text stringlengths 198 433k | conversation_id int64 0 109k |
|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n = input()
a = [a for a in input().split()]
b = [b for b in input().split()]
c = [c for c in input().split()]
print(set(a).difference(b))
print(set(b).difference(c))
```
No
| 86,100 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A and B are preparing themselves for programming contests.
B loves to debug his code. But before he runs the solution and starts debugging, he has to first compile the code.
Initially, the compiler displayed n compilation errors, each of them is represented as a positive integer. After some effort, B managed to fix some mistake and then another one mistake.
However, despite the fact that B is sure that he corrected the two errors, he can not understand exactly what compilation errors disappeared β the compiler of the language which B uses shows errors in the new order every time! B is sure that unlike many other programming languages, compilation errors for his programming language do not depend on each other, that is, if you correct one error, the set of other error does not change.
Can you help B find out exactly what two errors he corrected?
Input
The first line of the input contains integer n (3 β€ n β€ 105) β the initial number of compilation errors.
The second line contains n space-separated integers a1, a2, ..., an (1 β€ ai β€ 109) β the errors the compiler displayed for the first time.
The third line contains n - 1 space-separated integers b1, b2, ..., bn - 1 β the errors displayed at the second compilation. It is guaranteed that the sequence in the third line contains all numbers of the second string except for exactly one.
The fourth line contains n - 2 space-separated integers Ρ1, Ρ2, ..., Ρn - 2 β the errors displayed at the third compilation. It is guaranteed that the sequence in the fourth line contains all numbers of the third line except for exactly one.
Output
Print two numbers on a single line: the numbers of the compilation errors that disappeared after B made the first and the second correction, respectively.
Examples
Input
5
1 5 8 123 7
123 7 5 1
5 1 7
Output
8
123
Input
6
1 4 3 3 5 7
3 7 5 4 3
4 3 7 5
Output
1
3
Note
In the first test sample B first corrects the error number 8, then the error number 123.
In the second test sample B first corrects the error number 1, then the error number 3. Note that if there are multiple errors with the same number, B can correct only one of them in one step.
Submitted Solution:
```
n=input()
a1=input()
a2=input()
a3=input()
a1=a1.split()
a2=a2.split()
a3=a3.split()
l=[]
m=[]
for i in a1:
l.append(i)
if l.count(i) - a2.count(i)==1:
print (i)
if l.count(i) - a3.count(i)==1 and i in a2:
print (i)
```
No
| 86,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
n = int(input())
a = [list(map(int, input().split())) for i in range(n)]
k = 2
for i in range(1, n - 1):
x, h = a[i]
if x - h > a[i-1][0]:
k += 1
elif x + h < a[i+1][0]:
k += 1
a[i][0] += h
if n == 1:
print(1)
else:
print(k)
```
| 86,102 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
# author : prottoyfuad
# 30.09.2020 04:17:01(+0600)
import sys, math, time
start_time = time.time()
def read() :
return( int(input()) )
def read_args() :
return( map(int, input().split()) )
def read_array() :
return( list( map(int, input().split()) ) )
def read_string() :
return( input().strip() )
def hocus_pocus() :
n = read()
a = [0] * n
h = [0] * n
for i in range(n) :
a[i], h[i] = read_args()
ans = min(n, 2)
for i in range(1, n - 1) :
if a[i] - h[i] > a[i - 1] :
ans += 1
elif a[i] + h[i] < a[i + 1] :
ans += 1
a[i] += h[i]
print(ans)
if __debug__ :
input = sys.stdin.readline
else :
sys.stdin = open("D:\pro_code\py\in.txt","r");
sys.stdout = open("D:\pro_code\py\out.txt","w");
global tc
tt = 1
# tt = read()
for tc in range(tt) :
hocus_pocus()
if not __debug__ :
print("Time elapsed :", time.time() - start_time, "seconds")
sys.stdout.close()
```
| 86,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
a=2
n=int(input())
l=[]
for _ in range(n):
x,h=map(int,input().split())
l.append([x,h])
if n<3:print(n);exit()
t=l[0][0]
for i in range(1,n-1):
x,h=l[i][0],l[i][1]
if x-h>t:
a+=1;t=x
elif x+h<l[i+1][0]:
a+=1;t=x+h
else:
t=x
print(a)
```
| 86,104 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
n = int(input())
x = [0] * (n + 1)
h = [0] * (n + 1)
for i in range(n):
x[i], h[i] = (int(c) for c in input().split())
h[n] = 1
x[n] = x[n-1] + h[n-1] + 1
if n == 1:
print(1)
exit(0)
left = [0] * n
right = [0] * n
left[0] = 1
if x[0] + h[0] < x[1]:
right[0] = 1
for i in range(1, n):
if x[i] + h[i] < x[i + 1]:
right[i] = max(right[i-1], left[i-1]) + 1
else:
right[i] = max(right[i-1], left[i-1])
if x[i] - h[i] > x[i-1] + h[i-1]:
left[i] = max(right[i-1], left[i-1]) + 1
elif x[i] - h[i] > x[i-1]:
left[i] = left[i-1] + 1
else:
left[i] = max(right[i-1], left[i-1])
print(max(left[n-1], right[n-1]))
