text stringlengths 216 39.6k | conversation_id int64 219 108k | embedding list | cluster int64 11 11 |
|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
6
2
3
1
1
4
2
Output
Yes
Yes
Yes
No
Yes
No
Submitted Solution:
```
n = int(input())
M = 40
one = 0
rest = 2**M
for i in range(n):
x = int(input())
if x > M:
if not one:
if rest > 0:
print("Yes")
one = 1
rest -= 1
else:
print("No")
else:
print("Yes")
else:
val = 2**(M - x)
if val <= rest:
print("Yes")
rest -= val
else:
print("No")
```
No
| 41,146 | [
0.5009765625,
-0.061309814453125,
-0.1822509765625,
-0.130615234375,
-0.80322265625,
-0.4072265625,
0.07574462890625,
0.1724853515625,
0.05914306640625,
0.9306640625,
0.486328125,
0.16748046875,
0.03106689453125,
-0.71875,
-0.34521484375,
-0.10345458984375,
-0.466064453125,
-0.7998... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
6
2
3
1
1
4
2
Output
Yes
Yes
Yes
No
Yes
No
Submitted Solution:
```
c=p=0;b=[0]*1000000
for _ in range(int(input())):
x=int(input())
if (c==x and p!=x) or b[0]==1 or c>x:print('NO');continue
print('YES')
if x>=1000000:continue
p+=1;b[x]+=1
while b[x]>1:p-=1;b[x]-=2;b[x-1]+=1;x-=1
while b[c+1]==1 and c<999999:c+=1
```
No
| 41,147 | [
0.60107421875,
-0.1175537109375,
0.002132415771484375,
0.047607421875,
-0.46923828125,
-0.47314453125,
0.05755615234375,
0.37060546875,
0.009002685546875,
0.86279296875,
0.560546875,
0.16455078125,
-0.0139312744140625,
-0.685546875,
-0.49951171875,
-0.12493896484375,
-0.46337890625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
6
2
3
1
1
4
2
Output
Yes
Yes
Yes
No
Yes
No
Submitted Solution:
```
n = int(input())
bits = set()
count = 0
ans = []
for i in range(n):
x = int(input())
if x in bits:
y = x
while y in bits:
y -= 1
if y == 0 and count != x:
ans.append("No")
else:
bits.add(y)
if y == 0:
ans.append("Yes")
if i <= n-2:
ans.append("No\n"*(n-2-i) + "No")
break
count += 1
for i in range(y+1, x+1):
bits.remove(i)
count -= 1
ans.append("Yes")
else:
bits.add(x)
count += 1
ans.append("Yes")
print("\n".join(ans))
```
No
| 41,148 | [
0.415283203125,
-0.1151123046875,
-0.0204620361328125,
-0.12237548828125,
-0.7490234375,
-0.6357421875,
-0.1385498046875,
0.29541015625,
0.163330078125,
1.02734375,
0.175048828125,
-0.06268310546875,
0.012176513671875,
-0.79541015625,
-0.460693359375,
-0.30615234375,
-0.477294921875,... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
from collections import defaultdict,deque
import sys,heapq,bisect,math,itertools,string,queue,datetime
sys.setrecursionlimit(10**8)
INF = float('inf')
mod = 10**9+7
eps = 10**-7
def inpl(): return list(map(int, input().split()))
def inpl_str(): return list(input().split())
while True:
m,nmin,nmax = inpl()
if m == 0:
break
else:
pp = [int(input()) for _ in range(m)]
pp.sort(reverse=True)
ans = 0
ansgap = -1
for n in range(nmin,nmax+1):
gap = pp[n-1] - pp[n]
if ansgap <= gap:
ansgap = gap
ans = n
print(ans)
```
Yes
| 41,157 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
def solver():
m, nmin, nmax = map(int, input().split())
if m == 0 and nmin == 0 and nmax == 0:
return None
p = [int(input()) for _ in range(m)]
ans = 0
maxdelta = 0
for a in range(nmin, nmax+1):
delta = p[a-1] - p[a]
if delta >= maxdelta:
maxdelta = delta
ans = a
return ans
outputs = []
while True:
ans = solver()
if ans == None:
break
outputs.append(ans)
print('\n'.join(map(str, outputs)))
```
Yes
| 41,158 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
while 1:
m,nmi,nma=map(int,input().split())
if not m and not nmi and not nma:break
l=[int(input()) for _ in range(m)]
tg=[l[:i][-1]-l[i:][0] for i in range(nmi,nma+1)]
print(nma-tg[::-1].index(max(tg)))
```
Yes
| 41,159 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
# coding: utf-8
while 1:
m,n_min,n_max=map(int,input().split())
if m==0:
break
data=[]
for i in range(m):
data.append(int(input()))
data=sorted(data,reverse=True)
mx=(-1,-1)
for i in range(n_min,n_max+1):
if data[i-1]-data[i]>=mx[0]:
mx=(data[i-1]-data[i],i)
print(mx[1])
```
Yes
| 41,160 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
head = input().split(" ")
while head != "0 0 0":
num = int(head[0])
mini = int(head[1])
maxi = int(head[2])
scores = []
gaps = []
for _ in range(int(head[0])):
scores.append(int(input()))
for i in range(mini, maxi + 1):
oks = scores[:i]
ngs = scores[i:]
gap = oks[-1] - ngs[0]
gaps.append((i,gap))
filtered = list(filter(lambda x: x[1] == max(map(lambda x: x[1], gaps)), gaps))
sorted(filtered,key=lambda x: x[0])
print(filtered[-1][0])
```
No
| 41,161 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
while 1:
m,nmi,nma=map(int,input().split())
if not m and not nmi and not nma:break
l=[int(input()) for _ in range(m)]
tg=[l[:i][-1]-l[i:][0] for i in range(nmi,nma+1)]
print(tg.index(max(tg))+nmi)
```
No
| 41,162 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
head = input().split(" ")
while head != "0 0 0":
num = int(head[0])
mini = int(head[1])
maxi = int(head[2])
scores = []
gaps = []
for i in range(int(head[0])):
scores.append(int(input()))
for i in range(mini, maxi):
oks = scores[:i]
ngs = scores[i:]
gap = oks[-1] - ngs[0]
gaps.append((i,gap))
filtered = list(filter(lambda x: x[1] == max(gaps), gaps))
sorted(filtered, lambda x: x[0], True)
print(filtered[0][0])
```
No
| 41,163 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Entrance Examination
The International Competitive Programming College (ICPC) is famous for its research on competitive programming. Applicants to the college are required to take its entrance examination.
The successful applicants of the examination are chosen as follows.
* The score of any successful applicant is higher than that of any unsuccessful applicant.
* The number of successful applicants n must be between nmin and nmax, inclusive. We choose n within the specified range that maximizes the gap. Here, the gap means the difference between the lowest score of successful applicants and the highest score of unsuccessful applicants.
* When two or more candidates for n make exactly the same gap, use the greatest n among them.
Let's see the first couple of examples given in Sample Input below. In the first example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 82, 70, and 65. For n of two, three and four, the gaps will be 8, 12, and 5, respectively. We must choose three as n, because it maximizes the gap.
In the second example, nmin and nmax are two and four, respectively, and there are five applicants whose scores are 100, 90, 80, 75, and 65. For n of two, three and four, the gap will be 10, 5, and 10, respectively. Both two and four maximize the gap, and we must choose the greatest number, four.
You are requested to write a program that computes the number of successful applicants that satisfies the conditions.
Input
The input consists of multiple datasets. Each dataset is formatted as follows.
> m nmin nmax
> P1
> P2
> ...
> Pm
>
The first line of a dataset contains three integers separated by single spaces. m represents the number of applicants, nmin represents the minimum number of successful applicants, and nmax represents the maximum number of successful applicants. Each of the following m lines contains an integer Pi, which represents the score of each applicant. The scores are listed in descending order. These numbers satisfy 0 < nmin < nmax < m ≤ 200, 0 ≤ Pi ≤ 10000 (1 ≤ i ≤ m) and Pnmin > Pnmax+1. These ensure that there always exists an n satisfying the conditions.
The end of the input is represented by a line containing three zeros separated by single spaces.
Output
For each dataset, output the number of successful applicants in a line.
Sample Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output for the Sample Input
3
4
2
2
3
2
Example
Input
5 2 4
100
90
82
70
65
5 2 4
100
90
80
75
65
3 1 2
5000
4000
3000
4 2 3
10000
10000
8000
8000
4 2 3
10000
10000
10000
8000
5 2 3
100
80
68
60
45
0 0 0
Output
3
4
2
2
3
2
Submitted Solution:
```
while True:
P = list()
ans = list()
tans = list()
m, Mn, Mx = map(int, input().split())
if m == 0 and Mn == 0 and Mx == 0:
break
for i in range(0, m):
P.append(int(input()))
for j in range(Mn-1, Mx, 1):
ans.append(P[j]-P[j+1])
print(ans)
for k in range(0,len(ans)):
if ans[k] == max(ans):
tans.append(k)
print(tans)
print(max(tans)+Mn)
```
No
| 41,164 | [
0.279052734375,
-0.0760498046875,
-0.2030029296875,
0.0587158203125,
-0.63037109375,
-0.2283935546875,
-0.14501953125,
0.1578369140625,
-0.1236572265625,
0.9345703125,
0.69970703125,
0.278564453125,
0.1827392578125,
-1.09765625,
-0.2381591796875,
0.397216796875,
-0.53564453125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
class State:
def __init__(self, state, operations):
self.states_num = {state: 0}
self.states = [state]
self._path = {}
self.DIAMETER = 0
new_sts = [0]
while new_sts:
self.DIAMETER += 1
cur_sts, new_sts = new_sts, []
for i in cur_sts:
for op, f in enumerate(operations):
state = f(self.states[i])
if state not in self.states_num:
j = self.states_num[state] = len(self.states)
self.states.append(state)
new_sts.append(j)
else:
j = self.states_num[state]
self._path[(i, j)] = (i, op)
self.SIZE = n = len(self.states_num)
self.distance = dis = [[0 if i == j else 10 ** 6 for i in range(n)] for j in range(n)]
for i, j in self._path.keys():
self.distance[i][j] = 1
for k in range(n):
for i in range(n):
for j in range(n):
if dis[i][k] + dis[k][j] < dis[i][j]:
dis[i][j] = dis[i][k] + dis[k][j]
self._path[(i, j)] = self._path[(k, j)]
def path(self, st1, st2):
u = self.states_num[st1]
v = self.states_num[st2]
result = []
while u != v:
v, op = self._path[(u, v)]
result.append(op)
return result[::-1]
def min_path(self, st1, states):
result = [-1] * self.DIAMETER
choice = -1
for i, st2 in enumerate(states):
path = self.path(st1, st2)
if len(path) < len(result):
result = path
choice = i
return result, choice
# ############################## main
def op0(st):
# left top
return st[0], st[1] ^ 1, st[2] ^ 1, st[3] ^ 1
def op1(st):
# right top
return st[0] ^ 1, st[1], st[2] ^ 1, st[3] ^ 1
def op2(st):
# left bottom
return st[0] ^ 1, st[1] ^ 1, st[2], st[3] ^ 1
def op3(st):
# right bottom
return st[0] ^ 1, st[1] ^ 1, st[2] ^ 1, st[3]
zero = (0, 0, 0, 0)
ST = State(zero, (op0, op1, op2, op3))
right_zero = (zero, (0, 0, 1, 0), (1, 0, 0, 0), (1, 0, 1, 0))
bottom_zero = (zero, (0, 1, 0, 0), (1, 0, 0, 0), (1, 1, 0, 0))
def solve():
n, m = mpint()
matrix = [list(map(int, inp())) for _ in range(n)]
output = []
if m & 1:
# deal with right column
for i in range(n - 1):
y, x = i, m - 2
st = (matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1])
path, choice = ST.min_path(st, right_zero)
matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1] = right_zero[choice]
y += 1
x += 1
for op in path:
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
del pos[op << 1], pos[op << 1]
output.append(pos)
# for l in matrix:
# print(*l)
# print()
if n & 1:
# deal with bottom row
for j in range(m - 1 - (m & 1)):
y, x = n - 2, j
st = (matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1])
path, choice = ST.min_path(st, bottom_zero)
matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1] = bottom_zero[choice]
y += 1
x += 1
for op in path:
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
del pos[op << 1], pos[op << 1]
output.append(pos)
# for l in matrix:
# print(*l)
# print()
for i in range(n >> 1):
for j in range(m >> 1):
# left top position
y, x = i << 1, j << 1
st = (matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1])
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
for k in range(4):
matrix[pos[k * 2]][pos[k * 2 + 1]] = 0
y += 1
x += 1
for op in ST.path(st, zero):
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
del pos[op << 1], pos[op << 1]
output.append(pos)
assert len(output) <= n * m
# for l in matrix:
# print(*l)
# assert not any(l)
print(len(output))
for out in output:
print(*out)
def main():
# solve()
# print(solve())
for _ in range(itg()):
solve()
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
DEBUG = 0
URL = ''
if __name__ == '__main__':
if DEBUG == 1:
import requests # ImportError: cannot import name 'md5' from 'sys' (unknown location)
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
elif DEBUG == 2:
main()
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
main()
# Please check!
```
Yes
| 41,473 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import io
import os
#input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def solve():
n, m = map(int, input().split())
vals = []
for _ in range(n):
vals.append(list(map(int, input())))
ops = []
def flip(x1, y1, x2, y2, x3, y3):
for a, b in ([x1, y1], [x2, y2], [x3, y3]):
vals[b][a] = 1 - vals[b][a]
ops.append((y1+1, x1+1, y2+1, x2+1, y3+1, x3+1))
for y in range(n-2):
for x in range(m):
if x == m-1:
if vals[y][x]:
flip(x, y, x, y+1, x-1, y+1)
else:
if vals[y][x]:
flip(x, y, x, y+1, x+1, y+1)
for x in range(m-2):
if vals[n-2][x]:
flip(x, n-2, x+1, n-2, x+1, n-1)
if vals[n-1][x]:
flip(x, n-1, x+1, n-2, x+1, n-1)
# a b
# c d
a = m-2, n-2
b = m-1, n-2
c = m-2, n-1
d = m-1, n-1
def last():
return [vals[y][x] for x, y in [a, b, c, d]]
if sum(last()) == 4:
flip(*a, *b, *c)
if sum(last()) == 1:
i = last().index(1)
if i == 0:
flip(*a, *b, *c)
elif i == 1:
flip(*b, *d, *c)
elif i == 2:
flip(*a, *c, *d)
else:
flip(*b, *d, *c)
if sum(last()) == 2:
if last() == [1, 1, 0, 0]:
flip(*a, *c, *d)
elif last() == [1, 0, 1, 0]:
flip(*a, *b, *d)
elif last() == [1, 0, 0, 1]:
flip(*a, *b, *c)
elif last() == [0, 1, 1, 0]:
flip(*a, *b, *d)
elif last() == [0, 1, 0, 1]:
flip(*a, *b, *c)
else:
flip(*a, *b, *c)
if sum(last()) == 3:
i = last().index(0)
if i == 0:
flip(*b, *c, *d)
elif i == 1:
flip(*a, *c, *d)
elif i == 2:
flip(*a, *b, *d)
else:
flip(*a, *b, *c)
for row in vals:
assert(not any(row))
print(len(ops))
for op in ops:
print(*op)
t = int(input())
for _ in range(t):
solve()
```
Yes
| 41,474 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import math
a=int(input())
arr=[]
n,m=0,0
def flip(i):
if(i==0):
return 1
return 0
def p(ans):
print(len(ans))
for i in range(0,len(ans)):
for j in range(0,len(ans[0])):
ans[i][j]=ans[i][j]+1
for i in ans:
print(' '.join(map(str,i)))
def o1(temp):
temp.append([n-2,m-2,n-2,m-1,n-1,m-1])
arr[n-2][m-2]=flip(arr[n-2][m-2])
arr[n-2][m-1]=flip(arr[n-2][m-1])
arr[n-1][m-1]=flip(arr[n-1][m-1])
def o2(temp):
temp.append([n-1,m-1,n-2,m-1,n-1,m-2])
arr[n-1][m-1]=flip(arr[n-1][m-1])
arr[n-2][m-1]=flip(arr[n-2][m-1])
arr[n-1][m-2]=flip(arr[n-1][m-2])
def o3(temp):
temp.append([n-1,m-1,n-1,m-2,n-2,m-2])
arr[n-1][m-1]=flip(arr[n-1][m-1])
arr[n-1][m-2]=flip(arr[n-1][m-2])
arr[n-2][m-2]=flip(arr[n-2][m-2])
def o4(temp):
temp.append([n-2,m-2,n-1,m-2,n-2,m-1])
arr[n-2][m-2]=flip(arr[n-2][m-2])
arr[n-1][m-2]=flip(arr[n-1][m-2])
arr[n-2][m-1]=flip(arr[n-2][m-1])
for i in range(a):
n,m=map(int,input().split())
arr=[]
for i in range(n):
s=input()
arr.append([int(i) for i in s])
ans=[]
for i in range(0,n-2):
for j in range(0,m):
if(arr[i][j]==1 and j<m-1):
ans.append([i,j,i,j+1,i+1,j])
arr[i][j]=flip(arr[i][j])
arr[i][j+1]=flip(arr[i][j+1])
arr[i+1][j]=flip(arr[i+1][j])
elif(arr[i][j]==1):
ans.append([i,j,i+1,j,i+1,j-1])
arr[i][j]=flip(arr[i][j])
arr[i+1][j]=flip(arr[i+1][j])
arr[i+1][j-1]=flip(arr[i+1][j-1])
for j in range(0,m-2):
for i in range(n-2,n):
if(arr[i][j]==1):
if(i==n-2):
ans.append([i,j,i,j+1,i+1,j+1])
arr[i][j]=flip(arr[i][j])
arr[i][j+1]=flip(arr[i][j+1])
arr[i+1][j+1]=flip(arr[i+1][j+1])
else:
ans.append([i,j,i,j+1,i-1,j+1])
arr[i][j]=flip(arr[i][j])
arr[i][j+1]=flip(arr[i][j+1])
arr[i-1][j+1]=flip(arr[i-1][j+1])
su=0
for i in range(n-2,n):
for j in range(m-2,m):
su=su+arr[i][j]
if(su==0):
p(ans)
if(su==1):
if(arr[n-2][m-2]==1):
o1(ans)
o3(ans)
o4(ans)
p(ans)
if(arr[n-1][m-1]==1):
o1(ans)
o2(ans)
o3(ans)
p(ans)
if(arr[n-2][m-1]==1):
o2(ans)
o1(ans)
o4(ans)
p(ans)
if(arr[n-1][m-2]==1):
o4(ans)
o2(ans)
o3(ans)
p(ans)
if(su==2):
if(arr[n-2][m-2]==1 and arr[n-2][m-1]==1):
o2(ans)
o3(ans)
if(arr[n-2][m-2]==1 and arr[n-1][m-2]==1):
o1(ans)
o2(ans)
if(arr[n-1][m-1]==1 and arr[n-2][m-1]==1):
o3(ans)
o4(ans)
if(arr[n-1][m-1]==1 and arr[n-1][m-2]==1):
o4(ans)
o1(ans)
if(arr[n-1][m-1]==1 and arr[n-2][m-2]==1):
o2(ans)
o4(ans)
if(arr[n-2][m-1]==1 and arr[n-1][m-2]==1):
o1(ans)
o3(ans)
p(ans)
if(su==4):
o2(ans)
o1(ans)
o3(ans)
o4(ans)
p(ans)
if(su==3):
temp=[]
for i in range(n-2,n):
for j in range(m-2,m):
if(arr[i][j]==1):
temp.append(i)
temp.append(j)
ans.append(temp)
p(ans)
```
Yes
| 41,475 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import copy
def two_two_solver(mat, i, j, ans):
oneind = []
zeroind = []
for x in range(i, i+2):
for y in range(j, j+2):
if mat[x][y] == '1':
oneind += [[x, y]]
else:
zeroind += [[x, y]]
if len(oneind) == 0:
return
if len(oneind) == 4:
newoneind = [oneind[0][:]]
newzeroind = [oneind[1][:], oneind[2][:], oneind[3][:]]
ans += [[oneind[1][:], oneind[2][:], oneind[3][:]]]
oneind = copy.deepcopy(newoneind)
zeroind = copy.deepcopy(newzeroind)
if len(oneind) == 1:
newoneind = [zeroind[0][:], zeroind[1][:]]
newzeroind = [zeroind[2][:], oneind[0][:]]
ans += [[zeroind[0][:], zeroind[1][:], oneind[0][:]]]
oneind = copy.deepcopy(newoneind)
zeroind = copy.deepcopy(newzeroind)
if len(oneind) == 2:
newoneind = [zeroind[0][:], zeroind[1][:], oneind[0][:]]
newzeroind = [oneind[1][:]]
ans += [[zeroind[0][:], zeroind[1][:], oneind[1][:]]]
oneind = copy.deepcopy(newoneind)
zeroind = copy.deepcopy(newzeroind)
if len(oneind) == 3:
ans += [[oneind[0][:], oneind[1][:], oneind[2][:]]]
for x in range(i, i+2):
for y in range(j, j+2):
mat[x][y] = '0'
