d
if 1 <= j <=n and 1<= k <= n and 1<= l <= n :
S2 = set([i, j, k, l])
if len(S1 | S2) == n :
ans += n
elif len(S1 | S2) == n-1 :
ans += 1
print(ans)
``` | instruction | 0 | 87,125 | 7 | 174,250 |
No | output | 1 | 87,125 | 7 | 174,251 |
Provide a correct Python 3 solution for this coding contest problem.
We have 3N colored balls with IDs from 1 to 3N. A string S of length 3N represents the colors of the balls. The color of Ball i is red if S_i is `R`, green if S_i is `G`, and blue if S_i is `B`. There are N red balls, N green balls, and N blue balls.
Takahashi will distribute these 3N balls to N people so that each person gets one red ball, one blue ball, and one green ball. The people want balls with IDs close to each other, so he will additionally satisfy the following condition:
* Let a_j < b_j < c_j be the IDs of the balls received by the j-th person in ascending order.
* Then, \sum_j (c_j-a_j) should be as small as possible.
Find the number of ways in which Takahashi can distribute the balls. Since the answer can be enormous, compute it modulo 998244353. We consider two ways to distribute the balls different if and only if there is a person who receives different sets of balls.
Constraints
* 1 \leq N \leq 10^5
* |S|=3N
* S consists of `R`, `G`, and `B`, and each of these characters occurs N times in S.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of ways in which Takahashi can distribute the balls, modulo 998244353.
Examples
Input
3
RRRGGGBBB
Output
216
Input
5
BBRGRRGRGGRBBGB
Output
960 | instruction | 0 | 87,336 | 7 | 174,672 |
"Correct Solution:
```
"""
Writer: SPD_9X2
https://atcoder.jp/contests/agc037/tasks/agc037_b
最適な配り方とは何かを考えるべき
そうすれば自然なdpに落ちるんだろうな
RGBが出たら合体させないといけない？
→これは正しそう
RGGRBB
112212 →4+3=7
112221 →5+2=7
121212 →4+4=8
なので、RGBが出たらランダムに合わせればいいわけではない
2つでたら合体しなければいけない説
RG→合体
GR→合体
B →2通り
B →1通り
なので2通り？
それっぽいぞ
BBRGRRGRGGRBBGB
B1/B
B1/B2
R2/RB,B
G2/B
R2/RB
R2/R,RB
G2/R
R2/R2
G4/RG,R
G4/RG2
R4/RG2,R
B8/RG,R
B8/R
G8/RG
B8/
8通り…少なすぎるな
あ、5!を掛ければ答えじゃね
あってるっぽさそう？
"""
N = int(input())
S = input()
R = 0
G = 0
B = 0
RG = 0
RB = 0
GB = 0
ans = 1
mod = 998244353
for i in range(1,N+1):
ans *= i
ans %= mod
for i in range(3*N):
if S[i] == "R":
if GB > 0:
ans *= GB
GB -= 1
elif G > 0:
ans *= G
G -= 1
RG += 1
elif B > 0:
ans *= B
B -= 1
RB += 1
else:
R += 1
elif S[i] == "G":
if RB > 0:
ans *= RB
RB -= 1
elif R > 0:
ans *= R
R -= 1
RG += 1
elif B > 0:
ans *= B
B -= 1
GB += 1
else:
G += 1
else:
if RG > 0:
ans *= RG
RG -= 1
elif R > 0:
ans *= R
R -= 1
RB += 1
elif G > 0:
ans *= G
G -= 1
GB += 1
else:
B += 1
ans %= mod
print (ans)
``` | output | 1 | 87,336 | 7 | 174,673 |
Provide a correct Python 3 solution for this coding contest problem.
We have 3N colored balls with IDs from 1 to 3N. A string S of length 3N represents the colors of the balls. The color of Ball i is red if S_i is `R`, green if S_i is `G`, and blue if S_i is `B`. There are N red balls, N green balls, and N blue balls.