def three_two_solver(mat, i, j, ans):
oneind = []
zeroind = []
for x in range(i, i+2):
for y in range(j, j+2):
if mat[x][y] == '1':
oneind += [[x, y]]
else:
zeroind += [[x, y]]
oneind2 = []
zeroind2 = []
for x in range(i, i+2):
for y in range(j+1, j+3):
if mat[x][y] == '1':
oneind2 += [[x, y]]
else:
zeroind2 += [[x, y]]
if len(oneind2) < len(oneind):
two_two_solver(mat, i, j+1, ans)
two_two_solver(mat, i, j, ans)
else:
two_two_solver(mat, i, j, ans)
two_two_solver(mat, i, j+1, ans)
def two_three_solver(mat, i, j, ans):
oneind = []
zeroind = []
for x in range(i, i+2):
for y in range(j, j+2):
if mat[x][y] == '1':
oneind += [[x, y]]
else:
zeroind += [[x, y]]
oneind2 = []
zeroind2 = []
for x in range(i+1, i+3):
for y in range(j, j+2):
if mat[x][y] == '1':
oneind2 += [[x, y]]
else:
zeroind2 += [[x, y]]
if len(oneind2) < len(oneind):
two_two_solver(mat, i+1, j, ans)
two_two_solver(mat, i, j, ans)
else:
two_two_solver(mat, i, j, ans)
two_two_solver(mat, i+1, j, ans)
t = int(input())
for _ in range(t):
n, m = map(int, input().split())
mat = []
for i in range(n):
s = input()
mat += [[x for x in s]]
# print(mat)
ans = []
if n%2 == 0 and m%2 == 0:
for i in range(0, n-1, 2):
for j in range(0, m-1, 2):
two_two_solver(mat, i, j, ans)
elif n%2 == 0 and m%2 == 1:
for i in range(0, n-1, 2):
for j in range(0, m-3, 2):
two_two_solver(mat, i, j, ans)
three_two_solver(mat, i, m-3, ans)
elif n%2 == 1 and m%2 == 0:
for i in range(0, n-3, 2):
for j in range(0, m-1, 2):
two_two_solver(mat, i, j, ans)
for j in range(0, m-1, 2):
two_three_solver(mat, n-3, j, ans)
else:
if mat[n-1][m-1] == '1':
ans += [[[n-1, m-1], [n-2, m-1], [n-1, m-2]]]
mat[n-1][m-1] = '0'
if mat[n-2][m-1] == '1':
mat[n-2][m-1] = '0'
else:
mat[n-2][m-1] = '1'
if mat[n-1][m-2] == '1':
mat[n-1][m-2] = '0'
else:
mat[n-1][m-2] = '1'
if mat[n-1][m-2] == '1':
ans += [[[n-1, m-2], [n-2, m-2], [n-1, m-3]]]
mat[n-1][m-2] = '0'
if mat[n-2][m-2] == '1':
mat[n-2][m-2] = '0'
else:
mat[n-2][m-2] = '1'
if mat[n-1][m-3] == '1':
mat[n-1][m-3] = '0'
else:
mat[n-1][m-3] = '1'
if mat[n-1][m-3] == '1':
ans += [[[n-1, m-3], [n-2, m-3], [n-2, m-2]]]
mat[n-1][m-3] = '0'
if mat[n-2][m-3] == '1':
mat[n-2][m-3] = '0'
else:
mat[n-2][m-3] = '1'
if mat[n-2][m-2] == '1':
mat[n-2][m-2] = '0'
else:
mat[n-2][m-2] = '1'
for i in range(0, n-3, 2):
for j in range(0, m-3, 2):
two_two_solver(mat, i, j, ans)
for i in range(0, n-1, 2):
three_two_solver(mat, i, m-3, ans)
for j in range(0, m-1, 2):
two_three_solver(mat, n-3, j, ans)
# three_two_solver(mat,n-3, m-3,)
print(len(ans))
for x in ans:
# print(*x)
for y in x:
print(y[0]+1, y[1]+1, end=' ')
print('')
```
Yes
| 41,476 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
for _ in range(int(input())):
m, n = map(int, input().split())
A = [[0] * (n + 1)] + [[0] + list(map(int, list(input()))) for _ in range(m)]
ans = []
for i in range(1, m, 2):
for j in range(1, n):
tmp = [A[i][j], A[i + 1][j]]
if tmp == [0, 0]: continue
if tmp == [1, 1]:
ans.append([i, j, i, j + 1, i + 1, j])
A[i][j] ^= 1
A[i][j + 1] ^= 1
A[i + 1][j] ^= 1
elif tmp == [1, 0]:
ans.append([i, j, i, j + 1, i + 1, j])
A[i][j] ^= 1
A[i][j + 1] ^= 1
A[i + 1][j] ^= 1
ans.append([i + 1, j, i + 1, j + 1, i, j + 1])
A[i + 1][j] ^= 1
A[i + 1][j + 1] ^= 1
A[i][j + 1] ^= 1
elif tmp == [0, 1]:
ans.append([i, j, i, j + 1, i + 1, j])
A[i][j] ^= 1
A[i][j + 1] ^= 1
A[i + 1][j] ^= 1
ans.append([i, j, i, j + 1, i + 1, j + 1])
A[i][j] ^= 1
A[i][j + 1] ^= 1
A[i + 1][j + 1] ^= 1
tmp = [A[i][n], A[i + 1][n]]
if tmp == [0, 1]:
ans.append([i, n - 1, i, n, i + 1, n])
ans.append([i, n, i + 1, n, i + 1, n - 1])
ans.append([i, n - 1, i + 1, n - 1, i + 1, n])
elif tmp == [1, 0]:
ans.append([i, n - 1, i, n, i + 1, n])
ans.append([i, n, i + 1, n, i + 1, n - 1])
ans.append([i, n - 1, i + 1, n - 1, i, n - 1])
elif tmp == [1, 1]:
ans.append([i, n - 1, i, n, i + 1, n])
ans.append([i + 1, n - 1, i + 1, n, i, n])
A[i][n] = A[i + 1][n] = 0
if m % 2:
for j in range(1, n, 2):
tmp = [A[m][j], A[m][j + 1]]
if tmp == [1, 0]:
ans.append([m - 1, j, m, j, m, j + 1])
ans.append([m, j, m, j + 1, m - 1, j + 1])
ans.append([m - 1, j, m - 1, j + 1, m, j])
elif tmp == [0, 1]:
ans.append([m - 1, j, m, j, m, j + 1])
ans.append([m, j, m, j + 1, m - 1, j + 1])
ans.append([m - 1, j, m - 1, j + 1, m, j + 1])
elif tmp == [1, 1]:
ans.append([m - 1, j, m - 1, j + 1, m, j])
ans.append([m - 1, j, m - 1, j + 1, m, j + 1])
A[m][j] = A[m][j + 1] = 0
if n % 2 and m % 2 and A[-1][-1]:
ans.append([m, n, m - 1, n, m, n - 1])
ans.append([m - 1, n - 1, m - 1, n, m, n])
ans.append([m - 1, n - 1, m, n - 1, m, n])
A[-1][-1] = 0
print(len(ans))
for a in ans: print(*a)
```
No
| 41,477 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
class State:
def __init__(self, state, operations):
self.states_num = {state: 0}
self.states = [state]
self._path = {}
self.DIAMETER = 0
new_sts = [0]
while new_sts:
self.DIAMETER += 1
cur_sts, new_sts = new_sts, []
for i in cur_sts:
for op, f in enumerate(operations):
state = f(self.states[i])
if state not in self.states_num:
j = self.states_num[state] = len(self.states)
self.states.append(state)
new_sts.append(j)
else:
j = self.states_num[state]
self._path[(i, j)] = (i, op)
self.SIZE = n = len(self.states_num)
self.distance = dis = [[0 if i == j else 10 ** 6 for i in range(n)] for j in range(n)]
for i, j in self._path.keys():
self.distance[i][j] = 1
for k in range(n):
for i in range(n):
for j in range(n):
if dis[i][k] + dis[k][j] < dis[i][j]:
dis[i][j] = dis[i][k] + dis[k][j]
self._path[(i, j)] = self._path[(k, j)]
def path(self, st1, st2):
u = self.states_num[st1]
v = self.states_num[st2]
result = []
while u != v:
v, op = self._path[(u, v)]
result.append(op)
return result[::-1]
# ############################## main
def op0(st):
# left top
return st[0], st[1] ^ 1, st[2] ^ 1, st[3] ^ 1
def op1(st):
# right top
return st[0] ^ 1, st[1], st[2] ^ 1, st[3] ^ 1
def op2(st):
# left bottom
return st[0] ^ 1, st[1] ^ 1, st[2], st[3] ^ 1
def op3(st):
# right bottom
return st[0] ^ 1, st[1] ^ 1, st[2] ^ 1, st[3]
zero = (0, 0, 0, 0)
ST = State(zero, (op0, op1, op2, op3))
def solve():
n, m = mpint()
matrix = [list(map(int, inp())) for _ in range(n)]
output = []
for i in range(n):
for j in range(m):
# left top position
if matrix[i][j] == 0:
continue
y, x = i, j
if y + 1 >= n:
y = n - 2
if x + 1 >= m:
x = m - 2
st = (matrix[y][x], matrix[y][x + 1], matrix[y + 1][x], matrix[y + 1][x + 1])
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
for k in range(4):
matrix[pos[k * 2]][pos[k * 2 + 1]] = 0
y += 1
x += 1
for op in ST.path(st, zero):
pos = [y, x, y, x + 1, y + 1, x, y + 1, x + 1]
del pos[op << 1], pos[op << 1]
output.append(pos)
print(len(output))
for out in output:
print(*out)
def main():
# solve()
# print(solve())
for _ in range(itg()):
solve()
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
DEBUG = 0
URL = ''
if __name__ == '__main__':
if DEBUG == 1:
import requests # ImportError: cannot import name 'md5' from 'sys' (unknown location)
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
elif DEBUG == 2:
main()
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
main()
# Please check!
```
No
| 41,478 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
def revab(b):
for i in range(3):
rev(b[2*i],b[2*i+1])
def rev(x,y):
global c
if c[x-1][y-1]=="1":
c[x-1][y-1]="0"
else:
c[x-1][y-1]="1"
import sys
for _ in range(int(sys.stdin.readline())):
a= list(map(int,sys.stdin.readline().strip().split(" ")))
c=[]
for i in range(a[0]):
c.append(list(sys.stdin.readline().strip()))
opn=0
ops=[]
a[0]-=1
while a[0]>1:
for i in range(a[1]-1):
if c[a[0]][i]=="1":
opn+=1
ops.append([a[0]+1,i+1,a[0],i+1,a[0],i+2])
revab(ops[-1])
if c[a[0]][a[1]-1]=="1":
opn+=1
ops.append([a[0]+1,a[1],a[0],a[1]-1,a[0],a[1]])
revab(ops[-1])
a[0]-=1
a[1]-=1
while a[1]>1:
if c[0][a[1]]=="1" and c[1][a[1]]=="1":
opn+=1
ops.append([1,a[1]+1,2,a[1]+1,1,a[1]])
revab(ops[-1])
elif c[0][a[1]]=="1":
opn+=1
ops.append([1,a[1],2,a[1],1,a[1]+1])
revab(ops[-1])
elif c[1][a[1]]=="1":
opn+=1
ops.append([2,a[1],1,a[1],2,a[1]+1])
revab(ops[-1])
a[1]-=1
if c[0][0]=="0":
opn+=1
ops.append([1,2,2,1,2,2])
if c[1][0]=="0":
opn+=1
ops.append([1,2,1,1,2,2])
if c[0][1]=="0":
opn+=1
ops.append([1,1,2,1,2,2])
if c[1][1]=="0":
opn+=1
ops.append([1,2,2,1,1,1])
print(opn)
for i in ops:
print(" ".join(list(map(str,i))))
```
No
| 41,479 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
def correct(x, y):
a[x][y] = 0
ans.append([x+1, y+1, x+2, y+1, x+2, y+2])
a[x+1][y+1] = 1 - a[x+1][y+1]
a[x+1][y] = 1 - a[x+1][y]
def makeone(i):
a[-1][i] = a[-1][i+1] = a[-2][i] = a[-2][i+1] = 0
def one(i):
if a[-2][i]:
ans.append([n-1, i+2, n, i+1, n-1, i+1])
a[-2][i] = 0
a[-2][i+1] = a[-1][i] = 1
elif a[-2][i+1]:
ans.append([n-1, i+1, n, i+1, n-1, i+2])
a[-2][i+1] = 0
a[-2][i] = a[-1][i] = 1
elif a[-1][i]:
ans.append([n-1, i+1, n-1, i+2, n, i+1])
a[-1][i] = 0
a[-2][i] = a[-2][i+1] = 1
else:
ans.append([n-1, i+1, n-1, i+2, n, i+2])
a[-1][i+1] = 0
a[-2][i] = a[-2][i+1] = 1
two(i)
def two(i):
if a[-1][i]:
if a[-1][i+1]:
ans.append([n-1, i+1, n-1, i+2, n, i+1])
ans.append([n-1, i+1, n-1, i+2, n, i+2])
elif a[-2][i+1]:
ans.append([n-1, i+2, n, i+2, n, i+1])
ans.append([n-1, i+2, n, i+1, n, i+2])
elif a[-2][i+2]:
ans.append([n-1, i+1, n-1, i+2, n, i+2])
ans.append([n-1, i+1, n, i+1, n, i+2])
elif a[-2][i]:
if a[-2][i+1]:
ans.append([n-1, i+2, n, i+2, n, i+1])
ans.append([n-1, i+2, n, i+1, n, i+2])
elif a[-1][i+1]:
ans.append([n-1, i+2, n, i+2, n, i+1])
ans.append([n-1, i+1, n-1, i+2, n, i+1])
else:
ans.append([n-1, i+1, n-1, i+2, n, i+1])
ans.append([n-1, i+1, n, i+1, n, i+2])
makeone(i)
def final(i, c):
if c==1:
one(i)
elif c==4:
ans.append([n-1, i+2, n, i+1, n, i+2])
a[n-2][i+1] = a[n-1][i] = a[n-1][i+1] = 0
one(i)
elif c==3:
if not a[-2][i]:
ans.append([n-1, i+2, n, i+1, n, i+2])
elif not a[-2][i+1]:
ans.append([n-1, i+1, n, i+1, n, i+2])
elif not a[-1][i]:
ans.append([n-1, i+1, n-1, i+2, n, i+2])
else:
ans.append([n-1, i+1, n-1, i+2, n, i+1])
makeone(i)
elif c==2:
two(i)
def count(i):
c = 0
if a[-1][i]:
c += 1
if a[-1][i+1]:
c += 1
if a[-2][i]:
c += 1
if a[-2][i+1]:
c += 1
final(i, c)
for nt in range(int(input())):
n,m = map(int,input().split())
a, ans = [], []
for i in range(n):
a.append(list(map(int,list(input()))))
for i in range(n-2):
for j in range(m-1):
if a[i][j]==1:
correct(i, j)
if a[i][-1] == 1:
ans.append([i+1, m, i+2, m, i+2, m-1])
a[i][-1] = 0
a[i+1][-1] = 1 - a[i+1][-1]
a[i+1][-2] = 1 - a[i+1][-2]
# print (a)
if m%2:
if a[-1][-1] and a[-2][-1]:
for i in range(0, m-1, 2):
count(i)
count(m-2)
elif not a[-1][-1] and not a[-2][-1]:
for i in range(0, m-1, 2):
count(i)
else:
for i in range(0, m-3, 2):
count(i)
# print (a)
if a[-1][-1]:
ans.append([n-1, m-1, n, m-1, n, m])
a[-2][-2] = 1 - a[-2][-2]
a[-1][-2] = 1 - a[-1][-2]
a[-1][-1] = 0
count(m-3)
else:
ans.append([n-1, m-1, n, m-1, n-1, m])
a[-2][-2] = 1 - a[-2][-2]
a[-1][-2] = 1 - a[-1][-2]
a[-2][-1] = 0
count(m-3)
else:
for i in range(0, m, 2):
count(i)
# print (a)
print (len(ans))
for i in ans:
print (*i)
```
No
| 41,480 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def is_hora(lst):
for i in range(1,10):
if lst[i] >= 2:
if(is_4chunk(lst,i)): return True
return False
def is_4chunk(lst,atama):
searchlist = [0]*10
for i in range(10):
searchlist[i] = lst[i]
searchlist[atama] -= 2
for i in range(1,8):
if searchlist[i] < 0: return False
if searchlist[i] >=3: searchlist[i]-=3
if searchlist[i] ==1:
searchlist[i]-=1
searchlist[i+1]-=1
searchlist[i+2]-=1
elif searchlist[i] ==2:
searchlist[i]-=2
searchlist[i+1]-=2
searchlist[i+2]-=2
return (searchlist[8] == 0 or searchlist[8] == 3) and (searchlist[9] == 0 or searchlist[9] == 3)
while True:
try:
instr = input()
pai = [0]*10
ans = []
for i in range(13):
pai[(int(instr[i]))] += 1
for i in range(1,10):
if pai[i] == 4: continue
pai[i] += 1
if is_hora(pai): ans.append(i)
pai[i] -= 1
if len(ans)!=0:
print(" ".join(list(map(str,ans))))
else: print(0)
except EOFError: break
```
Yes
| 41,973 | [
0.7099609375,
0.2408447265625,
0.1126708984375,
0.220947265625,
-0.5888671875,
-0.318603515625,
-0.170166015625,
0.222412109375,
0.350830078125,
0.8662109375,
0.6806640625,
-0.0887451171875,
0.138916015625,
-0.7529296875,
-0.72412109375,
0.1448974609375,
-0.6787109375,
-0.923828125... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
import sys
def f(c):
if sum(c)in c:return 1
if 5 in c:return 0
if 4 in c:
k=c.index(4);c[k]-=3
if f(c):return 1
c[k]+=3
if 3 in c:
k=c.index(3);c[k]-=3
if f(c):return 1
c[k]+=3
for i in range(7):
if c[i]and c[i+1]and c[i+2]:
c[i]-=1;c[i+1]-=1;c[i+2]-=1
if f(c):return 1
c[i]+=1;c[i+1]+=1;c[i+2]+=1
n='123456789'
for e in sys.stdin:
e=list(e)
a=[i for i in n if f([(e+[i]).count(j)for j in n])]
if a:print(*a)
else:print(0)
```
Yes
| 41,974 | [
0.60693359375,
0.1392822265625,
0.0909423828125,
0.1788330078125,
-0.5927734375,
-0.41943359375,
-0.25439453125,
0.2454833984375,
0.257080078125,
0.90576171875,
0.6806640625,
-0.2034912109375,
0.2344970703125,
-0.70068359375,
-0.72314453125,
0.12225341796875,
-0.74609375,
-0.947265... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def check(nums):
for i in range(9):
if(nums[i]>4):return False
for head in range(9):
anums=nums[:]
if(anums[head]<2):continue
anums[head]-=2
for i in range(9):
if(anums[i]>=3):
anums[i]-=3
while(anums[i]>0):
if i>=7:break
ok=True
for j in range(i,i+3):
if anums[j]<=0:
ok=True
if not ok:
break
for j in range(i,i+3):
anums[j]-=1
if not any(anums):
return True
return False
while True:
st=""
try:
st=input()
if st=="":break
except:break
nums=[0 for i in range(9)]
for c in st:
nums[int(c)-1]+=1
anss=[]
for n in range(0,9):
nums[n]+=1
if(check(nums[:])):anss.append(n+1)
nums[n]-=1
if(len(anss)==0):anss.append(0)
print(" ".join(map(str,anss)))
```
Yes
| 41,975 | [
0.70166015625,
0.1390380859375,
0.0928955078125,
0.169921875,
-0.57470703125,
-0.4111328125,
-0.2164306640625,
0.20068359375,
0.36181640625,
0.9365234375,
0.693359375,
-0.22998046875,
0.2138671875,
-0.7177734375,
-0.75830078125,
0.13427734375,
-0.70166015625,
-0.8828125,
-0.47583... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def Solve(c,s):
if s:
if max(c)>4:return False
for i in range(9):
if c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
check=0
for i in range(4):check+=c.count(3*i)
if check==9:return True
else:
for i in range(7):
if c[i]>=1:
cc=c[:]
isneg=False
for j in range(3):
cc[i+j]-=1
if cc[i+j]<0:
isneg=True
break
if isneg==False and Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in=input()
except EOFError:break
for i in range(9):
l=(_in+str(i+1))[::1]
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
```
Yes
| 41,976 | [
0.64306640625,
0.1199951171875,
0.1185302734375,
0.1822509765625,
-0.5693359375,
-0.4267578125,
-0.22705078125,
0.231689453125,
0.274658203125,
0.94384765625,
0.68310546875,
-0.1810302734375,
0.2369384765625,
-0.69189453125,
-0.69189453125,
0.07708740234375,
-0.73291015625,
-0.8925... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def Solve(c,s):
if s:
for i in range(9):
if c[i]>4:return False
elif c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
if c.count(0)+c.count(3)==9:return True
else:
for i in range(9):
if c[i]<0:return False
elif c[i]>=1 and i<7:
cc=c[:]
for j in range(3):
cc[i+j]-=1
if Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in="9118992346175"
except EOFError:break
for i in range(9):
l=sorted((_in+str(i+1))[::1])
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
```
No
| 41,977 | [
0.64599609375,
0.11114501953125,
0.0860595703125,
0.197998046875,
-0.55859375,
-0.4169921875,
-0.24755859375,
0.238037109375,
0.29638671875,
0.95263671875,
0.68798828125,
-0.1910400390625,
0.235107421875,
-0.70654296875,
-0.70361328125,
0.086181640625,
-0.73095703125,
-0.8881835937... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
import sys
def f(c,h=0):
if 2 in c and sum(c)==2:return 1
if 5 in c:return 0 # sum(c)==2
if 4 in c:
k=c.index(4);c[k]-=3
if f(c,h+1):return 1
c[k]+=3
if 3 in c:
k=c.index(3);c[k]-=3
if f(c,h+1):return 1
c[k]+=3
for i in range(7):
if c[i]and c[i+1]and c[i+2]:
c[i]-=1;c[i+1]-=1;c[i+2]-=1
if f(c,h+1):return 1
c[i]+=1;c[i+1]+=1;c[i+2]+=1
n='123456789'
for e in sys.stdin:
e=list(e)
a=[i for i in n if f([(e+[i]).count(j)for j in n])]
print(*a if a else 0)
```
No
| 41,978 | [
0.609375,
0.1439208984375,
0.07806396484375,
0.1845703125,
-0.59912109375,
-0.41845703125,
-0.249267578125,
0.252685546875,
0.2724609375,
0.9052734375,
0.69140625,
-0.2088623046875,
0.24462890625,
-0.6904296875,
-0.724609375,
0.123291015625,
-0.73681640625,
-0.94384765625,
-0.500... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def Solve(c,s):
if s:
for i in range(9):
if c[i]>4:return False
elif c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
if c.count(0)+c.count(3)==9:return True
else:
for i in range(9):
if c[i]<0:return False
elif c[i]>=1 and i<7:
cc=c[:]
for j in range(3):
cc[i+j]-=1
if Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in=input()
except EOFError:break
for i in range(9):
l=sorted((_in+str(i+1))[::1])
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
```
No
| 41,979 | [
0.64599609375,
0.11114501953125,
0.0860595703125,
0.197998046875,
-0.55859375,
-0.4169921875,
-0.24755859375,
0.238037109375,
0.29638671875,
0.95263671875,
0.68798828125,
-0.1910400390625,
0.235107421875,
-0.70654296875,
-0.70361328125,
0.086181640625,
-0.73095703125,
-0.8881835937... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There is a puzzle to complete by combining 14 numbers from 1 to 9. Complete by adding another number to the given 13 numbers.