Takahashi will distribute these 3N balls to N people so that each person gets one red ball, one blue ball, and one green ball. The people want balls with IDs close to each other, so he will additionally satisfy the following condition:
* Let a_j < b_j < c_j be the IDs of the balls received by the j-th person in ascending order.
* Then, \sum_j (c_j-a_j) should be as small as possible.
Find the number of ways in which Takahashi can distribute the balls. Since the answer can be enormous, compute it modulo 998244353. We consider two ways to distribute the balls different if and only if there is a person who receives different sets of balls.
Constraints
* 1 \leq N \leq 10^5
* |S|=3N
* S consists of `R`, `G`, and `B`, and each of these characters occurs N times in S.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of ways in which Takahashi can distribute the balls, modulo 998244353.
Examples
Input
3
RRRGGGBBB
Output
216
Input
5
BBRGRRGRGGRBBGB
Output
960 | instruction | 0 | 87,337 | 7 | 174,674 |
"Correct Solution:
```
from bisect import*
e=enumerate
M=998244353
R,G,B=[],[],[]
n=int(input())
for i,c in e(input()):
if c=='R':R+=i,
elif c=='G':G+=i,
else:B+=i,
l,m,r=[],[],[]
for i,t in e(zip(R,G,B)):
a,b,c=sorted(t)
l+=a,
m+=b,
r+=c,
s=1
for i,b in e(m):
s=s*(bisect(l,b)-i)*(i-bisect(r,b)+1)*(i+1)%M
print(s)
``` | output | 1 | 87,337 | 7 | 174,675 |
Provide a correct Python 3 solution for this coding contest problem.
We have 3N colored balls with IDs from 1 to 3N. A string S of length 3N represents the colors of the balls. The color of Ball i is red if S_i is `R`, green if S_i is `G`, and blue if S_i is `B`. There are N red balls, N green balls, and N blue balls.
Takahashi will distribute these 3N balls to N people so that each person gets one red ball, one blue ball, and one green ball. The people want balls with IDs close to each other, so he will additionally satisfy the following condition:
* Let a_j < b_j < c_j be the IDs of the balls received by the j-th person in ascending order.
* Then, \sum_j (c_j-a_j) should be as small as possible.
Find the number of ways in which Takahashi can distribute the balls. Since the answer can be enormous, compute it modulo 998244353. We consider two ways to distribute the balls different if and only if there is a person who receives different sets of balls.
Constraints
* 1 \leq N \leq 10^5
* |S|=3N
* S consists of `R`, `G`, and `B`, and each of these characters occurs N times in S.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of ways in which Takahashi can distribute the balls, modulo 998244353.
Examples
Input
3
RRRGGGBBB
Output
216
Input
5
BBRGRRGRGGRBBGB
Output
960 | instruction | 0 | 87,338 | 7 | 174,676 |
"Correct Solution:
```
#!/usr/bin/env python3
MOD = 998244353
n = int(input())
s = input()
ans = 1
for i in range(n):
ans *= i + 1
ans %= MOD
cnt = [0] * 3
dic = {'R': 0, 'G': 1, 'B': 2}
for c in s:
p = dic[c]
m = min(cnt[p - 1], cnt[p - 2])
if m > 0:
ans *= m
cnt[p - 1] -= 1
cnt[p - 2] -= 1
else:
ans *= max(1, cnt[p - 1] + cnt[p - 2] - cnt[p])
cnt[p] += 1
ans %= MOD
print(ans)
``` | output | 1 | 87,338 | 7 | 174,677 |
Provide a correct Python 3 solution for this coding contest problem.
We have 3N colored balls with IDs from 1 to 3N. A string S of length 3N represents the colors of the balls. The color of Ball i is red if S_i is `R`, green if S_i is `G`, and blue if S_i is `B`. There are N red balls, N green balls, and N blue balls.