The conditions for completing the puzzle are
* You must have one combination of the same numbers.
* The remaining 12 numbers are 4 combinations of 3 numbers.
The combination of three numbers is either three of the same numbers or three consecutive numbers. However, sequences such as 9 1 2 are not considered consecutive numbers.
* The same number can be used up to 4 times.
Create a program that reads a string of 13 numbers and outputs all the numbers that can complete the puzzle in ascending order. If you cannot complete the puzzle by adding any number from 1 to 9, output 0.
For example, if the given string is 3456666777999
If there is a "2", 234 567 666 77 999
If there is a "3", then 33 456 666 777 999
If there is a "5", then 345 567 666 77 999
If there is an "8", then 345 666 678 77 999
And so on, the puzzle is complete when one of the numbers 2 3 5 8 is added. Note that "6" is fine, but it will be used for the 5th time, so it cannot be used in this example.
Input
The input consists of multiple datasets. For each dataset, 13 numbers are given on one line. The number of datasets does not exceed 50.
Output
For each dataset, the numbers that can complete the puzzle are output on one line in ascending order, separated by blanks.
Example
Input
3649596966777
6358665788577
9118992346175
9643871425498
7755542764533
1133557799246
Output
2 3 5 8
3 4
1 2 3 4 5 6 7 8 9
7 8 9
1 2 3 4 6 7 8
0
Submitted Solution:
```
def Solve(c,s):
if s:
if max(c)>4:return False
for i in range(9):
if c[i]>=2:
cc=c[:]
cc[i]-=2
if Solve(cc,False):return True
else:
check=0
for i in range(4):check+=c.count(3*i)
if check==9:return True
else:
for i in range(7):
if c[i]>=1:
cc=c[:]
for j in range(3):
cc[i+j]-=1
if cc[i+j]<0:break
if Solve(cc,False):return True
_in=""
while True:
ans=[]
try:_in="1123345567799"
except EOFError:break
for i in range(9):
l=(_in+str(i+1))[::1]
count=[l.count(str(j+1))for j in range(9)]
if Solve(count,True):ans.append(str(i+1))
print(0 if len(ans)==0 else " ".join(ans))
```
No
| 41,980 | [
0.64306640625,
0.1199951171875,
0.1185302734375,
0.1822509765625,
-0.5693359375,
-0.4267578125,
-0.22705078125,
0.231689453125,
0.274658203125,
0.94384765625,
0.68310546875,
-0.1810302734375,
0.2369384765625,
-0.69189453125,
-0.69189453125,
0.07708740234375,
-0.73291015625,
-0.8925... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Look for the Winner!
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.
The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates A, B, and C and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of A, B, and C are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for A, A is ensured to be the winner since A already has five votes and B or C can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.
Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
Input
The input consists of at most 1500 datasets, each consisting of two lines in the following format.
n
c1 c2 … cn
n in the first line represents the number of votes, and is a positive integer no greater than 100. The second line represents the n votes, separated by a space. Each ci (1 ≤ i ≤ n) is a single uppercase letter, i.e. one of 'A' through 'Z'. This represents the election candidate for which the i-th vote was cast. Counting shall be done in the given order from c1 to cn.
You should assume that at least two stand as candidates even when all the votes are cast for one candidate.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, unless the election ends in a tie, output a single line containing an uppercase letter c and an integer d separated by a space: c should represent the election winner and d should represent after counting how many votes the winner is identified. Otherwise, that is, if the election ends in a tie, output a single line containing `TIE'.
Sample Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output for the Sample Input
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Example
Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Submitted Solution:
```
from collections import Counter
while True:
N = int(input())
if not N:
break
C = input().split()
for i in range(N):
cnt = Counter(C[:i+1]).most_common()
if (len(cnt) == 1 and cnt[0][1] * 2 > N) or (len(cnt) > 1 and cnt[0][1] > cnt[1][1] + N - i - 1):
print(cnt[0][0], i + 1)
break
else:
print("TIE")
```
Yes
| 42,029 | [
0.247314453125,
-0.122314453125,
-0.385986328125,
0.09429931640625,
-0.54248046875,
-0.330810546875,
-0.0684814453125,
0.2900390625,
-0.00888824462890625,
0.57568359375,
0.463134765625,
-0.154296875,
-0.1722412109375,
-0.356689453125,
-0.80615234375,
-0.0307769775390625,
-0.336914062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Look for the Winner!
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.
The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates A, B, and C and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of A, B, and C are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for A, A is ensured to be the winner since A already has five votes and B or C can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.
Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
Input
The input consists of at most 1500 datasets, each consisting of two lines in the following format.
n
c1 c2 … cn
n in the first line represents the number of votes, and is a positive integer no greater than 100. The second line represents the n votes, separated by a space. Each ci (1 ≤ i ≤ n) is a single uppercase letter, i.e. one of 'A' through 'Z'. This represents the election candidate for which the i-th vote was cast. Counting shall be done in the given order from c1 to cn.
You should assume that at least two stand as candidates even when all the votes are cast for one candidate.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, unless the election ends in a tie, output a single line containing an uppercase letter c and an integer d separated by a space: c should represent the election winner and d should represent after counting how many votes the winner is identified. Otherwise, that is, if the election ends in a tie, output a single line containing `TIE'.
Sample Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output for the Sample Input
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Example
Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Submitted Solution:
```
printedlater = []
while True:
n = int(input())
if (n == 0): # the end of input
break
c = [x for x in input().split()]
voteNum = {}
for cand in list("ABCDEFGHIJKLMNOPQRSTUVWXYZ"):
voteNum[cand] = 0
while True:
if (len(c) == 0):
voteMax = max(voteNum.values())
winners = [x for x in voteNum.keys() if voteNum[x] == voteMax]
if (len(winners) == 1):
printedlater.append(winners[0] + " " + str(n))
else:
printedlater.append("TIE")
break
voteNum[c.pop(0)] += 1
vote2ndMax, voteMax = sorted(voteNum.values())[-2:]
if voteMax - vote2ndMax > len(c):
printedlater.append([x for x in voteNum.keys() if voteNum[x] == voteMax][0] + " " + str(n - len(c)))
break
for line in printedlater:
print(line)
```
Yes
| 42,030 | [
0.247314453125,
-0.122314453125,
-0.385986328125,
0.09429931640625,
-0.54248046875,
-0.330810546875,
-0.0684814453125,
0.2900390625,
-0.00888824462890625,
0.57568359375,
0.463134765625,
-0.154296875,
-0.1722412109375,
-0.356689453125,
-0.80615234375,
-0.0307769775390625,
-0.336914062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Look for the Winner!
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.
The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates A, B, and C and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of A, B, and C are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for A, A is ensured to be the winner since A already has five votes and B or C can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.
Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
Input
The input consists of at most 1500 datasets, each consisting of two lines in the following format.
n
c1 c2 … cn
n in the first line represents the number of votes, and is a positive integer no greater than 100. The second line represents the n votes, separated by a space. Each ci (1 ≤ i ≤ n) is a single uppercase letter, i.e. one of 'A' through 'Z'. This represents the election candidate for which the i-th vote was cast. Counting shall be done in the given order from c1 to cn.
You should assume that at least two stand as candidates even when all the votes are cast for one candidate.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, unless the election ends in a tie, output a single line containing an uppercase letter c and an integer d separated by a space: c should represent the election winner and d should represent after counting how many votes the winner is identified. Otherwise, that is, if the election ends in a tie, output a single line containing `TIE'.
Sample Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output for the Sample Input
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Example
Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Submitted Solution:
```
ans_list = []
while True:
n = int(input())
if n == 0:
break
C = input().split()
al = [0]*26
ans = "TIE"
for i, c in enumerate(C):
al[ord(c)-65] += 1
tmp = sorted(al,reverse=True)
one = tmp[0]
two = tmp[1]
remain = n - i - 1
if one > two + remain:
win = chr(al.index(one) + 65)
ans = "{} {}".format(win, i+1)
break
ans_list.append(ans)
for ans in ans_list:
print(ans)
```
Yes
| 42,031 | [
0.247314453125,
-0.122314453125,
-0.385986328125,
0.09429931640625,
-0.54248046875,
-0.330810546875,
-0.0684814453125,
0.2900390625,
-0.00888824462890625,
0.57568359375,
0.463134765625,
-0.154296875,
-0.1722412109375,
-0.356689453125,
-0.80615234375,
-0.0307769775390625,
-0.336914062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Look for the Winner!
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.
The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates A, B, and C and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of A, B, and C are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for A, A is ensured to be the winner since A already has five votes and B or C can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.
Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
Input
The input consists of at most 1500 datasets, each consisting of two lines in the following format.
n
c1 c2 … cn
n in the first line represents the number of votes, and is a positive integer no greater than 100. The second line represents the n votes, separated by a space. Each ci (1 ≤ i ≤ n) is a single uppercase letter, i.e. one of 'A' through 'Z'. This represents the election candidate for which the i-th vote was cast. Counting shall be done in the given order from c1 to cn.
You should assume that at least two stand as candidates even when all the votes are cast for one candidate.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, unless the election ends in a tie, output a single line containing an uppercase letter c and an integer d separated by a space: c should represent the election winner and d should represent after counting how many votes the winner is identified. Otherwise, that is, if the election ends in a tie, output a single line containing `TIE'.
Sample Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output for the Sample Input
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Example
Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Submitted Solution:
```
while True:
n = int(input())
if n == 0:
break
C = input().split()
D = {}
for x in [chr(i) for i in range(65, 65+26)]:
D[x] = 0
for i in range(n):
D[C[i]] += 1
E = sorted(D.items(), key=lambda x: x[1], reverse=True)
if E[0][1] > E[1][1] + n - i - 1:
print(E[0][0], i + 1)
break
if E[0][1] == E[1][1]:
print("TIE")
```
Yes
| 42,032 | [
0.247314453125,
-0.122314453125,
-0.385986328125,
0.09429931640625,
-0.54248046875,
-0.330810546875,
-0.0684814453125,
0.2900390625,
-0.00888824462890625,
0.57568359375,
0.463134765625,
-0.154296875,
-0.1722412109375,
-0.356689453125,
-0.80615234375,
-0.0307769775390625,
-0.336914062... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Look for the Winner!
The citizens of TKB City are famous for their deep love in elections and vote counting. Today they hold an election for the next chairperson of the electoral commission. Now the voting has just been closed and the counting is going to start. The TKB citizens have strong desire to know the winner as early as possible during vote counting.
The election candidate receiving the most votes shall be the next chairperson. Suppose for instance that we have three candidates A, B, and C and ten votes. Suppose also that we have already counted six of the ten votes and the vote counts of A, B, and C are four, one, and one, respectively. At this moment, every candidate has a chance to receive four more votes and so everyone can still be the winner. However, if the next vote counted is cast for A, A is ensured to be the winner since A already has five votes and B or C can have at most four votes at the end. In this example, therefore, the TKB citizens can know the winner just when the seventh vote is counted.
Your mission is to write a program that receives every vote counted, one by one, identifies the winner, and determines when the winner gets ensured.
Input
The input consists of at most 1500 datasets, each consisting of two lines in the following format.
n
c1 c2 … cn
n in the first line represents the number of votes, and is a positive integer no greater than 100. The second line represents the n votes, separated by a space. Each ci (1 ≤ i ≤ n) is a single uppercase letter, i.e. one of 'A' through 'Z'. This represents the election candidate for which the i-th vote was cast. Counting shall be done in the given order from c1 to cn.
You should assume that at least two stand as candidates even when all the votes are cast for one candidate.
The end of the input is indicated by a line containing a zero.
Output
For each dataset, unless the election ends in a tie, output a single line containing an uppercase letter c and an integer d separated by a space: c should represent the election winner and d should represent after counting how many votes the winner is identified. Otherwise, that is, if the election ends in a tie, output a single line containing `TIE'.
Sample Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output for the Sample Input
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Example
Input
1
A
4
A A B B
5
L M N L N
6
K K K K K K
6
X X X Y Z X
10
A A A B A C A C C B
10
U U U U U V V W W W
0
Output
A 1
TIE
TIE
K 4
X 5
A 7
U 8
Submitted Solution:
```
def secMax(a):
mx = 0
smx = 0
for i in range(0,len(a)):
if a[i] > mx:
smx = mx
mx = a[i]
elif a[i] > smx:
smx = a[i]
else:
return smx
while True:
n = int(input())
if n == 0:
break
c = list(map(str,input().split()))
##alphabet = char('a')##char型ねぇよって言われた(´・ω・`)
alph = ["A","B","C","D","E","F","G","H",
"I","J","K","L","M","N","O","P","Q",
"R","S","T","U","V","W","X","Y","Z"]
abcCount = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]
for i in range(0, n):
for j in range(0, 26):
if c[i] == alph[j]:
abcCount[j] = abcCount[j] + 1
break
print(abcCount)
if max(abcCount) > n-1-i+secMax(abcCount):
print(alph[abcCount.index(max(abcCount))], sum(abcCount))
break
else:
print("TIE")
```
No
| 42,033 | [
0.247314453125,
-0.122314453125,
-0.385986328125,
0.09429931640625,
-0.54248046875,
-0.330810546875,
-0.0684814453125,
0.2900390625,
-0.00888824462890625,
0.57568359375,
0.463134765625,
-0.154296875,
-0.1722412109375,
-0.356689453125,
-0.80615234375,
-0.0307769775390625,
-0.336914062... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
import sys
input = sys.stdin.readline
N,M=map(int,input().split())
A=[int(x) for x in input().split()]
K=[0]*(N+1)
H=set()
S=''
for a in A:
H.add(a)
K[a]+=1
if len(H)==N:
S+='1'
H=set()
for i in range(1,N+1):
K[i]-=1
if K[i]>=1:
H.add(i)
else:
S+='0'
print(S)
```
| 42,141 | [
0.444091796875,
-0.2763671875,
-0.1126708984375,
0.130126953125,
-0.488037109375,
-0.30615234375,
-0.2000732421875,
0.2298583984375,
0.158203125,
0.90625,
0.82666015625,
-0.207763671875,
0.55517578125,
-0.98583984375,
-0.25,
0.2119140625,
-0.873046875,
-1.009765625,
-0.8618164062... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
n,m=map(int,input().split())
arr=list(map(int,input().split()))
d={}
res = ""
for i in arr:
if i in d:
d[i]+=1
else:
d[i]=1
is_round = (len(d) == n)
res+=str(int(is_round))
if is_round:
d2= {}
for k, v in d.items():
if v > 1:
d2[k] = v - 1
d = d2
print(res)
```
| 42,142 | [
0.43896484375,
-0.261962890625,
-0.1954345703125,
0.13623046875,
-0.4970703125,
-0.32275390625,
-0.14501953125,
0.2362060546875,
0.173583984375,
0.9140625,
0.87109375,
-0.203857421875,
0.57275390625,
-1.0390625,
-0.246826171875,
0.239501953125,
-0.8837890625,
-0.96337890625,
-0.8... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
N, M = map(int, input().split())
A = [int(a) for a in input().split()]
X = [0] * N
s = 0
sc = N
def cts():
global s
global sc
# print(X)
s = min(X)
sc = 0
# print("CTS1", s, sc)
for i in range(N):
if X[i] == s:
sc += 1
# print("CTS2", s, sc)
for i in range(M):
X[A[i]-1] += 1
if X[A[i]-1] == s+1:
sc -= 1
if sc == 0:
cts()
if s == 1:
print(1, end = "")
for j in range(N):
X[j] -= 1
s -= 1
else:
print(0, end = "")
print("")
```
| 42,143 | [
0.427490234375,
-0.30126953125,
-0.132568359375,
0.11846923828125,
-0.460205078125,
-0.35009765625,
-0.2235107421875,
0.23779296875,
0.147216796875,
0.8955078125,
0.85986328125,
-0.202880859375,
0.52197265625,
-1.0185546875,
-0.26513671875,
0.233642578125,
-0.88916015625,
-0.980957... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
def main():
t = 1
for _ in range(t):
n,m = map(int,input().split())
tab = list(map(int,input().split()))
se = {}
inc = 0
for x in range(m):
if tab[x] in se:
if se[tab[x]]==0:
inc+=1
se[tab[x]]+=1
else:
se[tab[x]]=1
inc += 1
if inc==n:
for x in range(1,n+1):
se[x] -= 1
if se[x] == 0:
inc -= 1
print(1,end="")
else:
print(0,end="")
print()
main()
```
| 42,144 | [
0.44091796875,
-0.30322265625,
-0.10943603515625,
0.1607666015625,
-0.475341796875,
-0.31689453125,
-0.202392578125,
0.2288818359375,
0.2015380859375,
0.9111328125,
0.8203125,
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0.56494140625,
-0.99853515625,
-0.2459716796875,
0.2352294921875,
-0.89599609375,
-0.99462... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
import sys
input=sys.stdin.readline
f=lambda :list(map(int, input().split()))
n, m=f()
cnt=[0]*(n+1)
num=[0]*(m+1)
ans=[0]*m
for i, v in enumerate(f()):
cnt[v]+=1
num[cnt[v]]+=1
ans[i]=1 if num[cnt[v]]==n else 0
print(''.join(map(str, ans)))
```
| 42,145 | [
0.436767578125,
-0.2666015625,
-0.10784912109375,
0.138671875,
-0.485595703125,
-0.2880859375,
-0.209228515625,
0.2249755859375,
0.1651611328125,
0.9296875,
0.8330078125,
-0.2183837890625,
0.5390625,
-0.97998046875,
-0.303955078125,
0.208740234375,
-0.86865234375,
-1.01171875,
-0... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
n, m = map(int, input().split())
a = list(map(int, input().split()))
mn = 1
cnt, cnt2 = [0 for i in range(n + 1)], [0 for i in range(m + 1)]
for i in range(m):
cnt[a[i]] += 1
cnt2[cnt[a[i]]] += 1
if cnt2[mn] == n:
print(1, end = "")
mn += 1
else:
print(0, end = "")
print()
```
| 42,146 | [
0.4267578125,
-0.29150390625,
-0.17236328125,
0.117431640625,
-0.455322265625,
-0.333740234375,
-0.1815185546875,
0.20458984375,
0.184326171875,
0.9306640625,
0.83984375,
-0.1824951171875,
0.5576171875,
-1.0087890625,
-0.26416015625,
0.23779296875,
-0.875,
-0.9736328125,
-0.84716... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
n,m = map(int,input().split())
l1 = list(map(int,input().split()))
freq = {}
diff = 0
ans = ""
def update():
global diff
for i in range(1,n+1):
freq[i] -= 1
if(freq[i] == 0):
diff -= 1
return
for i in range(m):
atual = l1[i]
if((atual not in freq) or freq[atual] == 0):
diff += 1
freq[atual] = 1
else:
freq[atual] += 1
if(diff==n):
ans += "1"
update()
else:
ans += "0"
print(ans)
```
| 42,147 | [
0.462646484375,
-0.203125,
-0.1915283203125,
0.1497802734375,
-0.447509765625,
-0.3310546875,
-0.2222900390625,
0.1363525390625,
0.1602783203125,
0.966796875,
0.80029296875,
-0.1776123046875,
0.57568359375,
-1.041015625,
-0.25439453125,
0.235595703125,
-0.88232421875,
-1.03125,
-... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Tags: data structures, implementation
Correct Solution:
```
from collections import Counter
c = Counter()
n,m = list(map(int, input().split()))
t = list(map(int, input().split()))
s = ''
for i in t:
if i in c:
c[i]+=1
else:
c[i]=1
#print(i, len(c), c)
if len(c)==n:
for i in range(1,n+1):
if c[i]==1:
del c[i]
else:
c[i]-=1
#print('Deleted', c)
s += '1'
else:
s += '0'
print(s)
```
| 42,148 | [
0.427978515625,
-0.288330078125,
-0.16259765625,
0.1109619140625,
-0.479248046875,
-0.31103515625,
-0.178466796875,
0.2098388671875,
0.18310546875,
0.900390625,
0.81494140625,
-0.224853515625,
0.58935546875,
-1.0234375,
-0.275634765625,
0.2373046875,
-0.92041015625,
-0.978515625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
from fractions import gcd
from datetime import date
from math import factorial
import functools
from heapq import*
from collections import deque
import collections
import math
from collections import defaultdict, Counter
import sys
sys.setrecursionlimit(10 ** 7)
MOD = 10 ** 9 + 7
inf = float("inf")
def main():
n, m = map(int, input().split())
a = list(map(int, input().split()))
l = set()
ans = ""
d = defaultdict(int)
c = 0
for v in a:
if d[v] == 0:
c += 1
d[v] += 1
if c == n:
ans += "1"
for ke in d.keys():
d[ke] -= 1
if d[ke] == 0:
c -= 1
else:
ans += "0"
print(ans)
if __name__ == '__main__':
main()
```
Yes
| 42,149 | [
0.61328125,
-0.1768798828125,
-0.215087890625,
0.06646728515625,
-0.56201171875,
-0.07379150390625,
-0.1815185546875,
0.2113037109375,
0.164794921875,
0.96337890625,
0.80517578125,
-0.2235107421875,
0.4462890625,
-0.9296875,
-0.1793212890625,
0.1473388671875,
-0.85205078125,
-0.974... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
## Soru 1
# nk = list(map(int, input().split()))
# n = nk[0]
# k = nk[1]
#
# lst = list(map(int, input().split()))
#
# e = 0
# s = 0
# for i in lst:
# if i == 1:
# e += 1
# else:
# s += 1
# larg = 0
# for b in range(1,k+1):
# a = 0
# x = e
# y = s
# while a*k + b <= n:
# if lst[a*k+b-1] == 1:
# x -= 1
# else:
# y -= 1
# a += 1
# if abs(x-y)>larg:
# larg = abs(x-y)
# print(larg)
## Soru 2
nm = list(map(int, input().split()))
n = nm[0]
lst = list(map(int, input().split()))
s = set()
result = ""
dct = {}
for i in lst:
x = len(s)
s.add(i)
if x == len(s):
dct[i] = dct.get(i, 0) + 1
result = result + "0"
elif len(s) == n:
result = result + "1"
s = set(dct.keys())
for i in s:
dct[i] -= 1
if dct[i] == 0:
dct.pop(i)
else:
result = result + "0"
print(result)
## Soru 3
# import math
# nr = list(map(int, input().split()))
# n = nr[0]
# inner_r = nr[1]
# alfa = 180/n
# alfa = math.radians(alfa)
# sinus = math.sin(alfa)
# outer_r = sinus*inner_r / (1-sinus)
# print(outer_r)
```
Yes
| 42,150 | [
0.5869140625,
-0.1898193359375,
-0.2198486328125,
0.057647705078125,
-0.60986328125,
-0.222900390625,
-0.2646484375,
0.30078125,
0.1124267578125,
0.94580078125,
0.7587890625,
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0.4912109375,
-0.931640625,
-0.1741943359375,
0.1258544921875,
-0.8310546875,
-0.90673828... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
N, M = map(int, input().split())
a = list(map(lambda x: int(x)-1, input().split()))
problems = [0]*N
count = 0
ans = []
append = ans.append
for n in a:
if not problems[n]:
count += 1
problems[n] += 1
append(1 if count == N else 0)
if count == N:
count = 0
for i in range(N):
problems[i] -= 1
if problems[i] > 0:
count += 1
print(*ans, sep="")
```
Yes
| 42,151 | [
0.5791015625,
-0.17578125,
-0.2178955078125,
0.0712890625,
-0.5869140625,
-0.2135009765625,
-0.263916015625,
0.2890625,
0.10845947265625,
0.9541015625,
0.76025390625,
-0.1256103515625,
0.46826171875,
-0.91748046875,
-0.1915283203125,
0.11529541015625,
-0.83837890625,
-0.91943359375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
n, m = map(int, input().split())
ls, d,c,currmin =list(map(lambda x: int(x)-1, input().split())), [0]*n,0,0
for i in range(min(n, m)-1):
d[ls[i]]+=1
print(0,end="")
for i in range(min(n, m)-1, m):
d[ls[i]]+=1
while currmin < n and d[currmin] > c: currmin+=1
if currmin == n:
currmin,c = 0,c+1
print(1,end="")
else:
print(0,end="")
print()
```
Yes
| 42,152 | [
0.58056640625,
-0.1641845703125,
-0.221435546875,
0.0654296875,
-0.59765625,
-0.2093505859375,
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0.287109375,
0.08197021484375,
0.94775390625,
0.7724609375,
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0.4541015625,
-0.94482421875,
-0.2003173828125,
0.1004638671875,
-0.82763671875,
-0.93359375... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
a,b=map(int,input().split())
c=list(map(int,input().split()))
d=[i for i in range(1,a+1)]
e=set(d)
f=[]
h=''
for i in c:
f.append(i)
f.sort()
g=set(f)
if g==e:
h+='1'
f=[]
g=set(f)
else:
h+='0'
print(h)
```
No
| 42,153 | [
0.5791015625,
-0.184326171875,
-0.220458984375,
0.0650634765625,
-0.59423828125,
-0.182861328125,
-0.248291015625,
0.28955078125,
0.10137939453125,
0.92431640625,
0.76708984375,
-0.131103515625,
0.481689453125,
-0.98046875,
-0.19873046875,
0.08782958984375,
-0.85107421875,
-0.92236... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
"""
____ _ _____
/ ___|___ __| | ___| ___|__ _ __ ___ ___ ___
| | / _ \ / _` |/ _ \ |_ / _ \| '__/ __/ _ \/ __|
| |__| (_) | (_| | __/ _| (_) | | | (_| __/\__ \
\____\___/ \__,_|\___|_| \___/|_| \___\___||___/
"""
"""
░░██▄░░░░░░░░░░░▄██
░▄▀░█▄░░░░░░░░▄█░░█░
░█░▄░█▄░░░░░░▄█░▄░█░
░█░██████████████▄█░
░█████▀▀████▀▀█████░
▄█▀█▀░░░████░░░▀▀███
██░░▀████▀▀████▀░░██
██░░░░█▀░░░░▀█░░░░██
███▄░░░░░░░░░░░░▄███
░▀███▄░░████░░▄███▀░
░░░▀██▄░▀██▀░▄██▀░░░
░░░░░░▀██████▀░░░░░░
░░░░░░░░░░░░░░░░░░░░
"""
import sys
import math
import collections
import operator as op
from collections import deque
from math import gcd, inf, sqrt
from bisect import bisect_right, bisect_left
#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
from functools import reduce
from sys import stdin, stdout, setrecursionlimit
setrecursionlimit(2**20)
def factorial(n):
if n == 0:
return 1
return n * factorial(n - 1)
def ncr(n, r):
r = min(r, n - r)
numer = reduce(op.mul, range(n, n - r, -1), 1)
denom = reduce(op.mul, range(1, r + 1), 1)
return numer // denom # or / in Python 2
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return len(set(factors))
def isPowerOfTwo(x):
return (x and (not(x & (x - 1))))
def factors(n):
return list(set(reduce(list.__add__, ([i, n // i] for i in range(1, int(n**0.5) + 1) if n % i == 0))))
def printp(p):
for i in p:
print(i)
MOD = 10**9 + 7
T = 1
# T = int(stdin.readline())
for _ in range(T):
# n, k = list(map(int, stdin.readline().split()))
# s1 = list(stdin.readline().strip('\n'))
# s = str(stdin.readline().strip('\n'))
# n = int(stdin.readline())
n, k = list(map(int, stdin.readline().split()))
# s = list(stdin.readline().strip('\n'))
a = list(map(int, stdin.readline().split()))
d = {}
ans = ''
for i in range(k):
if a[i] not in d:
d[a[i]] = 1
else:
d[a[i]] += 1
if len(d) != n:
ans += '0'
else:
ans += '1'
d = {}
print(ans)
```
No
| 42,154 | [
0.59619140625,
-0.151611328125,
-0.27001953125,
0.11663818359375,
-0.63818359375,
-0.22314453125,
-0.26953125,
0.36083984375,
0.10552978515625,
0.875,
0.78515625,
-0.0924072265625,
0.45849609375,
-0.97314453125,
-0.2239990234375,
0.056243896484375,
-0.74560546875,
-0.9345703125,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
#Try 1+1
#Unsuccessful
from collections import defaultdict
n,m = map(int, input().split())
List = [int(x) for x in input().split()]
Dict = defaultdict(bool)
Numbers = defaultdict(int)
Ans = str()
count = 0
for i in range(m):
if(Dict[List[i]]==False):
Dict[List[i]]=True
count+=1
Numbers[List[i]]+=1
else:
Numbers[List[i]]+=1
if(count==n):
Ans+="1"
count = 0
Dict = defaultdict(int)
for i in range(1,n+1):
Numbers[i]-=1
if(Numbers[i]>0):
Dict[i] = True
else:
Ans+="0"
print(Ans)
```
No
| 42,155 | [
0.52490234375,
-0.186767578125,
-0.2568359375,
0.026458740234375,
-0.5830078125,
-0.1358642578125,
-0.276611328125,
0.240234375,
0.173583984375,
0.984375,
0.74755859375,
-0.197021484375,
0.45947265625,
-0.9296875,
-0.239501953125,
0.07928466796875,
-0.8662109375,
-0.865234375,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.
To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can't just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.
At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.
You are given a sequence of problems' difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 10^5) — the number of difficulty levels and the number of problems Arkady created.
The second line contains m integers a_1, a_2, …, a_m (1 ≤ a_i ≤ n) — the problems' difficulties in the order Arkady created them.
Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.
Examples
Input
3 11
2 3 1 2 2 2 3 2 2 3 1
Output
00100000001
Input
4 8
4 1 3 3 2 3 3 3
Output
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.
Submitted Solution:
```
a = list(input().split(" "))
b = list(input().split(" "))
aux = set()
for x in range(len(b)):
aux.add(b[x])
if len(aux)==int(a[0]):
a.append("1")
aux = set()
continue
a.append("0")
if(int(a[1])!=0):
del a[0]
del a[0]
b = ""
for x in a:
b += x + " "
print((b))
```
No
| 42,156 | [
0.59912109375,
-0.168701171875,
-0.205810546875,
0.05328369140625,
-0.6171875,
-0.216552734375,
-0.2322998046875,
0.285888671875,
0.12939453125,
0.94189453125,
0.76318359375,
-0.09136962890625,
0.461669921875,
-0.9599609375,
-0.1951904296875,
0.096923828125,
-0.84326171875,
-0.9277... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
def flip(x,y):
global mat
if mat[x][y]:
mat[x][y]=0
else:
mat[x][y]=1
def func(x1,y1,x2,y2):
global mat,k
ones,z=[],[]
if mat[x1][y1]==1:
ones.append((x1+1,y1+1))
else:
z.append((x1+1,y1+1))
if mat[x2][y1]==1:
ones.append((x2+1,y1+1))
else:
z.append((x2+1,y1+1))
if mat[x1][y2]==1:
ones.append((x1+1,y2+1))
else:
z.append((x1+1,y2+1))
if mat[x2][y2]==1:
ones.append((x2+1,y2+1))
else:
z.append((x2+1,y2+1))
while len(ones)!=0:
if len(ones)==4:
k+=1
t=[]
for x,y in ones[:-1]:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
out.append(t)
elif len(ones)==3:
k+=1
t=[]
for x,y in ones:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
out.append(t)
elif len(ones)==2:
k+=1
t=[]
for x,y in ones[:-1]:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
for x,y in z:
mat[x-1][y-1]=1
t.append(x)
t.append(y)
out.append(t)
else:
k+=1
t=[]
for x,y in ones:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
for x,y in z[:-1]:
mat[x-1][y-1]=1
t.append(x)
t.append(y)
out.append(t)
ones,z=[],[]
if mat[x1][y1]==1:
ones.append((x1+1,y1+1))
else:
z.append((x1+1,y1+1))
if mat[x2][y1]==1:
ones.append((x2+1,y1+1))
else:
z.append((x2+1,y1+1))
if mat[x1][y2]==1:
ones.append((x1+1,y2+1))
else:
z.append((x1+1,y2+1))
if mat[x2][y2]==1:
ones.append((x2+1,y2+1))
else:
z.append((x2+1,y2+1))
for _ in range(int(input())):
n,m=map(int,input().split())
mat=[]
for i in range(n):
mat.append(list(map(int,list(input()))))
k=0
out=[]
i=0
if n%2:
j=0
if m%2:
func(n-2,m-2,n-1,m-1)
while j<m-1:
if mat[n-1][j]==1:
flip(n-1,j)
flip(n-2,j)
flip(n-2,j+1)
k+=1
out.append([n,j+1,n-1,j+1,n-1,j+2])
if mat[n-1][j+1]==1:
flip(n-1,j+1)
flip(n-2,j)
flip(n-2,j+1)
k+=1
out.append([n,j+2,n-1,j+1,n-1,j+2])
j+=2
while i<n-1:
j=0
if m%2:
if mat[i][m-1]:
flip(i,m-1)
flip(i,m-2)
flip(i+1,m-2)
k+=1
out.append([i+1,m,i+1,m-1,i+2,m-1])
if mat[i+1][m-1]:
flip(i+1,m-1)
flip(i,m-2)
flip(i+1,m-2)
k+=1
out.append([i+2,m,i+1,m-1,i+2,m-1])
while j<m-1:
func(i,j,i+1,j+1)
j+=2
i+=2
print(k)
for i in out:
print(*i)
```
Yes
| 42,301 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def main():
for _ in range(int(input())):
n, m = [int(x) for x in input().split()]
t = [[int(x) for x in input()] for i in range(n)]
ans = []
def op(a, b, c):
t[a[0]][a[1]] ^= 1
t[b[0]][b[1]] ^= 1
t[c[0]][c[1]] ^= 1
ans.append([a[0]+1, a[1]+1, b[0]+1, b[1]+1, c[0]+1, c[1]+1])
for i in range(n-1, 1, -1):
for j in range(m):
if t[i][j]:
if j + 1 < m:
op((i, j), (i-1, j), (i-1, j+1))
else:
op((i, j), (i-1, j), (i-1, j-1))
for j in range(m-1, 1, -1):
for i in range(2):
if t[i][j]:
op((i, j), (0, j-1), (1, j-1))
for i in range(2):
for j in range(2):
tot = t[0][0] + t[0][1] + t[1][0] + t[1][1]
if (tot + t[i][j]) % 2:
op((1-i, j), (1-i, 1-j), (i, 1-j))
print(len(ans))
for op in ans:
print(*op)
main()
```
Yes
| 42,302 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
for _ in range(0, int(input())):
n, m = map(int, input().split())
mat = [[char for char in input()] for _ in range(0, n)]
answer = []
def update(changes):
answer.append((changes[0], changes[1], changes[2]))
for x, y in changes:
if mat[x][y] == '0':
mat[x][y] = '1'
else:
mat[x][y] = '0'
def change(a, b, c, vec0, vec1):
answer.append((a, b, c))
for x, y in [a, b, c]:
if mat[x][y] == '0':
vec0.remove((x, y))
vec1.append((x, y))
mat[x][y] = '1'
else:
vec0.append((x, y))
vec1.remove((x, y))
mat[x][y] = '0'
return vec0, vec1
if n > 2 or m > 2:
for x in range(0, n, 2):
for j in range(0, m - 1, 1):
i = min(n - 2, x)
if mat[i][j] == '0' and mat[i + 1][j] == '0':
continue
changes = []
if mat[i][j] == '1':
changes.append((i, j))
if mat[i + 1][j] == '1':
changes.append((i + 1, j))
if len(changes) < 3:
changes.append((i, j + 1))
if len(changes) < 3:
changes.append((i + 1, j + 1))
update(changes)
for x in range(0, n - 2):
i = x
j = m - 2
if mat[i][j] == '0' and mat[i][j + 1] == '0':
continue
changes = []
if mat[i][j] == '1':
changes.append((i, j))
if mat[i][j + 1] == '1':
changes.append((i, j + 1))
if len(changes) < 3:
changes.append((i + 1, j))
if len(changes) < 3:
changes.append((i + 1, j + 1))
update(changes)
a = n - 2
b = m - 2
vec0 = [(x, y) for x in range(a, a + 2) for y in range(b, b + 2) if mat[x][y] == '0']
vec1 = [(x, y) for x in range(a, a + 2) for y in range(b, b + 2) if mat[x][y] == '1']
cnt = 0
while len(vec1) > 0:
if len(vec1) == 4:
vec0, vec1 = change(vec1[0], vec1[1], vec1[2], vec0, vec1)
elif len(vec1) == 3:
vec0, vec1 = change(vec1[0], vec1[1], vec1[2], vec0, vec1)
else:
vec0, vec1 = change(vec0[0], vec0[1], vec1[0], vec0, vec1)
print(len(answer))
for x, y, z in answer:
print(x[0] + 1, x[1] + 1, y[0] + 1, y[1] + 1, z[0] + 1, z[1] + 1)
```
Yes
| 42,303 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#######################################
def solve(area,ans,n,m):
if(n%2==1):
n-=1
if(m%2==1):
m-=1
for q in range(0,n,2):
for r in range(0,m,2):
g1=[area[q][r],q+1,r+1]
g2=[area[q+1][r],q+2,r+1]
g3=[area[q][r+1],q+1,r+2]
g4=[area[q+1][r+1],q+2,r+2]
ass=[g1,g2,g3,g4]
ass.sort()
cnt=0
for i in range(len(ass)):
if(ass[i][0]==1):
cnt+=1
if(cnt==4):
t1=[]
t1.extend([ass[0][1],ass[0][2]])
t1.extend([ass[1][1],ass[1][2]])
t1.extend([ass[2][1],ass[2][2]])
for i in range(3):
ass[i][0]=0
ans.append(t1)
t1=[]
cnt=1
if(cnt==3):
t1=[]
for i in range(len(ass)):
if(ass[i][0]==1):
t1.extend([ass[i][1],ass[i][2]])
ans.append(t1)
continue;
cnt=1
if(cnt==2):
t1=[]
for i in range(len(ass)):
if(ass[i][0]==0):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=-1
for i in range(len(ass)):
if(ass[i][0]==1):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=0
break;
ans.append(t1)
t1=[]
for i in range(len(ass)):
if(abs(ass[i][0])==1):
t1.extend([ass[i][1],ass[i][2]])
ans.append(t1)
continue;
if(cnt==1):
t1=[]
for i in range(len(ass)):
if(ass[i][0]==0):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=-1
if(len(t1)==4):
break;
for i in range(len(ass)):
if(ass[i][0]==1):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=0
break;
for i in range(len(ass)):
ass[i][0]=abs(ass[i][0])
ans.append(t1)
cnt=2
if(cnt==2):
t1=[]
for i in range(len(ass)):
if(ass[i][0]==0):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=-1
for i in range(len(ass)):
if(ass[i][0]==1):
t1.extend([ass[i][1],ass[i][2]])
ass[i][0]=0
break;
ans.append(t1)
t1=[]
for i in range(len(ass)):
if(abs(ass[i][0])==1):
t1.extend([ass[i][1],ass[i][2]])
ans.append(t1)
continue;
a=int(input())
for i in range(a):
n,m=map(int,input().split())
ans=[]
area=[]
for i in range(n):
s=input()
g1=[]
for j in range(len(s)):
g1.append(int(s[j]))
area.append(g1)
if(n%2==0 and m%2==0):
solve(area,ans,n,m)
print(len(ans))
for j in range(len(ans)):
print(*ans[j])
continue;
if( m%2==1):
ans=[]
# x varies from 0 to m
# y varies from 0 to n
for j in range(n):
if(area[j][m-1]==1):
area[j][m-1]=0
t1=[]
t1.extend([j+1,m])
t1.extend([j+1,m-1])
area[j][m-2]=1-area[j][m-2]
if(j-1>=0):
area[j-1][m-2]=1-area[j-1][m-2]
t1.extend([j,m-1])
else:
area[j+1][m-2]=1-area[j+1][m-2]
t1.extend([j+2,m-1])
ans.append(t1)
if(n%2==0):
solve(area,ans,n,m)
print(len(ans))
for j in range(len(ans)):
print(*ans[j])
if(n%2==1):
for j in range(m):
if(area[n-1][j]==1):
t1=[]
area[n-1][j]=0
t1.extend([n,j+1])
t1.extend([n-1,j+1])
area[n-2][j]=1-area[n-2][j]
if(j-1>=0):
area[n-2][j-1]=1-area[n-2][j-1]
t1.extend([n-1,j])
else:
area[n-2][j+1]=1-area[n-2][j+1]
t1.extend([n-1,j+2])
ans.append(t1)
solve(area,ans,n,m)
print(len(ans))
for j in range(len(ans)):
print(*ans[j])
```
Yes
| 42,304 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
import sys
import argparse
import copy
solution = { i : [5 for i in range(0,17)] for i in range(0,16) }
dx = [1,-1,0,0,0]
dy = [0,0,1,-1,0]
def within(x,y):
return x <= 1 and y <= 1 and x >= 0 and y >= 0
def decode(x):
ret = ""
for y in x:
ret += ''.join(y)
return int(ret, 2)
def encode(x):
u = bin(x)[2:].zfill(4)
ret = []
a = []
for i in range(0,2):
a.append(int(u[i]))
b = []
for i in range(2,4):
b.append(int(u[i]))
ret.append(a)
ret.append(b)
return ret
def findall(x, steps):
de = decode(x)
if len(solution[de]) > len(steps):
solution[de] = steps
else:
return
for i in range(0,2):
for j in range(0,2):
nx = copy.deepcopy(x)
s = copy.deepcopy(steps)
for l in range(0,5):
if within(i+dx[l], j + dy[l]):
nx[i+dx[l]][j + dy[l]] = '0' if nx[i+dx[l]][j + dy[l]] == '1' else '1'
s.append((i+dx[l],j + dy[l]))
findall(nx, s)
return
solx = { i : [] for i in range(0,16) }
soly = { i : [] for i in range(0,16) }
def pickingtwoandthree():
zerosx = [[0,0],[0,0]]
for x in solution:
y = encode(x)
ret = 0
for i in range(0,2):
for j in range(0,2):
ret += y[i][j]
if ret == 0 or ret == 2 or ret == 3:
solx[x] = zerosx
soly[x] = solution[x]
def solvingonepast():
targetx = [[0,0],[0,0]]
onesx = [['0','0'], ['0', '0']]
for i in range(0,2):
for j in range(0,2):
if i == 0 and j == 0:
continue
onesx[i][j] = '1'
for k in range(0,2):
for l in range(0,2):
targetx[k][l] = 0
action = []
for k in range(0,2):
for l in range(0,2):
if k == 0 and l == 0:
continue
action.append((k,l))
if onesx[k][l] == '1':
targetx[k][l] = 0
else:
targetx[k][l] = 1
x = decode(onesx)
solx[x] = targetx
soly[x] = action
onesx[i][j] = '0'
onesx[0][0] = '1'
x = decode(onesx)
solx[x] = [[0,1],[1,0]]
soly[x] = [(0,0), (0,1), (1,0)]
def solvingone():
targetx = [[0,0],[0,1]]
onesx = [['0','0'], ['0', '0']]
for i in range(0,2):
for j in range(0,2):
onesx[i][j] = '1'
action = []
for k in range(0,2):
for l in range(0,2):
if onesx[k][l] == '0':
action.append((k,l))
action.append((0,0))
action.append((1,0))
action.append((0,1))
x = decode(onesx)
solx[x] = targetx
soly[x] = action
onesx[i][j] = '0'
soly[1] = []
def solvingfour():
targetx = [[0,0],[0,1]]
onesx = [['1','1'], ['1', '1']]
x = decode(onesx)
solx[x] = targetx
soly[x] = [(0,0), (0,1), (1,0)]
def cadd(c0, c1):
return (c0[0]+c1[0], c0[1] + c1[1])
def main():
findall([['0','0'],['0','0']],[])
solvingone()
solvingfour()
pickingtwoandthree()
for _ in range(int(input())):
n,m = list(map(int,input().split()))
arr = []
for i in range(0,n):
x = input()
x = list(x)
arr.append(list(map(int,x)))
ans = []
for i in range(0,n-2):
for j in range(0, m):
nj = j + 1
if j == m-1:
nj = j - 1
if arr[i][j] == 1:
arr[i][j] ^= 1
arr[i+1][j] ^= 1
arr[i+1][nj] ^= 1
ans.append((i,j))
ans.append((i+1,j))
ans.append((i+1,nj))
for i in range(0,m-2):
if arr[n-1][i] == 1 and arr[n-2][i] == 1:
arr[n-1][i] ^= 1
arr[n-2][i] ^= 1
arr[n-2][i+1] ^= 1
ans.append((n-1,i))
ans.append((n-2,i))
ans.append((n-2,i+1))
elif arr[n-1][i] == 1:
arr[n-1][i] ^= 1
arr[n-1][i+1] ^= 1
arr[n-2][i+1] ^= 1
ans.append((n-1,i))
ans.append((n-1,i+1))
ans.append((n-2,i+1))
elif arr[n-2][i] == 1:
arr[n-2][i] ^= 1
arr[n-1][i+1] ^= 1
arr[n-2][i+1] ^= 1
ans.append((n-2,i))
ans.append((n-1,i+1))
ans.append((n-2,i+1))
i = n-2
j = n-2
tmp = []
for k in range(0,2):
for l in range(0,2):
tmp.append(str(arr[i+k][j+l]))
tmpx = decode(tmp)
actions = solution[tmpx]
for a in actions:
ans.append(cadd((i,j),a))
print(int(len(ans)/3))
for x in range(0,len(ans), 3):
nx = x + 3
line = ""
for y in range(x, nx):
line += str(ans[y][0]+1)
line += " "
line += str(ans[y][1]+1)
line += " "
print(line)
if __name__ == "__main__":
parser = argparse.ArgumentParser()
parser.add_argument('--file', dest='filename', default=None)
args = parser.parse_args()
if args.filename is not None:
sys.stdin = open(args.filename)
main()
```
No
| 42,305 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
from sys import stdin,stdout
"""
n=int(stdin.readline().strip())
n,m=map(int,stdin.readline().strip().split())
s=list(map(int,stdin.readline().strip().split()))
"""
def solve(x,y):
c=0
for i in range(x,x+2):
for j in range(y,y+2):
if s[i][j]=='1':
c+=1
if c==0:
return 1
if c==4 or c==3:
aux=[]
z=0
for i in range(x,x+2):
for j in range(y,y+2):
if s[i][j]=='1' and z<=2:
aux.append(i+1)
aux.append(j+1)
s[i][j]='0'
z+=1
ans.append(aux.copy())
return 0
if c==1 or c==2:
y1=0
z=0
aux=[]
for i in range(x,x+2):
for j in range(y,y+2):
if s[i][j]=='1' and y1==0:
aux.append(i+1)
aux.append(j+1)
s[i][j]='0'
y1+=1
continue
if s[i][j]=='0' and z<=1:
aux.append(i+1)
aux.append(j+1)
s[i][j]='1'
z+=1
continue
ans.append(aux.copy())
return 0
def clean1(x,y):
if x+1==n:
return 0
c=0
aux=[]
if(s[x][y+1]=='1'):
aux.append(x+1)
aux.append(y+2)
s[x][y+1]='0'
c+=1
if(s[x+1][y+1]=='1'):
aux.append(x+2)
aux.append(y+2)
s[x+1][y+1]='0'
c+=1
if c==0:
return
if c<=2:
aux.append(x+2)
aux.append(y+1)
if s[x+1][y]=='1':
s[x+1][y]='0'
else:
s[x+1][y]='1'
if c<=1:
aux.append(x+1)
aux.append(y+1)
if s[x][y]=='1':
s[x][y]='0'
else:
s[x][y]='1'
ans.append(aux.copy())
def clean2(x,y):
if y+1==m:
return 0
c=0
aux=[]
if(s[x+1][y]=='1'):
aux.append(x+2)
aux.append(y+1)
s[x+1][y]='0'
c+=1
if(s[x+1][y+1]=='1'):
aux.append(x+2)
aux.append(y+2)
s[x+1][y+1]='0'
c+=1
if c==0:
return
if c<=2:
aux.append(x+1)
aux.append(y+1)
if s[x][y]=='1':
s[x][y]='0'
else:
s[x][y]='1'
if c<=1:
aux.append(x+1)
aux.append(y+2)
if s[x][y+1]=='1':
s[x][y+1]='0'
else:
s[x][y+1]='1'
ans.append(aux.copy())
T=int(stdin.readline().strip())
for caso in range(T):
n,m=map(int,stdin.readline().strip().split())
s=[list(stdin.readline().strip()) for i in range(n)]
ans=[]
x=n-2
y=m-2
q=solve(x,y)
while q!=1:
q=solve(x,y)
x=n-2
y=m-2
if m%2!=0:
for i in range(0,n,2):
clean1(i,m-2)
if n%2!=0:
for i in range(0,m,2):
clean2(n-2,i)
for i in range(0,n-1,2):
for j in range(0,m-1,2):
x=solve(i,j)
while x!=1:
x=solve(i,j)
stdout.write("%d\n" % len(ans))
for i in ans:
stdout.write("%d %d %d %d %d %d\n" % (i[0],i[1],i[2],i[3],i[4],i[5]))
```
No
| 42,306 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @Time : 2020/12/18 16:31
# @Author : zx
# @File : Binary Table (Hard Version).py
# @Email : zx081325@163.com
T = int(input())
for _ in range(T):
n, m = map(int, input().split())
s, res = [], []
for i in range(n):
s.append(list(input()))
for i in range(n - 1):
for j in range(m - 1):
if s[i][j] == '1' and s[i + 1][j] == '0':
res.append([i + 1, j + 1, i + 1, j + 2, i + 2, j + 2])
s[i][j] = '0'
s[i][j + 1] = '0' if s[i][j + 1] == '1' else '1'
s[i + 1][j + 1] = '0' if s[i + 1][j + 1] == '1' else '1'
elif s[i][j] == '0' and s[i + 1][j] == '1':
res.append([i + 2, j + 1, i + 1, j + 2, i + 2, j + 2])
s[i + 1][j] = '0'
s[i][j + 1] = '0' if s[i][j + 1] == '1' else '1'
s[i + 1][j + 1] = '0' if s[i + 1][j + 1] == '1' else '1'
elif s[i][j] == '1' and s[i + 1][j] == '1':
res.append([i + 1, j + 1, i + 1, j + 2, i + 2, j + 1])
s[i][j] = '0'
s[i + 1][j] = '0'
s[i][j + 1] = '0' if s[i][j + 1] == '1' else '1'
for i in range(n - 1):
if s[i][m - 1] == '1':
res.append([i + 1, m - 1, i + 1, m, i + 2, m - 1])
res.append([i + 1, m - 1, i + 2, m, i + 2, m - 1])
s[i + 1][m - 1] = '0' if s[i + 1][m - 1] == '1' else '1'
if s[n - 1][m - 1] == '1':
res.append([n, m, n, m - 1, n - 1, m - 1])
res.append([n - 1, m - 1, n - 1, m, n, m])
res.append([n, m - 1, n - 1, m, n, m])
print(len(res))
for arr in res:
print(" ".join('%s' %d for d in arr))
```
No
| 42,307 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
This is the hard version of the problem. The difference between the versions is in the number of possible operations that can be made. You can make hacks if and only if you solved both versions of the problem.
You are given a binary table of size n × m. This table consists of symbols 0 and 1.
You can make such operation: select 3 different cells that belong to one 2 × 2 square and change the symbols in these cells (change 0 to 1 and 1 to 0).
Your task is to make all symbols in the table equal to 0. You are allowed to make at most nm operations. You don't need to minimize the number of operations.
It can be proved, that it is always possible.
Input
The first line contains a single integer t (1 ≤ t ≤ 5000) — the number of test cases. The next lines contain descriptions of test cases.
The first line of the description of each test case contains two integers n, m (2 ≤ n, m ≤ 100).
Each of the next n lines contains a binary string of length m, describing the symbols of the next row of the table.
It is guaranteed, that the sum of nm for all test cases does not exceed 20000.
Output
For each test case print the integer k (0 ≤ k ≤ nm) — the number of operations.
In the each of the next k lines print 6 integers x_1, y_1, x_2, y_2, x_3, y_3 (1 ≤ x_1, x_2, x_3 ≤ n, 1 ≤ y_1, y_2, y_3 ≤ m) describing the next operation. This operation will be made with three cells (x_1, y_1), (x_2, y_2), (x_3, y_3). These three cells should be different. These three cells should belong to some 2 × 2 square.
Example
Input
5
2 2
10
11
3 3
011
101
110
4 4
1111
0110
0110
1111
5 5
01011
11001
00010
11011
10000
2 3
011
101
Output
1
1 1 2 1 2 2
2
2 1 3 1 3 2
1 2 1 3 2 3
4
1 1 1 2 2 2
1 3 1 4 2 3
3 2 4 1 4 2
3 3 4 3 4 4
4
1 2 2 1 2 2
1 4 1 5 2 5
4 1 4 2 5 1
4 4 4 5 3 4
2
1 3 2 2 2 3
1 2 2 1 2 2
Note
In the first test case, it is possible to make only one operation with cells (1, 1), (2, 1), (2, 2). After that, all symbols will be equal to 0.
In the second test case:
* operation with cells (2, 1), (3, 1), (3, 2). After it the table will be:
011
001
000
* operation with cells (1, 2), (1, 3), (2, 3). After it the table will be:
000
000
000
In the fifth test case:
* operation with cells (1, 3), (2, 2), (2, 3). After it the table will be:
010
110
* operation with cells (1, 2), (2, 1), (2, 2). After it the table will be:
000
000
Submitted Solution:
```
def func(x1,y1,x2,y2):
global mat,k
ones,z=[],[]
if mat[x1][y1]==1:
ones.append((x1+1,y1+1))
else:
z.append((x1+1,y1+1))
if mat[x2][y1]==1:
ones.append((x2+1,y1+1))
else:
z.append((x2+1,y1+1))
if mat[x1][y2]==1:
ones.append((x1+1,y2+1))
else:
z.append((x1+1,y2+1))
if mat[x2][y2]==1:
ones.append((x2+1,y2+1))
else:
z.append((x2+1,y2+1))
while len(ones)!=0:
if len(ones)==4:
k+=1
t=[]
for x,y in ones[:-1]:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
out.append(t)
elif len(ones)==3:
k+=1
t=[]
for x,y in ones:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
out.append(t)
elif len(ones)==2:
k+=1
t=[]
for x,y in ones[:-1]:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
for x,y in z:
mat[x-1][y-1]=1
t.append(x)
t.append(y)
out.append(t)
else:
k+=1
t=[]
for x,y in ones:
mat[x-1][y-1]=0
t.append(x)
t.append(y)
for x,y in z[:-1]:
mat[x-1][y-1]=1
t.append(x)
t.append(y)
out.append(t)
ones,z=[],[]
if mat[x1][y1]==1:
ones.append((x1+1,y1+1))
else:
z.append((x1+1,y1+1))
if mat[x2][y1]==1:
ones.append((x2+1,y1+1))
else:
z.append((x2+1,y1+1))
if mat[x1][y2]==1:
ones.append((x1+1,y2+1))
else:
z.append((x1+1,y2+1))
if mat[x2][y2]==1:
ones.append((x2+1,y2+1))
else:
z.append((x2+1,y2+1))
for _ in range(int(input())):
n,m=map(int,input().split())
mat=[]
for i in range(n):
mat.append(list(map(int,list(input()))))
k=0
out=[]
i=0
if n%2:
j=0
if m%2:
func(n-2,m-2,n-1,m-1)
while j<m-1:
if mat[n-1][j]==1:
k+=1
out.append([n,j+1,n-1,j,n-1,j+1])
j+=1
while i<n-1:
j=0
if m%2 and mat[i][m-1]==1:
mat[i][m-1]=0
k+=1
out.append([i+1,m,i+1,m-1,i+2,m-1])
while j<m-1:
func(i,j,i+1,j+1)
j+=2
i+=2
print(k)
for i in out:
print(*i)
```
No
| 42,308 | [
0.1798095703125,
-0.024658203125,
-0.08465576171875,
-0.19775390625,
-0.5888671875,
-0.3466796875,
-0.10693359375,
-0.03936767578125,
0.1805419921875,
0.85986328125,
0.88330078125,
-0.0015954971313476562,
0.31787109375,
-0.75927734375,
-0.3125,
0.01065826416015625,
-0.66796875,
-0.... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
# -*- coding:utf-8 -*-
"""
created by shuangquan.huang at 1/7/20
"""
import collections
import time
import os
import sys
import bisect
import heapq
from typing import List
def solve(N, A, M, bugs, MOD):
dp = [[[0 for _ in range(bugs + 1)] for _ in range(M + 1)] for _ in range(2)]
for j in range(1, M + 1):
if A[0] * j <= bugs:
dp[1][j][A[0] * j] = 1
dp[1][0][0] = 1
dp[0][0][0] = 1
for i in range(2, N + 1):
for j in range(1, M + 1):
for b in range(bugs + 1):
s = dp[i % 2][j - 1][b - A[i - 1]] if b >= A[i - 1] else 0
s += dp[(i + 1) % 2][j][b]
s %= MOD
dp[i % 2][j][b] = s
return sum(dp[N % 2][M]) % MOD
N, M, B, MOD = map(int, input().split())
A = [int(x) for x in input().split()]
print(solve(N, A, M, B, MOD))
```
| 42,486 | [
0.260986328125,
-0.07940673828125,
-0.07080078125,
0.277099609375,
-0.315673828125,
-0.80517578125,
0.0131988525390625,
-0.08673095703125,
0.4013671875,
0.9658203125,
0.173095703125,
-0.1517333984375,
0.1583251953125,
-0.463134765625,
-0.424072265625,
-0.06671142578125,
-0.603515625,... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
def main():
n, m, b, mod = map(int, input().split())
b += 1
nxt = [[0] * b for _ in range(m)]
row_zero = [0] * b
row_zero[0] = 1
for a in list(map(int, input().split())):
cur, nxt = nxt, []
src0 = row_zero
for src1 in cur:
src0 = [(x + src0[i]) % mod if i >= 0 else x for i, x in enumerate(src1, -a)]
nxt.append(src0)
print(sum(nxt[-1]) % mod)
if __name__ == '__main__':
main()
```
| 42,487 | [
0.281494140625,
0.004329681396484375,
-0.1361083984375,
0.12432861328125,
-0.3232421875,
-0.8828125,
-0.011077880859375,
0.00449371337890625,
0.33251953125,
1.0048828125,
0.2274169921875,
-0.09320068359375,
0.19091796875,
-0.483154296875,
-0.3076171875,
-0.08258056640625,
-0.57275390... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
a = list(map(int, input().split()))
n = a[0]
m = a[1]
b = a[2]
mod = a[3]
ac = list(map(int,input().split()))
ac = [0] + ac
dp = [[[0 for k in range(b+1)] for _ in range(m+1)] for z in range(2)]
for i in range(n+1) :
for x in range(b+1) :
dp[i%2][0][x] = 1
for i in range(1,n+1) :
for j in range(1,m+1) :
for x in range(b+1) :
if ac[i] <= x :
dp[i%2][j][x] = (dp[(i-1)%2][j][x] + dp[i%2][j-1][x-ac[i]] ) % mod
else :
dp[i%2][j][x] = dp[(i-1)%2][j][x] % mod
print(dp[n%2][m][b])
```
| 42,488 | [
0.2978515625,
-0.054168701171875,
-0.09722900390625,
0.12091064453125,
-0.328125,
-0.87158203125,
-0.0221405029296875,
0.0164794921875,
0.318603515625,
0.962890625,
0.2210693359375,
-0.0980224609375,
0.1766357421875,
-0.50048828125,