Takahashi will distribute these 3N balls to N people so that each person gets one red ball, one blue ball, and one green ball. The people want balls with IDs close to each other, so he will additionally satisfy the following condition:
* Let a_j < b_j < c_j be the IDs of the balls received by the j-th person in ascending order.
* Then, \sum_j (c_j-a_j) should be as small as possible.
Find the number of ways in which Takahashi can distribute the balls. Since the answer can be enormous, compute it modulo 998244353. We consider two ways to distribute the balls different if and only if there is a person who receives different sets of balls.
Constraints
* 1 \leq N \leq 10^5
* |S|=3N
* S consists of `R`, `G`, and `B`, and each of these characters occurs N times in S.
Input
Input is given from Standard Input in the following format:
N
S
Output
Print the number of ways in which Takahashi can distribute the balls, modulo 998244353.
Examples
Input
3
RRRGGGBBB
Output
216
Input
5
BBRGRRGRGGRBBGB
Output
960 | instruction | 0 | 87,339 | 7 | 174,678 |
"Correct Solution:
```
from functools import reduce
N = int(input())
S = input()
MOD = 998244353
class ModInt:
def __init__(self, x):
self.x = x % MOD
def __str__(self):
return str(self.x)
__repr__ = __str__
def __add__(self, other):
return (
ModInt(self.x + other.x) if isinstance(other, ModInt) else
ModInt(self.x + other)
)
def __sub__(self, other):
return (
ModInt(self.x - other.x) if isinstance(other, ModInt) else
ModInt(self.x - other)
)
def __mul__(self, other):
return (
ModInt(self.x * other.x) if isinstance(other, ModInt) else
ModInt(self.x * other)
)
def __truediv__(self, other):
return (
ModInt(
self.x * pow(other.x, MOD - 2, MOD)
) if isinstance(other, ModInt) else
ModInt(self.x * pow(other, MOD - 2, MOD))
)
def __pow__(self, other):
return (
ModInt(
pow(self.x, other.x, MOD)
) if isinstance(other, ModInt) else
ModInt(pow(self.x, other, MOD))
)
def __radd__(self, other):
return ModInt(other + self.x)
def __rsub__(self, other):
return ModInt(other - self.x)
def __rmul__(self, other):
return ModInt(other * self.x)
def __rtruediv__(self, other):
return ModInt(other * pow(self.x, MOD - 2, MOD))
def __rpow__(self, other):
return ModInt(pow(other, self.x, MOD))
# 先頭のボールから順に、割り当て可能な人のうち最もボールを持っている人に割り当てることを繰り返す
def f(s, res, z, r, g, b, rg, gb, br):
return (
(
(res * gb, z, r, g, b, rg, gb - 1, br) if gb else
(res * g, z, r, g - 1, b, rg + 1, gb, br) if g else
(res * b, z, r, g, b - 1, rg, gb, br + 1) if b else
(res * z, z - 1, r + 1, g, b, rg, gb, br)
) if s == 'R' else
(
(res * br, z, r, g, b, rg, gb, br - 1) if br else
(res * r, z, r - 1, g, b, rg + 1, gb, br) if r else
(res * b, z, r, g, b - 1, rg, gb + 1, br) if b else
(res * z, z - 1, r, g + 1, b, rg, gb, br)
) if s == 'G' else
(
(res * rg, z, r, g, b, rg - 1, gb, br) if rg else
(res * r, z, r - 1, g, b, rg, gb, br + 1) if r else
(res * g, z, r, g - 1, b, rg, gb + 1, br) if g else
(res * z, z - 1, r, g, b + 1, rg, gb, br)
)
)
ans, *_ = reduce(
lambda acc, s: f(s, *acc),
S,
(ModInt(1), N, 0, 0, 0, 0, 0, 0)
)
print(ans)
``` | output | 1 | 87,339 | 7 | 174,679 |
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