-0.324462890625,
-0.056915283203125,
-0.6044921875,... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
n, m, b, mod = map(int, input().split())
a = list(map(int, input().split()))
dp = [[0 for col in range(b + 1)]for row in range(m + 1)]
dp[0][0] = 1
for i in range(n):
for j in range(1, m + 1):
for k in range(a[i], b + 1):
dp[j][k] = (dp[j][k] + dp[j - 1][k - a[i]]) % mod
ans = 0
for i in range(b + 1):
ans = (ans + dp[m][i]) % mod
print(ans)
```
| 42,489 | [
0.3017578125,
-0.055145263671875,
-0.08221435546875,
0.12939453125,
-0.336669921875,
-0.86328125,
-0.036712646484375,
0.0225677490234375,
0.311279296875,
0.99853515625,
0.19580078125,
-0.09637451171875,
0.1640625,
-0.439697265625,
-0.3037109375,
-0.046722412109375,
-0.58935546875,
... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
n, m, b, mod = list(map(int, input().split()))
a = list(map(int, input().split()))
A = [[0 for i in range(m + 1)] for j in range(b+1)]
A[0][0] = 1
for i in range(n):
for j in range(a[i], b + 1):
for k in range(m):
A[j][k + 1] = (A[j][k + 1] + A[j - a[i]][k]) % mod
#print(A)
ans = 0
for i in range(b + 1):
ans = (ans + A[i][m]) % mod
print(ans)
```
| 42,490 | [
0.30712890625,
-0.06121826171875,
-0.126708984375,
0.0963134765625,
-0.35498046875,
-0.84912109375,
-0.0119171142578125,
0.04193115234375,
0.3134765625,
1.01953125,
0.2100830078125,
-0.099609375,
0.1876220703125,
-0.47216796875,
-0.31201171875,
-0.02935791015625,
-0.60009765625,
-0... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
import sys,os,io
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def ii():
return int(input())
def li():
return list(map(int,input().split()))
n,m,b,mod = li()
a = li()
dp = [[0 for i in range(b+1)] for j in range(m+1)]
dp[0][0]=1
for i in range(n):
for j in range(1,m+1):
for k in range(a[i],b+1):
dp[j][k]+=dp[j-1][k-a[i]]
dp[j][k]%=mod
ans = 0
for j in range(b+1):
ans+= dp[m][j]
print(ans%mod)
```
| 42,491 | [
0.3095703125,
-0.11566162109375,
-0.15869140625,
0.23681640625,
-0.35595703125,
-0.88623046875,
0.0270843505859375,
0.07318115234375,
0.4794921875,
1.0498046875,
0.2310791015625,
-0.0711669921875,
0.269287109375,
-0.47021484375,
-0.341552734375,
-0.040771484375,
-0.6220703125,
-1.0... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m,b,mod=map(int,input().split())
a=list(map(int,input().split()))
dp=[[0 for _ in range(b+1)] for _ in range(m+1)]
dp[0][0]=1
for i in a:
for j in range(1,m+1):
for k in range(i,b+1):
dp[j][k]+=dp[j-1][k-i]
dp[j][k]%=mod
su=0
for i in dp[m]:
su+=i
su%=mod
print(su)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 42,492 | [
0.28662109375,
-0.055908203125,
-0.1651611328125,
0.128662109375,
-0.350341796875,
-0.865234375,
-0.032379150390625,
0.0182037353515625,
0.364990234375,
0.990234375,
0.208740234375,
-0.10162353515625,
0.2088623046875,
-0.439697265625,
-0.356201171875,
-0.06170654296875,
-0.634765625,... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m,b,mod = map(int,input().split())
dp = [[0]*(b+1) for _ in range(m+1)]
bugs = list(map(int,input().split()))
dp[0][0] = 1
for j in bugs:
for i in range(1,m+1):
for k in range(j,b+1):
dp[i][k] += dp[i-1][k-j]
dp[i][k] %= mod
ans = 0
for i in dp[m]:
ans += i
ans %= mod
print(ans)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == '__main__':
main()
```
| 42,493 | [
0.29833984375,
-0.10247802734375,
-0.1009521484375,
0.129150390625,
-0.34716796875,
-0.85693359375,
-0.0262451171875,
0.0458984375,
0.33349609375,
0.98046875,
0.2139892578125,
-0.08392333984375,
0.180908203125,
-0.40087890625,
-0.331298828125,
-0.0911865234375,
-0.6552734375,
-0.88... | 11 |
Provide tags and a correct Python 2 solution for this coding contest problem.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Tags: dp
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
raw_input = stdin.readline
pr = stdout.write
def in_arr():
return map(int,raw_input().split())
def pr_num(n):
stdout.write(str(n)+'\n')
def pr_arr(arr):
for i in arr:
stdout.write(str(i)+' ')
stdout.write('\n')
range = xrange # not for python 3.0+
#main code
n,m,b,mod=in_arr()
l=in_arr()
dp=[[[0 for i in range(b+1)] for j in range(m+1)] for k in range(2)]
dp[0][0][0]=1
for tp in range(1,n+1):
i=tp%2
for j in range(m+1):
for k in range(b+1):
dp[i][j][k]=dp[i^1][j][k]
if j>0 and k>=l[tp-1]:
dp[i][j][k]+=dp[i][j-1][k-l[tp-1]]
dp[i][j][k]%=mod
ans=0
for i in range(k+1):
ans=(ans+dp[n%2][m][i])%mod
pr_num(ans)
```
| 42,494 | [
0.214599609375,
0.026336669921875,
-0.0853271484375,
0.19384765625,
-0.400634765625,
-0.9306640625,
-0.1854248046875,
0.036712646484375,
0.34765625,
1.048828125,
0.2294921875,
-0.200927734375,
0.220458984375,
-0.351806640625,
-0.39013671875,
-0.08673095703125,
-0.6484375,
-0.933593... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
n, m, b, mod = map(int, input().split())
a = list(map(int, input().split()))
dp = [[0 for i in range(m + 1)] for j in range(b + 1)]
dp[0][0] = 1
for i in range(n):
for j in range(a[i], b + 1):
for k in range(m):
dp[j][k + 1] = (dp[j][k + 1] + dp[j - a[i]][k]) % mod
res = 0
for i in range(b + 1):
res = (res + dp[i][m]) % mod
print(res)
```
Yes
| 42,495 | [
0.386474609375,
-0.03509521484375,
-0.2042236328125,
0.04345703125,
-0.448974609375,
-0.69580078125,
-0.026824951171875,
0.1531982421875,
0.273193359375,
1.0439453125,
0.197021484375,
-0.09814453125,
0.1708984375,
-0.54443359375,
-0.327392578125,
-0.177001953125,
-0.599609375,
-0.9... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
n, m, b, mod = list(map(int, input().split()))
a = list(map(int, input().split()))
A = [[0 for i in range(m + 1)] for j in range(b + 1)]
A[0][0] = 1
for i in range(n):
for j in range(a[i], b + 1):
for k in range(m):
A[j][k + 1] = (A[j][k + 1] + A[j - a[i]][k]) % mod
ans = 0
for i in range(b + 1):
ans = (ans + A[i][m]) % mod
print(ans)
```
Yes
| 42,496 | [
0.396484375,
-0.03729248046875,
-0.2100830078125,
0.0283050537109375,
-0.4609375,
-0.6943359375,
-0.0246124267578125,
0.168212890625,
0.266357421875,
1.0458984375,
0.20068359375,
-0.0921630859375,
0.173583984375,
-0.54736328125,
-0.32275390625,
-0.1727294921875,
-0.607421875,
-0.96... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
def main():
n, m, b, mod = map(int, input().split())
row_zero = [1] + [0] * b
b += 1
dp = [[0] * b for _ in range(m)]
for a in list(map(int, input().split())):
cur = row_zero
for nxt in dp:
for i, u in zip(range(a, b), cur):
nxt[i] = (nxt[i] + u) % mod
cur = nxt
print(sum(dp[-1]) % mod)
if __name__ == '__main__':
main()
```
Yes
| 42,497 | [
0.421630859375,
0.01203155517578125,
-0.1861572265625,
0.0229949951171875,
-0.447265625,
-0.6904296875,
-0.01021575927734375,
0.125244140625,
0.2296142578125,
1.0634765625,
0.21142578125,
-0.08038330078125,
0.12115478515625,
-0.52197265625,
-0.303955078125,
-0.2232666015625,
-0.55468... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
import sys
input = sys.stdin.readline
n, m, b, mod = map(int, input().split())
a = list(map(int, input().split()))
ways = [[[0] * (b + 1) for j in range(m + 1)] for i in range(2)]
for nn in range(n + 1):
ways[nn % 2][0][0] = 1
for nn in range(1, n + 1):
i = nn & 1
for mm in range(1, m + 1):
for bb in range(a[nn - 1], b + 1):
ways[i][mm][bb] = ways[i ^ 1][mm][bb] + ways[i][mm - 1][bb - a[nn - 1]]
if ways[i][mm][bb] >= mod:
ways[i][mm][bb] -= mod
print(sum(ways[n & 1][m][bb] for bb in range(b + 1)) % mod)
```
No
| 42,498 | [
0.36181640625,
0.0006155967712402344,
-0.255126953125,
0.0799560546875,
-0.54052734375,
-0.69775390625,
-0.0014562606811523438,
0.1353759765625,
0.2132568359375,
1.0634765625,
0.156494140625,
-0.096435546875,
0.235107421875,
-0.60595703125,
-0.3896484375,
-0.1898193359375,
-0.5346679... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m,b,mod=map(int,input().split())
a=list(map(int,input().split()))
dp=[[0 for _ in range(b+1)] for _ in range(m+1)]
dp[0][0]=1
for i in a:
for j in range(1,m+1):
for k in range(j,b+1):
dp[j][k]+=dp[j-1][k-i]
dp[j][k]%=mod
su=0
for i in dp[m]:
su+=i
su%=mod
print(su)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 42,499 | [
0.4140625,
-0.0386962890625,
-0.2440185546875,
0.056060791015625,
-0.4560546875,
-0.69775390625,
0.0022640228271484375,
0.1707763671875,
0.335693359375,
1.064453125,
0.2127685546875,
-0.1356201171875,
0.1668701171875,
-0.49462890625,
-0.378662109375,
-0.146484375,
-0.57666015625,
-... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
def main():
n, m, b, mod = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
sol = [[0 for j in range(b+1)] for i in range(m+1)]
sol[0][0] = 1
for i in range(n):
for j in range(1,m+1):
for k in range(a[i], b+1):
sol[j][k] = (sol[j][k] + sol[j - 1][k - a[i]]) % mod
print( sum(sol[m]) )
if __name__ == '__main__':
main()
```
No
| 42,500 | [
0.423583984375,
-0.025421142578125,
-0.199951171875,
0.0155029296875,
-0.45947265625,
-0.66357421875,
0.0133209228515625,
0.165771484375,
0.310302734375,
1.0283203125,
0.2152099609375,
-0.1124267578125,
0.150634765625,
-0.55126953125,
-0.35009765625,
-0.2005615234375,
-0.603515625,
... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly ai bugs in every line of code that he writes.
Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v1 lines of the given task, then the second programmer writes v2 more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.
Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.
Input
The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 109 + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.
The next line contains n space-separated integers a1, a2, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.
Output
Print a single integer — the answer to the problem modulo mod.
Examples
Input
3 3 3 100
1 1 1
Output
10
Input
3 6 5 1000000007
1 2 3
Output
0
Input
3 5 6 11
1 2 1
Output
0
Submitted Solution:
```
import sys
def choose(n, k):
result = 1
for i in range(k):
result *= n-i
result = result // (i+1)
return result
nn, mm, bb, mod = [int(i) for i in sys.stdin.readline().split()]
bugs_per_line = [int(i) for i in sys.stdin.readline().split()]
bugs_per_line.sort()
bugs_per_line_counts = [(x, bugs_per_line.count(x)) for x in set(bugs_per_line)]
memo = {}
def solve(n, m, b):
# returns number of solutions if we only use n lowest bug counts, write m lines, have at most b bugs
# or -1 if no solution
if n == 0:
if m == 0:
return 1
else:
return -1
if (n, m, b) in memo:
return memo[(n, m, b)]
result = 0
solution_exists = False
if bugs_per_line_counts[n-1][0] == 0:
maxlines = m
else:
maxlines = b // bugs_per_line_counts[n-1][0]
for i in range(maxlines+1):
query = solve(n-1, m-i, b-(i*bugs_per_line_counts[n-1][0]))
if (query != -1):
solution_exists = True
result += query * choose(i+bugs_per_line_counts[n-1][1]-1, bugs_per_line_counts[n-1][1]-1)
result %= mod
else:
if n == 1: continue
if solution_exists:
memo[(n, m, b)] = result
return result
else:
return -1;
if solution_exists:
memo[(n, m, b)] = result
return result
else:
return -1;
print(solve(len(bugs_per_line_counts), mm, bb))
```
No
| 42,501 | [
0.366455078125,
-0.0814208984375,
-0.1959228515625,
0.0723876953125,
-0.495849609375,
-0.66064453125,
-0.0260467529296875,
0.20166015625,
0.288818359375,
1.01171875,
0.210693359375,
-0.1502685546875,
0.1390380859375,
-0.51953125,
-0.4326171875,
-0.2391357421875,
-0.6083984375,
-0.9... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
from operator import itemgetter
class SegTree: # 0-index !!!
"""
fx: モノイドXでの二項演算
ex: モノイドXでの単位元
init(seq, fx, ex): 配列seqで初期化 O(N)
update(i, x): i番目の値をxに更新 O(logN)
query(l, r): 区間[l,r)をfxしたものを返す O(logN)
get(i): i番目の値を返す
show(): 配列を返す
"""
def __init__(self, seq, fx, ex):
self.n = len(seq)
self.fx = fx
self.ex = ex
self.size = 1<<(self.n - 1).bit_length()
self.tree = [ex] * (self.size * 2)
# build
for i, x in enumerate(seq, start=self.size):
self.tree[i] = x
for i in reversed(range(1, self.size)):
self.tree[i] = self.fx(self.tree[2 * i], self.tree[2 * i + 1])
def set(self, i, x): # O(log(n))
i += self.size
self.tree[i] = x
while i:
i >>= 1
self.tree[i] = self.fx(self.tree[2 * i], self.tree[2 * i + 1])
def update(self, i, x):
i += self.size
self.tree[i] = x
while i > 1:
i >>= 1
self.tree[i] = self.fx(self.tree[2 * i], self.tree[2 * i + 1])
def query(self, l, r): # l = r の場合はたぶんバグるので注意
tmp_l = self.ex
tmp_r = self.ex
l += self.size
r += self.size
while l < r:
if l & 1:
tmp_l = self.fx(tmp_l, self.tree[l])
l += 1
if r & 1:
tmp_r = self.fx(self.tree[r - 1], tmp_r) # 交換法則を仮定しない(順序大事に)
l >>= 1
r >>= 1
return self.fx(tmp_l, tmp_r)
def get(self, i):
return self.tree[self.size + i]
def show(self):
return self.tree[self.size: self.size + self.n]
# ---------------------- #
n = int(input())
LR = [tuple(int(x) for x in input().split()) for _ in range(n)]
LR.sort(key=itemgetter(0))
L = []
R = []
for l, r in LR:
L.append(l)
R.append(r)
seg = SegTree(R, fx=min, ex=10**18)
ans = 0
for i in range(1, n):
contest1 = seg.query(0, i) - L[i - 1] + 1
contest2 = seg.query(i, n) - L[n - 1] + 1
ans = max(ans, contest1 + contest2)
LR.sort(key=lambda item: item[1] - item[0], reverse=True)
lmax = max(l for l, _ in LR[1:])
rmin = min(r for _, r in LR[1:])
ans = max(ans, LR[0][1] - LR[0][0] + 1 + max(rmin - lmax + 1, 0))
print(ans)
```
| 42,672 | [
0.5615234375,
0.072265625,
-0.0222015380859375,
0.35107421875,
-0.58251953125,
-0.619140625,
-0.061004638671875,
0.408203125,
0.1783447265625,
0.7353515625,
0.0897216796875,
-0.10260009765625,
0.01049041748046875,
-0.4990234375,
-0.335693359375,
-0.2529296875,
-0.619140625,
-0.7080... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
from bisect import *
from collections import *
from fractions import gcd
from math import factorial
from itertools import *
from heapq import *
N=int(input())
LR=[list(map(int,input().split())) for i in range(N)]
for i in range(N):
LR[i][1]+=1
value=max([R-L for L,R in LR])
#LR=sorted(LR,key=lambda x:x[1])
LR=sorted(LR,key=lambda x:x[0]*10**9+x[1])
a=2*(10**9)
minRlst=[a for i in range(N)]
minRlst[-1]=LR[-1][1]
maxLlst=[0 for i in range(N)]
maxLlst[0]=LR[0][0]
for i in range(N-2,-1,-1):
minRlst[i]=min(minRlst[i+1],LR[i][1])
for i in range(1,N):
maxLlst[i]=max(minRlst[i-1],LR[i][0])
L1,R1=LR[0]
L2,R2=LR[-1][0],minRlst[1]
maxR=min([R for L,R in LR])
for i in range(0,N-1):
L,R=LR[i]
a=max(R-L,0)
b=max(maxR-LR[-1][0],0)
value=max(value,a+b)
L1=L
R1=min(R1,R)
R2=minRlst[i+1]
a=max(R1-L1,0)
b=max(R2-L2,0)
value=max(value,a+b)
print(value)
```
| 42,673 | [
0.5791015625,
-0.030914306640625,
-0.07257080078125,
0.312255859375,
-0.54052734375,
-0.57666015625,
0.1690673828125,
0.08905029296875,
0.1512451171875,
0.8212890625,
0.1917724609375,
-0.10638427734375,
0.100830078125,
-0.6484375,
-0.32763671875,
-0.2318115234375,
-0.64892578125,
-... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
import sys
import math
from collections import defaultdict
sys.setrecursionlimit(10**7)
def input():
return sys.stdin.readline()[:-1]
mod = 10**9 + 7
def I(): return int(input())
def II(): return map(int, input().split())
def III(): return list(map(int, input().split()))
def Line(N,num):
if N<=0:
return [[]]*num
elif num==1:
return [I() for _ in range(N)]
else:
read_all = [tuple(II()) for _ in range(N)]
return map(list, zip(*read_all))
#################
N = I()
L,R = Line(N,2)
l = max(L)
li = L.index(l)
r = min(R)
ri = R.index(r)
ans = max(r-l+1,0) + max([0 if (i==li or i==ri) else R[i]-L[i]+1 for i in range(N)])
if li==ri:
print(ans)
exit()
x = []
for i in range(N):
if i==li or i==ri:
continue
else:
x1 = max(R[i]-l+1, 0)
x2 = max(r-L[i]+1, 0)
x.append((x1,x2))
x.sort(key=lambda s:(-s[0],s[1]))
amin = R[li]-l+1
bmin = r-L[ri]+1
one = [amin]
two = [bmin]
for x0 in x:
amin = min(amin,x0[0])
one.append(amin)
for x0 in x[::-1]:
bmin = min(bmin,x0[1])
two.append(bmin)
two = list(reversed(two))
for i in range(len(one)):
temp = one[i]+two[i]
if temp>ans:
ans = temp
print(ans)
```
| 42,674 | [
0.54736328125,
0.0753173828125,
0.024810791015625,
0.257568359375,
-0.69677734375,
-0.56689453125,
-0.01399993896484375,
0.25830078125,
0.13525390625,
0.80029296875,
0.1737060546875,
-0.031280517578125,
0.13330078125,
-0.44873046875,
-0.29931640625,
-0.1973876953125,
-0.5869140625,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
#!/usr/bin/env python3
import sys
def solve(N: int, L: "List[int]", R: "List[int]"):
LR= list(zip(L,R))
LR.sort(key=lambda x:x[0])
r_left_max = [] ##index 以下のrightのmin
for lr in LR:
if r_left_max:
r_left_max.append(min(r_left_max[-1],lr[1]))
else:
r_left_max.append(lr[1])
r_right_max = []
for lr in LR[::-1]:
if r_right_max:
r_right_max.append(min(r_right_max[-1],lr[1]))
else:
r_right_max.append(lr[1])
answer = 0
for i in range(1,N):
a = 0
a += max(r_left_max[i-1]-LR[i-1][0]+1,0)
a += max(r_right_max[N-i-1]-LR[-1][0]+1,0)
answer = max(a,answer)
## 一つとその他の場合。indexがiのものを一つにする
for i in range(N):
a =0
a+= LR[i][1]-LR[i][0]+1
if i == N-1:
min_upper_limit = r_left_max[N-2]
a+= max(min_upper_limit-LR[N-2][0]+1,0)
elif i == 0:
min_upper_limit = r_right_max[N-2]
a+= max(min_upper_limit-LR[-1][0]+1,0)
else:
min_upper_limit = min(r_left_max[i-1],r_right_max[N-i-2])
a+= max(min_upper_limit-LR[-1][0]+1,0)
answer = max(a,answer)
print(answer)
return
def main():
def iterate_tokens():
for line in sys.stdin:
for word in line.split():
yield word
tokens = iterate_tokens()
N = int(next(tokens)) # type: int
L = [int()] * (N) # type: "List[int]"
R = [int()] * (N) # type: "List[int]"
for i in range(N):
L[i] = int(next(tokens))
R[i] = int(next(tokens))
solve(N, L, R)
if __name__ == '__main__':
main()
```
| 42,675 | [
0.580078125,
0.0264129638671875,
-0.0643310546875,
0.31396484375,
-0.6337890625,
-0.64208984375,
0.0009493827819824219,
0.330322265625,
0.177734375,
0.7392578125,
0.1383056640625,
-0.084228515625,
0.041290283203125,
-0.5341796875,
-0.380126953125,
-0.255126953125,
-0.61181640625,
-... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
from sys import stdin
def main():
N, *LR = map(int, stdin.buffer.read().split())
L, R = LR[::2], LR[1::2]
Lp = max(L)
Rq = min(R)
A, B = zip(*sorted((max(0, r - Lp + 1), -max(0, Rq - l + 1)) for l, r in zip(L, R)))
mi = float("inf")
ma = max(r - l + 1 for l, r in zip(L, R)) + max(0, Rq - Lp + 1)
for a, b in zip(A[1:], B):
mi = min(mi, -b)
ma = max(ma, mi + a)
print(ma)
main()
```
| 42,676 | [
0.6044921875,
0.04217529296875,
0.0121612548828125,
0.273681640625,
-0.66357421875,
-0.62646484375,
-0.0411376953125,
0.28564453125,
0.054840087890625,
0.77880859375,
0.182373046875,
-0.08966064453125,
0.036834716796875,
-0.5380859375,
-0.331787109375,
-0.2763671875,
-0.5751953125,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
def solve(N, LRs):
from operator import itemgetter
L = 0
R = 1
IND = 2
LRs.sort(key=itemgetter(L)) # L昇順ソート
LRs = tuple(LR + (ind,) for ind, LR in enumerate(LRs))
min_r = min(LRs, key=itemgetter(R)) # R最小の区間 == 右端が一番左の区間
max_l = max(LRs, key=itemgetter(L)) # L最大の区間 == 左端が一番右の区間
ans = 0
if min_r[IND] == max_l[IND]:
# L最大とR最小の問題が一致する場合
fixed = max(0, min_r[R] - max_l[L] + 1)
max_ = max(0, max(x[R] - x[L] + 1 for x in LRs if x[IND] != max_l[IND]))
ans = max(ans, fixed + max_)
else:
# L最大とR最小の問題が異なる場合
# 両方の問題を同日実施する場合
fixed = max(0, min_r[R] - max_l[L] + 1)
max_ = max(0, max(x[R] - x[L] + 1 for x in LRs if x[IND] != max_l[IND]))
ans = max(ans, fixed + max_)
r_mins = []
t = 10 ** 9 + 1
for LR in reversed(LRs):
if LR[IND] != min_r[IND]:
t = min(t, LR[R])
r_mins.append(t)
r_mins.reverse()
# 別日に実施する場合
for x in LRs:
# min_rと同日に実施する問題で、L最大の問題x
if x[IND] == max_l[IND]:
continue
fixed = max(0, min_r[R] - x[L] + 1)
r_ = r_mins[x[IND] + 1]
opp = max(0, r_ - max_l[L] + 1)
ans = max(ans, fixed + opp)
return ans
if __name__ == '__main__':
N = int(input())
LRs = [tuple(map(int, input().split())) for _ in range(N)]
print(solve(N, LRs))
# 残った区間
# <-->, <--->
# <>は、いずれかの区間の左右端なので、n通り**4ある
# 右端が一番左にあるような区間に注目すると
# どの区間と組み合わせても、右端は変わらない（emptyになることはある）
# 右端が一番左 / 左端が一番右 に注目する
# 例外 == 両者一致 == 含まれるケース
# <----->
# <--->
# <-> ... この区間を含むとこの区間だけになるので、最長区間とそれ以外(これを含む）に分ける
# 右端が一番左 / 左端が一番右 を同じ区間に入れる場合と違う区間に入れる場合に分ける
# 同じ区間に入れる場合は、例外同様、最長区間とその他に分ける
# 右端が一番左 / 左端が一番右 を別日に実施する場合、
# x---o (右端は固定）
# xに決めたとき、 o---y yを最長になるようにしたい
# x---o (右端は固定）
# <--- xより左に左端があるもの : 別グループ（左端が一番右の区間以下確定）に入れると、
# 区間長維持または減少なので、x----oと同じグループに入れて、別グループの長さを維持したい
# <--- oより右に左端があるもの : 別グループに入れないとx---o区間を取るという決め方に反する
```
| 42,677 | [
0.54052734375,
0.022918701171875,
-0.1541748046875,
0.240234375,
-0.544921875,
-0.732421875,
0.056793212890625,
0.304931640625,
0.2271728515625,
0.75439453125,
0.2083740234375,
-0.036407470703125,
0.04638671875,
-0.54541015625,
-0.47021484375,
-0.34375,
-0.591796875,
-0.72265625,
... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
N, *LR = map(int, open(0).read().split())
L, R = LR[::2], LR[1::2]
Lp = max(L)
Rq = min(R)
A = sorted(max(0, r - Lp + 1) for r in R)
B = sorted(-max(0, Rq - l + 1) for l in L)
mi = float("inf")
ma = max(r - l + 1 for l, r in zip(L, R)) + max(0, Rq - Lp + 1)
for a, b in zip(A[1:], B):
mi = min(mi, -b)
ma = max(ma, mi + a)
print(ma)
```
| 42,678 | [
0.669921875,
0.012603759765625,
-0.049346923828125,
0.2152099609375,
-0.5810546875,
-0.6376953125,
0.059906005859375,
0.27197265625,
0.09832763671875,
0.751953125,
0.24365234375,
-0.040802001953125,
0.06329345703125,
-0.5615234375,
-0.322265625,
-0.251708984375,
-0.62255859375,
-0.... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
"Correct Solution:
```
N, *LR = map(int, open(0).read().split())
L, R = LR[::2], LR[1::2]
max_l = max(L)
min_r = min(R)
A = sorted(max(0, r - max_l + 1) for r in R)
B = sorted(-max(0, min_r - l + 1) for l in L)
mi = float("inf")
ma = max(r - l + 1 for l, r in zip(L, R)) + max(0, min_r - max_l + 1)
for i in range(N - 1):
mi = min(mi, -B[i])
ma = max(ma, mi + A[i + 1])
print(ma)
```
| 42,679 | [
0.67578125,
0.0230712890625,
-0.043975830078125,
0.1976318359375,
-0.5947265625,
-0.650390625,
0.08306884765625,
0.28466796875,
0.0703125,
0.74169921875,
0.251953125,
-0.053131103515625,
0.08294677734375,
-0.58154296875,
-0.328857421875,
-0.259521484375,
-0.6298828125,
-0.775390625... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
# coding: utf-8
def solve(*args: str) -> str:
n = int(args[0])
LR = [tuple(map(int, a.split())) for a in args[1:]]
L, R = zip(*LR)
ret = 0
# lp, rq = max(L), min(R)
lp, rq = 0, 1+10**9
for l, r in LR:
lp, rq = max(lp, l), min(rq, r)
ret = max(0, 1+rq-lp) + max(0, max(1+r-l for l, r in LR))
AB = [[max(0, 1+r-lp), max(0, 1+rq-l)] for l, r in LR]
AB.sort(key=lambda x: (x[0], -x[1]))
A, B = map(list, zip(*AB))
# for i in range(1, n):
# ret = max(ret, min(A[i:]) + min(B[:i]))
b_min = 1+10**9
for i in range(n-1):
b_min = min(b_min, B[i])
ret = max(ret, b_min + A[i+1])
return str(ret)
if __name__ == "__main__":
print(solve(*(open(0).read().splitlines())))
```
Yes
| 42,680 | [
0.6083984375,
-0.066650390625,
-0.00897216796875,
0.3447265625,
-0.7451171875,
-0.67236328125,
0.079833984375,
0.273193359375,
0.0416259765625,
0.75927734375,
0.192626953125,
0.036285400390625,
0.00981903076171875,
-0.513671875,
-0.283447265625,
-0.272216796875,
-0.5732421875,
-0.7... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
import sys
N, = map(int, input().split())
mx, md, my = 0, 0, 10**9
ps = []
for _ in range(N):
x, y = map(int, input().split())
mx = max(x, mx)
my = min(y, my)
md = max(y-x+1, md)
ps.append([x,y])
ps.sort()
for p in ps:
# p[1] *= -1
if p == [mx, my]:
print(my-mx+1+md)
sys.exit()
for i, (x, y) in enumerate(ps):
if y == my:
mmx = x
del ps[i]
break
from collections import deque
rs = []
dq = deque()
for i in range(N-1):
while True:
if not len(dq):
dq.append(i)
break
lxi = dq[-1]
if ps[i][1] >= ps[lxi][1]:
dq.append(i)
break
dq.pop()
r = max(my-mx+1, 0) + md
r = max(r, max(my - mmx + 1, 0) + max(ps[dq.popleft()][1] - mx + 1, 0))
#SegTree(pys, N-1, 10**9, min)
for i in range(N-2):
x = max(ps[i][0], mmx)
if len(dq):
if i+1 == dq[0]:
y = ps[dq.popleft()][1] # mx
#y = min(p[1] for p in ps[i+1:])
# print(r)
r = max(r, max(my - x + 1, 0) + max(y - mx + 1, 0))
print(r)
```
Yes
| 42,681 | [
0.578125,
-0.056610107421875,
0.06939697265625,
0.3291015625,
-0.71240234375,
-0.53173828125,
-0.0188751220703125,
0.297119140625,
0.04290771484375,
0.86962890625,
0.24365234375,
0.04132080078125,
0.0293426513671875,
-0.469970703125,
-0.267578125,
-0.310791015625,
-0.53125,
-0.7773... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
import heapq
n = int(input())
ll, rr = [], []
task = []
ans = 0
for _ in range(n):
l, r = map(int, input().split())
heapq.heappush(ll, -l)
heapq.heappush(rr, r)
task.append([l, r])
maxl1, minr1 = -heapq.heappop(ll), heapq.heappop(rr)
maxl2, minr2 = -heapq.heappop(ll), heapq.heappop(rr)
for i in range(n):
ma = maxl1 if not task[i][0] == maxl1 else maxl2
mi = minr1 if not task[i][1] == minr1 else minr2
ans = max(ans, task[i][1] - task[i][0] + max(0, mi - ma + 1) + 1)
for k in range(1):
task.sort(key = lambda x : x[k])
score = [0] * (n - 1)
l1, r1 = task[0][0], task[0][1]
l2, r2 = task[n - 1][0], task[n - 1][1]
i, j = 0, n - 1
for _ in range(n - 1):
l1, r1 = max(l1, task[i][0]), min(r1, task[i][1])
l2, r2 = max(l2, task[j][0]), min(r2, task[j][1])
j -= 1
score[i] += max(0, r1 - l1 + 1)
score[j] += max(0, r2 - l2 + 1)
i += 1
ans = max(ans, max(score))
print(ans)
```
Yes
| 42,682 | [
0.58154296875,
0.01427459716796875,
-0.07916259765625,
0.446044921875,
-0.5234375,
-0.459228515625,
0.0926513671875,
0.1160888671875,
0.2276611328125,
0.88330078125,
0.139892578125,
-0.10748291015625,
0.006084442138671875,
-0.64697265625,
-0.389404296875,
-0.285888671875,
-0.46142578... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
import sys
stdin = sys.stdin
ri = lambda: int(rs())
rl = lambda: list(map(int, stdin.readline().split())) # applies to only numbers
rs = lambda: stdin.readline().rstrip() # ignores trailing space
N = ri()
L = []
R = []
for _ in range(N):
l, r = rl()
L.append(l)
R.append(r)
max_l = max(L)
min_R = min(R)
ab = [(max(0, r - max_l + 1), max(0, min_R - l + 1)) for l, r in zip(L, R)]
ab.sort()
answer = max(r - l + 1 for l, r in zip(L, R)) + max(0, min_R - max_l + 1)
mi = float('inf')
for i in range(N-1):
mi = min(mi, ab[i][1])
answer = max(answer, mi + ab[i+1][0])
print(answer)
```
Yes
| 42,683 | [
0.48974609375,
0.05609130859375,
0.03790283203125,
0.303955078125,
-0.69091796875,
-0.572265625,
-0.0171966552734375,
0.2391357421875,
-0.063720703125,
0.875,
0.208251953125,
0.0074920654296875,
-0.093994140625,
-0.446533203125,
-0.314697265625,
-0.394775390625,
-0.54736328125,
-0.... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
def cmnrng(x, y):
if x[0] < y[0]:
a = x
b = y
else:
a = y
b = x
if a[1] < b[0]:
return([])
elif a[1] > b[1]:
return(b)
else:
return([b[0], a[1]])
def lenrng(l):
if l == []:
return 0
else:
return l[1] - l[0] + 1
N = int(input())
LR = [list(map(int, input().split())) for _ in range(N)]
G1 = LR[0]
common = [0] * N
for i in range(N):
cmn = cmnrng(LR[i], G1)
common[i] = lenrng(cmn)
idxG2 = common.index(min(common))
G2 = LR[idxG2]
for i in range(1, N):
if i == idxG2:
pass
else:
tmpG1 = cmnrng(G1, LR[i])
tmpG2 = cmnrng(G2, LR[i])
case1 = lenrng(tmpG1) + lenrng(G2)
case2 = lenrng(tmpG2) + lenrng(G1)
if case1 >= case2:
G1 = tmpG1
else:
G2 = tmpG2
print(lenrng(G1) + lenrng(G2))
```
No
| 42,684 | [
0.55224609375,
-0.078857421875,
0.00763702392578125,
0.392333984375,
-0.6455078125,
-0.64599609375,
0.071044921875,
0.314453125,
0.0809326171875,
0.8203125,
0.25244140625,
0.051849365234375,
0.04931640625,
-0.51123046875,
-0.285400390625,
-0.3427734375,
-0.60107421875,
-0.777832031... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
n = int(input())
lrlr = [0, -1e10, 0, -1e10]
for i in range(n):
l, r = map(int, input().split())
LRLR = [max(lrlr[0],l), min(abs(lrlr[1]),r), max(lrlr[2],l), min(abs(lrlr[3]),r)]
p1 = LRLR[1]-LRLR[0]+1 + lrlr[3]-lrlr[2]+1
p2 = lrlr[1]-lrlr[0]+1 + LRLR[3]-LRLR[2]+1
if p1>p2:
lrlr[0] = LRLR[0]
lrlr[1] = LRLR[1]
else:
lrlr[2] = LRLR[2]
lrlr[3] = LRLR[3]
print(lrlr[1]-lrlr[0]+lrlr[3]-lrlr[2]+2)
```
No
| 42,685 | [
0.56005859375,
-0.105712890625,
-0.037078857421875,
0.271484375,
-0.5927734375,
-0.63720703125,
0.0810546875,
0.27392578125,
-0.0283660888671875,
0.82861328125,
0.2384033203125,
0.048583984375,
-0.05419921875,
-0.52294921875,
-0.310791015625,
-0.283447265625,
-0.62646484375,
-0.770... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
#!/usr/bin/env python3
from collections import defaultdict,deque
from heapq import heappush, heappop
from bisect import bisect_left, bisect_right
import sys, random, itertools, math
sys.setrecursionlimit(10**5)
input = sys.stdin.readline
sqrt = math.sqrt
def LI(): return list(map(int, input().split()))
def LF(): return list(map(float, input().split()))
def LI_(): return list(map(lambda x: int(x)-1, input().split()))
def II(): return int(input())
def IF(): return float(input())
def LS(): return list(map(list, input().split()))
def S(): return list(input().rstrip())
def IR(n): return [II() for _ in range(n)]
def LIR(n): return [LI() for _ in range(n)]
def FR(n): return [IF() for _ in range(n)]
def LFR(n): return [LI() for _ in range(n)]
def LIR_(n): return [LI_() for _ in range(n)]
def SR(n): return [S() for _ in range(n)]
def LSR(n): return [LS() for _ in range(n)]
mod = 1000000007
inf = float('INF')
#A
def A():
s = S()
n = len(s)
Llis = [0] * (n + 1)
Rlis = [0] * (n + 1)
for i in range(n):
if s[i] == "<":
Llis[i + 1] = 1
else:
Rlis[i] = 1
for i in range(n):
if Llis[i + 1] == 0:
continue
Llis[i + 1] += Llis[i]
for i in range(n):
if Rlis[-i - 2] == 0:
continue
Rlis[-i - 2] += Rlis[-i - 1]
ans = 0
for i in range(n + 1):
ans += max(Rlis[i], Llis[i])
print(ans)
return
#B
def B():
n = II()
LR = LIR(n)
LR.sort(key=lambda x: x[1] - x[0])
ans = [[LR[0][0], LR[0][1] + 1], [LR[1][0], LR[1][1] + 1]]
for i in range(2, n):
l, r = LR[i]
r += 1
a12 = min(ans[0][1], ans[1][1]) - max(ans[0][0], ans[1][0])
a12 = (a12 >= 0) * a12 + r - l
a1lr = min(ans[0][1], r) - max(ans[0][0], l)
a1lr = (a1lr >= 0) * a1lr + ans[1][1] - ans[1][0]
a2lr = min(ans[1][1], r) - max(ans[1][0], l)
a2lr = (a2lr >= 0) * a2lr + ans[0][1] - ans[0][0]
res = [(a12, 0), (a1lr, 1), (a2lr, 2)]
res.sort()
if res[2][1] == 0:
ans[0] = [max(ans[0][0], ans[1][0]), min(ans[0][1], ans[1][1])]
ans[1] = [l, r]
elif res[2][1] == 1:
ans[0] = [max(ans[0][0], l), min(ans[0][1], r)]
else:
ans[1] = [max(ans[1][0], l), min(ans[1][1], r)]
if ans[0][1] < ans[0][0]:
ans[0] = [inf, -inf]
if ans[1][1] < ans[1][0]:
ans[1] = [inf, -inf]
# print(ans,a12,a1lr,a2lr,l,r,res)
a = ans[0][1] - ans[0][0]
b = ans[1][1] - ans[1][0]
if a < 0:
a = 0
if b < 0:
b = 0
print(a+b)
return
#C
def C():
return
#D
def D():
return
#E
def E():
return
#F
def F():
return
#G
def G():
return
#H
def H():
return
#Solve
if __name__ == '__main__':
B()
```
No
| 42,686 | [
0.4853515625,
0.0264739990234375,
0.010528564453125,
0.400146484375,
-0.751953125,
-0.52685546875,
0.0545654296875,
0.12359619140625,
0.152587890625,
0.8134765625,
0.1700439453125,
-0.11383056640625,
0.074951171875,
-0.5673828125,
-0.39013671875,
-0.331298828125,
-0.59716796875,
-0... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
10^9 contestants, numbered 1 to 10^9, will compete in a competition. There will be two contests in this competition.
The organizer prepared N problems, numbered 1 to N, to use in these contests. When Problem i is presented in a contest, it will be solved by all contestants from Contestant L_i to Contestant R_i (inclusive), and will not be solved by any other contestants.
The organizer will use these N problems in the two contests. Each problem must be used in exactly one of the contests, and each contest must have at least one problem.
The joyfulness of each contest is the number of contestants who will solve all the problems in the contest. Find the maximum possible total joyfulness of the two contests.
Constraints
* 2 \leq N \leq 10^5
* 1 \leq L_i \leq R_i \leq 10^9
* All values in input are integers.
Input
Input is given from Standard Input in the following format:
N
L_1 R_1
L_2 R_2
\vdots
L_N R_N
Output
Print the maximum possible total joyfulness of the two contests.
Examples
Input
4
4 7
1 4
5 8
2 5
Output
6
Input
4
1 20
2 19
3 18
4 17
Output
34
Input
10
457835016 996058008
456475528 529149798
455108441 512701454
455817105 523506955
457368248 814532746
455073228 459494089
456651538 774276744
457667152 974637457
457293701 800549465
456580262 636471526
Output
540049931
Submitted Solution:
```
import sys
input = sys.stdin.readline
ri = lambda: int(input())
rs = lambda: input().rstrip()
ril = lambda: list(map(int, input().split()))
rsl = lambda: input().rstrip().split()
ris = lambda n: [ri() for _ in range(n)]
rss = lambda n: [rs() for _ in range(n)]
rils = lambda n: [ril() for _ in range(n)]
rsls = lambda n: [rsl() for _ in range(n)]
n = ri()
ls = rils(n)
ls.sort(key=lambda x: x[1])
res = max(r - l + 1 for l, r in ls)
lmax0 = [ls[0][0]]
rmin0 = [ls[0][1]]
for i in range(1, n - 1):
lmax0.append(max(lmax0[-1], ls[i][0]))
rmin0.append(min(rmin0[-1], ls[i][1]))
lmax1 = [ls[-1][0]]
rmin1 = [ls[-1][1]]
for i in range(1, n - 1):
lmax1.append(max(lmax1[-1], ls[n - i - 1][0]))
rmin1.append(min(rmin1[-1], ls[n - i - 1][1]))
s0 = [max(0, r - l + 1) for l, r in zip(lmax0, rmin0)]
s1 = [max(0, r - l + 1) for l, r in zip(lmax1, rmin1)]
for x, y in zip(s0, reversed(s1)):
res = max(res, x + y)
print(res)
```
No
| 42,687 | [
0.459716796875,
0.1400146484375,
0.007404327392578125,
0.331298828125,
-0.69482421875,
-0.51416015625,
-0.09124755859375,
0.2037353515625,
-0.01092529296875,
0.85595703125,
0.24365234375,
0.0010499954223632812,
-0.0928955078125,
-0.456298828125,
-0.328125,
-0.464111328125,
-0.5146484... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Example
Input
4 3
1000 1
2000 2
3000 3
Output
2 3 4 4
Submitted 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
dd = [(0,-1),(1,0),(0,1),(-1,0)]
ddn = [(0,-1),(1,-1),(1,0),(1,1),(0,1),(-1,-1),(-1,0),(-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,q = LI()
qa = sorted([LI() for _ in range(q)])
u = [0] * (n+2)
d = [0] * (n+2)
for x,i in qa:
u[i+1] = u[i] + 1
d[i] = d[i+1] + 1
rr = []
for i in range(1,n+1):
rr.append(u[i]+d[i]+1)
return '\n'.join(map(str,rr))
print(main())
```
No
| 42,853 | [
0.281005859375,
-0.0036220550537109375,
-0.034515380859375,
0.05291748046875,
-0.875,
-0.11248779296875,
-0.2462158203125,
0.1522216796875,
0.191650390625,
1.208984375,
0.279541015625,
-0.2115478515625,
0.0235137939453125,
-0.47119140625,
-0.595703125,
-0.053802490234375,
-0.63818359... | 11 |
Provide a correct Python 3 solution for this coding contest problem.
B： 階層的計算機 (Hierarchical Calculator)
Problem
Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.
She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".
Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:
* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,
where |s| is the length of the s.
Input
N
a_1 a_2 $\cdots$ a_N
Constraints
* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.
Output
Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).
Sample Input 1
4
2 0 -2 1
Sample Output for Input 1
1
1
She evaluates the first one and gets the maximum value 2.
Sample Input 2
3
2 -2 -2
Sample Output for Input 2
3
1
2
3
She evaluates all of them and gets the maximum value 8.
Sample Input 3
2
-1 0
Sample Output for Input 3
0
She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.
Sample Input 4
5
-1 2 1 -2 -1
Sample Output for Input 4
3
1
2
4
She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.
Example
Input
4
2 0 -2 1
Output
1
1
"Correct Solution:
```
# -*- coding: utf-8 -*-
from collections import Counter
def inpl(): return tuple(map(int, input().split()))
N = int(input())
A = list(map(int, input().split()))
C = Counter(A)
if C[-2] % 2 == 0:
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2]
print(len(ans))
if len(ans):
print(*ans, sep="\n")
elif C[-1] > 0:
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2] + [A.index(-1) + 1]
print(len(ans))
if len(ans):
print(*sorted(ans), sep="\n")
else:
d = N -A[::-1].index(-2)
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2]
del ans[ans.index(d)]
print(len(ans))
if len(ans):
print(*ans, sep="\n")
```
| 42,859 | [
0.1280517578125,
0.310791015625,
0.5966796875,
0.15869140625,
-0.266845703125,
-0.066650390625,
-0.1297607421875,
0.1910400390625,
0.0562744140625,
0.76171875,
0.7197265625,
-0.05743408203125,
-0.11920166015625,
-1.1337890625,
-0.65966796875,
0.0953369140625,
-0.473388671875,
-0.69... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B： 階層的計算機 (Hierarchical Calculator)
Problem
Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.
She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".
Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:
* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,
where |s| is the length of the s.
Input
N
a_1 a_2 $\cdots$ a_N
Constraints
* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.
Output
Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).
Sample Input 1
4
2 0 -2 1
Sample Output for Input 1
1
1
She evaluates the first one and gets the maximum value 2.
Sample Input 2
3
2 -2 -2
Sample Output for Input 2
3
1
2
3
She evaluates all of them and gets the maximum value 8.
Sample Input 3
2
-1 0
Sample Output for Input 3
0
She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.
Sample Input 4
5
-1 2 1 -2 -1
Sample Output for Input 4
3
1
2
4
She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.
Example
Input
4
2 0 -2 1
Output
1
1
Submitted Solution:
```
N = int(input())
src = list(map(int,input().split()))
ans = set()
for i,a in enumerate(src):
if a > 1: ans.add(i)
for i in range(N-1):
if i in ans: continue
for j in range(i+1,N):
if j in ans: continue
if src[i]*src[j] > 1:
ans.add(i)
ans.add(j)
print(len(ans))
for a in list(sorted(ans)):
print(a+1)
```
No
| 42,860 | [
0.131591796875,
0.2037353515625,
0.5771484375,
0.104736328125,
-0.291015625,
-0.06695556640625,
-0.08154296875,
0.271484375,
0.001537322998046875,
0.80078125,
0.74365234375,
0.0938720703125,
-0.134521484375,
-1.01171875,
-0.6630859375,
0.01454925537109375,
-0.434326171875,
-0.61572... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B： 階層的計算機 (Hierarchical Calculator)
Problem
Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.
She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".
Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:
* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,
where |s| is the length of the s.
Input
N
a_1 a_2 $\cdots$ a_N
Constraints
* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.
Output
Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).
Sample Input 1
4
2 0 -2 1
Sample Output for Input 1
1
1
She evaluates the first one and gets the maximum value 2.
Sample Input 2
3
2 -2 -2
Sample Output for Input 2
3
1
2
3
She evaluates all of them and gets the maximum value 8.
Sample Input 3
2
-1 0
Sample Output for Input 3
0
She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.
Sample Input 4
5
-1 2 1 -2 -1
Sample Output for Input 4
3
1
2
4
She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.
Example
Input
4
2 0 -2 1
Output
1
1
Submitted Solution:
```
N = int(input())
src = list(map(int,input().split()))
ans = set()
for i,a in enumerate(src):
if a > 1:
ans.add(i)
min2_count = len([a for a in ans if a < -1])
for i in range(N-1):
if i in ans: continue
if src[i] == 1 and (ans and min(ans) > i):
ans.add(i)
continue
if min2_count%2 and src[i] == -1 and (ans and min(ans) > i):
ans.add(i)
continue
for j in range(i+1,N):
if j in ans: continue
if src[i]*src[j] > 1:
ans.add(i)
ans.add(j)
elif src[i]*src[j] == 1 and (ans and min(ans) > i):
ans.add(i)
ans.add(j)
print(len(ans))
for a in list(sorted(ans)):
print(a+1)
```
No
| 42,861 | [
0.131591796875,
0.2037353515625,
0.5771484375,
0.104736328125,
-0.291015625,
-0.06695556640625,
-0.08154296875,
0.271484375,
0.001537322998046875,
0.80078125,
0.74365234375,
0.0938720703125,
-0.134521484375,
-1.01171875,
-0.6630859375,
0.01454925537109375,
-0.434326171875,
-0.61572... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B： 階層的計算機 (Hierarchical Calculator)
Problem
Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.
She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".
Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:
* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,
where |s| is the length of the s.
Input
N
a_1 a_2 $\cdots$ a_N
Constraints
* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.
Output
Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).
Sample Input 1
4
2 0 -2 1
Sample Output for Input 1
1
1
She evaluates the first one and gets the maximum value 2.
Sample Input 2
3
2 -2 -2
Sample Output for Input 2
3
1
2
3
She evaluates all of them and gets the maximum value 8.
Sample Input 3
2
-1 0
Sample Output for Input 3
0
She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.
Sample Input 4
5
-1 2 1 -2 -1
Sample Output for Input 4
3
1
2
4
She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.
Example
Input
4
2 0 -2 1
Output
1
1
Submitted Solution:
```
N = int(input())
src = list(map(int,input().split()))
ans = set()
for i,a in enumerate(src):
if a > 1:
ans.add(i)
for i in range(N-1):
if i in ans: continue
if src[i] == 1 and (ans and min(ans) > i):
ans.add(i)
continue
for j in range(i+1,N):
if j in ans: continue
if src[i]*src[j] > 1:
ans.add(i)
ans.add(j)
elif src[i]*src[j] == 1 and (ans and min(ans) > i):
ans.add(i)
ans.add(j)
print(len(ans))
for a in list(sorted(ans)):
print(a+1)
```
No
| 42,862 | [
0.131591796875,
0.2037353515625,
0.5771484375,
0.104736328125,
-0.291015625,
-0.06695556640625,
-0.08154296875,
0.271484375,
0.001537322998046875,
0.80078125,
0.74365234375,
0.0938720703125,
-0.134521484375,
-1.01171875,
-0.6630859375,
0.01454925537109375,
-0.434326171875,
-0.61572... | 11 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
B： 階層的計算機 (Hierarchical Calculator)
Problem
Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.
She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".
Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:
* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,
where |s| is the length of the s.
Input
N
a_1 a_2 $\cdots$ a_N
Constraints
* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.
Output
Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).
Sample Input 1
4
2 0 -2 1
Sample Output for Input 1
1
1
She evaluates the first one and gets the maximum value 2.
Sample Input 2
3
2 -2 -2
Sample Output for Input 2
3
1
2
3
She evaluates all of them and gets the maximum value 8.
Sample Input 3
2
-1 0
Sample Output for Input 3
0
She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.
Sample Input 4
5
-1 2 1 -2 -1
Sample Output for Input 4
3
1
2
4
She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.
Example
Input
4
2 0 -2 1
Output
1
1
Submitted Solution:
```
# -*- coding: utf-8 -*-
from collections import Counter
def inpl(): return tuple(map(int, input().split()))
N = int(input())
A = list(map(int, input().split()))
C = Counter(A)
if C[-2] % 2 == 0:
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2]
print(len(ans))
print(*ans, sep="\n")
elif C[-1] > 0:
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2] + [A.index(-1) + 1]
print(len(ans))
print(*sorted(ans), sep="\n")
else:
del A[-A[::-1].index(-2) - 1]
ans = [i+1 for i, a in enumerate(A) if abs(a) == 2]
print(len(ans))
print(*ans, sep="\n")
```
No
| 42,863 | [
0.131591796875,
0.2037353515625,
0.5771484375,
0.104736328125,
-0.291015625,
-0.06695556640625,
-0.08154296875,
0.271484375,
0.001537322998046875,
0.80078125,
0.74365234375,
0.0938720703125,
-0.134521484375,
-1.01171875,
-0.6630859375,
0.01454925537109375,
-0.434326171875,
-0.61572... | 11 |
Provide tags and a correct Python 2 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
from itertools import permutations, combinations
from fractions import gcd
import heapq
raw_input = stdin.readline
pr = stdout.write
mod=998244353
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return tuple(map(int,stdin.read().split()))
range = xrange # not for python 3.0+
# main code
n,k=li()
l1=li()
d1=Counter(l1)
d=[0]*k
d[k-1]=d1[k]
for i in range(k-2,-1,-1):
d[i]=d[i+1]+d1[i+1]
l=li()
ans=0
for i in range(k):
ans=max(ans,(d[i]/l[i])+(d[i]%l[i]!=0))
arr=[[] for i in range(ans)]
l1.sort()
for i in range(n):
arr[i%ans].append(l1[i])
pn(ans)
for i in arr:
print len(i),
for j in i:
print j,
print
```
| 43,107 | [
0.408447265625,
0.037567138671875,
0.2291259765625,
0.343994140625,
-1.05859375,
-0.638671875,
-0.1268310546875,
0.33203125,
0.002376556396484375,
0.91357421875,
0.8349609375,
-0.488037109375,
0.41162109375,
-0.8271484375,
-0.52978515625,
0.2012939453125,
-0.466064453125,
-0.668945... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
[n, k] = [int(x) for x in stdin.readline().split()]
m = [int(x) for x in stdin.readline().split()]
k = [int(x) for x in stdin.readline().split()]
m.sort()
summ = 0
minRes = 1
res = []
for i in range(n):
res.append([])
idx = 1
maxLen = 0
for x in m[::-1]:
summ += 1
count = summ // k[x - 1] + (1 if summ % k[x - 1] else 0)
if count > minRes:
minRes = count
res[minRes - 1].append(str(x))
if idx == 1 and len(res[idx-1]) > 1:
idx = minRes
else:
res[idx-1].append(str(x))
if idx == 1:
maxLen = len(res[idx-1])
if minRes > 1:
idx += 1
if len(res[idx-1]) == maxLen:
if idx == minRes:
idx = 1
else:
idx += 1
# print(x)
# print(res)
# print(idx)
# print(maxLen)
# print(minRes)
stdout.write(str(minRes) + "\n")
for i in range(minRes):
stdout.write(str(len(res[i])) + ' ' + ' '.join(res[i]) + "\n")
```
| 43,108 | [
0.46337890625,
0.00904083251953125,
0.1859130859375,
0.2880859375,
-1.0322265625,
-0.63134765625,
-0.1175537109375,
0.35400390625,
-0.019866943359375,
0.8388671875,
0.81787109375,
-0.50048828125,
0.424560546875,
-0.88916015625,
-0.5166015625,
0.09521484375,
-0.468017578125,
-0.6132... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
n,k=map(int,input().split())
sz=[int(x) for x in input().split()]
c=[int(x) for x in input().split()]
sz.sort()
sz=sz[::-1]
ans=[]
h=[]
ans.append(h)
for i in range(0,len(sz)):
ele=sz[i]
low=0
high=len(ans)-1
ind=-1
while(low<=high):
mid=(low+high)>>1
if(c[ele-1]-len(ans[mid])>0):
ind=mid
high=mid-1
else:
low=mid+1
if(ind==-1):
h2=[]
ans.append(h2)
ans[ind].append(ele)
print(len(ans))
for i in ans:
print(len(i),*i)
```
| 43,109 | [
0.46337890625,
0.00904083251953125,
0.1859130859375,
0.2880859375,
-1.0322265625,
-0.63134765625,
-0.1175537109375,
0.35400390625,
-0.019866943359375,
0.8388671875,
0.81787109375,
-0.50048828125,
0.424560546875,
-0.88916015625,
-0.5166015625,
0.09521484375,
-0.468017578125,
-0.6132... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
import io,os
input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
def main():
n,k=map(int,input().split())
M=list(map(int,input().split()))
M.sort(reverse=True)
C=list(map(int,input().split()))
Ans=[]
def search(Ans,m,judge):
ng=0
ok=len(Ans)
while ok-ng>1:
mid=(ng+ok)//2
if len(Ans[mid])<judge:
ok=mid
else:
ng=mid
return ok
for m in M:
judge=C[m-1]
if (not Ans) or len(Ans[-1])>=judge:
Ans.append([m])
elif len(Ans[0])<judge:
Ans[0].append(m)
else:
idx=search(Ans,m,judge)
Ans[idx].append(m)
sys.stdout.write(str(len(Ans))+'\n')
for ans in Ans:
sys.stdout.write(str(len(ans))+' ')
sys.stdout.write(' '.join(map(str,ans))+'\n')
if __name__=='__main__':
main()
```
| 43,110 | [
0.46337890625,
0.00904083251953125,
0.1859130859375,
0.2880859375,
-1.0322265625,
-0.63134765625,
-0.1175537109375,
0.35400390625,
-0.019866943359375,
0.8388671875,
0.81787109375,
-0.50048828125,
0.424560546875,
-0.88916015625,
-0.5166015625,
0.09521484375,
-0.468017578125,
-0.6132... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
from sys import stdin, stdout
import heapq as hq
from collections import defaultdict
import math
t = 1
for tc in range(t):
n,k = list(map(int, stdin.readline().split()))
sizes=list(map(int, stdin.readline().split()))
limits=list(map(int, stdin.readline().split()))
num=[0 for x in range(k)]
count=defaultdict(int)
for size in sizes:
count[size]+=1
csum=0
#print(sizes)
#print(num)
for key in sorted(count.keys())[::-1]:
cur=key+1
while cur-1<len(num) and num[cur-1]==0:
num[cur-1]=csum
cur+=1
csum+=count[key]
num[key-1]=csum
#print(num)
cur=0
while cur < len(num) and num[cur] == 0:
num[cur] = csum
cur+=1
maxfrac=0
#print(num)
#print(limits)
for top,denom in zip(num,limits):
frac=math.ceil(top/denom)
if frac>maxfrac:
maxfrac=frac
result=defaultdict(list)
i=0
for key in sorted(count.keys())[::-1]:
number=count[key]
while number>0:
result[i].append(key)
number-=1
i+=1
i%=maxfrac
#print(count)
#print(result)
stdout.write(str(maxfrac) + "\n")
for i in range(maxfrac):
res=len(result[i])
stdout.write(str(res) + " ")
for val in result[i]:
stdout.write(str(val) + " ")
stdout.write("\n")
```
| 43,111 | [
0.46337890625,
0.00904083251953125,
0.1859130859375,
0.2880859375,
-1.0322265625,
-0.63134765625,
-0.1175537109375,
0.35400390625,
-0.019866943359375,
0.8388671875,
0.81787109375,
-0.50048828125,
0.424560546875,
-0.88916015625,
-0.5166015625,
0.09521484375,
-0.468017578125,
-0.6132... | 11 |
Provide tags and a correct Python 3 solution for this coding contest problem.
So you decided to hold a contest on Codeforces. You prepared the problems: statements, solutions, checkers, validators, tests... Suddenly, your coordinator asks you to change all your tests to multiple testcases in the easiest problem!
Initially, each test in that problem is just an array. The maximum size of an array is k. For simplicity, the contents of arrays don't matter. You have n tests — the i-th test is an array of size m_i (1 ≤ m_i ≤ k).
Your coordinator asks you to distribute all of your arrays into multiple testcases. Each testcase can include multiple arrays. However, each testcase should include no more than c_1 arrays of size greater than or equal to 1 (≥ 1), no more than c_2 arrays of size greater than or equal to 2, ..., no more than c_k arrays of size greater than or equal to k. Also, c_1 ≥ c_2 ≥ ... ≥ c_k.
So now your goal is to create the new testcases in such a way that:
* each of the initial arrays appears in exactly one testcase;
* for each testcase the given conditions hold;
* the number of testcases is minimum possible.
Print the minimum possible number of testcases you can achieve and the sizes of arrays included in each testcase.
Input
The first line contains two integers n and k (1 ≤ n, k ≤ 2 ⋅ 10^5) — the number of initial tests and the limit for the size of each array.
The second line contains n integers m_1, m_2, ..., m_n (1 ≤ m_i ≤ k) — the sizes of the arrays in the original tests.
The third line contains k integers c_1, c_2, ..., c_k (n ≥ c_1 ≥ c_2 ≥ ... ≥ c_k ≥ 1); c_i is the maximum number of arrays of size greater than or equal to i you can have in a single testcase.
Output
In the first line print a single integer ans (1 ≤ ans ≤ n) — the minimum number of testcases you can achieve.
Each of the next ans lines should contain the description of a testcase in the following format:
t a_1 a_2 ... a_{t} (1 ≤ t≤ n) — the testcase includes t arrays, a_i is the size of the i-th array in that testcase.
Each of the initial arrays should appear in exactly one testcase. In particular, it implies that the sum of t over all ans testcases should be equal to n.
Note that the answer always exists due to c_k ≥ 1 (and therefore c_1 ≥ 1).
If there are multiple answers, you can output any one of them.
Examples
Input
4 3
1 2 2 3
4 1 1
Output
3
1 2
2 1 3
1 2
Input
6 10
5 8 1 10 8 7
6 6 4 4 3 2 2 2 1 1
Output
2
3 8 5 7
3 10 8 1
Input
5 1
1 1 1 1 1
5
Output
1
5 1 1 1 1 1
Input
5 1
1 1 1 1 1
1
Output
5
1 1
1 1
1 1
1 1
1 1
Note
In the first example there is no way to distribute the tests into less than 3 testcases. The given answer satisfies the conditions: each of the testcases includes no more than 4 arrays of size greater than or equal to 1 and no more than 1 array of sizes greater than or equal to 2 and 3.
Note that there are multiple valid answers for this test. For example, testcases with sizes [[2], [1, 2], [3]] would also be correct.
However, testcases with sizes [[1, 2], [2, 3]] would be incorrect because there are 2 arrays of size greater than or equal to 2 in the second testcase.
Note the difference between the third and the fourth examples. You can include up to 5 arrays of size greater than or equal to 1 in the third example, so you can put all arrays into a single testcase. And you can have only up to 1 array in the fourth example. Thus, every array should be included in a separate testcase.
Tags: binary search, constructive algorithms, data structures, greedy, sortings, two pointers
Correct Solution:
```
import sys
from math import *
def eprint(*args, **kwargs):
print(*args, file=sys.stderr, **kwargs)
zz=1
if zz:
input=sys.stdin.readline
else:
sys.stdin=open('input.txt', 'r')
sys.stdout=open('output2.txt','w')
#print("LOL")
n,k=map(int,input().split())
a=[int(x) for x in input().split()]
c=[int(x) for x in input().split()]
d=[0 for i in range(k+1)]
for i in a:
d[i]+=1
for i in range(k-1,0,-1):
d[i]+=d[i+1]
cc=0
for i in range(1,k+1):
cc=max(cc,int(ceil(d[i]/c[i-1])))
if cc==0:
cc=1
z=cc
a.sort(reverse=True)
u=[[] for i in range(cc+1)]
for j in range(len(a)):
u[j%cc].append(a[j])
#print(j,t,tt) """
f=0
eprint(a)
yt=1
for i in u:
if len(i)==0:
f+=1
print(len(u)-f)
for i in u:
if len(i)!=0:
print(len(i),*i)
```
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0.1859130859375,
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-1.0322265625,
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0.35400390625,
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0.8388671875,
0.81787109375,
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0.424560546875,
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-0.5166015625,
0.09521484375,
-0.468017578125,
-0.6132... | 11 |
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