```
| 86,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
import sys
input = sys.stdin.readline
n = int(input())
arr = [list(map(int, input().split())) for _ in range(n)]
if n <= 2:
print(n)
else:
ans = 0
for i in range(1,n-1):
if arr[i][0] - arr[i][1] > arr[i-1][0]:
ans +=1
elif arr[i][0] + arr[i][1] < arr[i+1][0]:
ans +=1
arr[i][0] += arr[i][1]
print(ans+2)
```
| 86,106 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
n = int(input())
h = []
x = []
for i in range(n):
a, b = map(int, input().split())
x.append(a)
h.append(b)
last = x[0]
t = 1
for i in range(1, n - 1):
if last < x[i] - h[i]:
t += 1
last = x[i]
elif x[i] + h[i] < x[i + 1]:
t += 1
last = x[i] + h[i]
else:
last = x[i]
if n > 1:
t += 1
print(t)
```
| 86,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
n = int(input())
p = n
x = []
h = []
while(p>0):
p = p-1
a,b = map(int,input().split())
x.append(a)
h.append(b)
c = 0
table = [0]*(n)
table[0] = -1
table[n-1] = 1
for i in range(1,n-1):
if table[i-1] == -1:
if x[i]-x[i-1]>h[i]:
table[i] = -1
c = c+1
else:
if x[i+1]-x[i]>h[i]:
table[i] = 1
c = c+1
elif table[i-1] == 0:
if x[i]-x[i-1] >h[i]:
table[i] = -1
c = c+1
else:
if x[i+1]-x[i]>h[i]:
c = c+1
table[i] = 1
else:
if x[i]-x[i-1]-h[i-1]>h[i]:
table[i] = -1
c = c+1
else:
if x[i+1]-x[i]>h[i]:
table[i] = 1
c = c+1
if n<2:
print(1)
else:
print(c+2)
```
| 86,108 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Tags: dp, greedy
Correct Solution:
```
n=int(input())
a=[]
for i in range(n):
x,h=map(int,input().split())
a.append([x,h])
num=1
i=1
while i<n-1:
if a[i][1]<a[i][0]-a[i-1][0]:
num+=1
else:
if a[i][1]<a[i+1][0]-a[i][0]:
num+=1
a[i][0]+=a[i][1]
i+=1
if n>=2:
num+=1
print(num)
```
| 86,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
n = int(input())
arr = [list(map(int, input().split())) for _ in range(n)]
count = 1
for i in range(1, n-1):
x = arr[i][0] - arr[i][1]
y = arr[i][0] + arr[i][1]
if x > arr[i-1][0]:
count += 1
elif y < arr[i+1][0]:
count += 1
arr[i][0] += arr[i][1]
print(count + 1 if n > 1 else 1)
```
Yes
| 86,110 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
from sys import stdin
N = int(stdin.readline().rstrip())
trees = [None for n in range(N)]
for n in range(N):
trees[n] = [int(x) for x in stdin.readline().rstrip().split()]
start = -1000000000
count = 1
for n in range(N - 1):
tree = trees[n]
if tree[0] - tree[1] > start:
count += 1
start = tree[0]
elif tree[0] + tree[1] < trees[n+1][0]:
count += 1
start = tree[0] + tree[1]
else:
start = tree[0]
print(count)
```
Yes
| 86,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
t = int(input())
l = [[int(x) for x in input().split()]for i in range(t)]
if t >= 3:
count = 2
end = l[0][0]
begin = l[2][0]
for i in range(1,t-2):
if l[i][0] - l[i][1] > end:
count += 1
end = l[i][0]
elif l[i][0] + l[i][1] < begin:
count += 1
end = l[i][0] + l[i][1]
else:
end = l[i][0]
begin = l[i+2][0]
if l[-2][0] - l[-2][1] > end or l[-2][0] + l[-2][1] < l[-1][0]:
print(count+1)
else:
print(count)
else:
print(t)
```
Yes
| 86,112 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
n=int(input())
d=[[-1e9-10, 0]]
for i in range(n):
d.append(list(map(int, input().split())))
d.append([2*1e9+2, 0])
c=-1e9-10
ans=0
for i in range(1, n+1):
x=d[i][0]
h=d[i][1]
if x-h>c:
ans+=1
c=x
elif x+h<d[i+1][0]:
ans+=1
c=x+h
else: c=x
print(ans)
```
Yes
| 86,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
def main():
n = int(input())
a = []
b = []
for _ in range(n):
x,y = map(int,input().split())
a.append(x)
b.append(y)
start = a[0]
cnt = 2
if n==1:
cnt = 1
for i in range(1,len(a)-1):
if start<a[i]-b[i]:
cnt+=1
start = a[i]
elif a[i+1]>(a[i]+b[i]):
cnt+=1
start=a[i]+b[i]
print(cnt)
main()
# t= int(input())
# while t:
# main()
# t-=1
```
No
| 86,114 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
n=int(input())
l=[]
m=[]
c=0
for _ in range(n):
a,b=map(int,input().split())
l.append(a)
m.append(b)
for i in range(1,n-1):
if(m[i]<abs(l[i]-l[i+1])):
c=c+1
l[i]+=m[i]
elif(m[i]<abs(l[i]-l[i-1])):
c=c+1
print(c+2)
```
No
| 86,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
n=int(input())
cor = []
for _ in range(n):
x,h = map(int,input().split())
cor.append([x,h])
ans=2
for i in range(1,n-1):
if cor[i][0]-cor[i-1][0] > cor[i][1]:
ans+=1
elif cor[i+1][0]-cor[i][0] > cor[i][1]:
ans+=1
cor[i][0]+=cor[i][1]
print(ans)
```
No
| 86,116 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Little Susie listens to fairy tales before bed every day. Today's fairy tale was about wood cutters and the little girl immediately started imagining the choppers cutting wood. She imagined the situation that is described below.
There are n trees located along the road at points with coordinates x1, x2, ..., xn. Each tree has its height hi. Woodcutters can cut down a tree and fell it to the left or to the right. After that it occupies one of the segments [xi - hi, xi] or [xi;xi + hi]. The tree that is not cut down occupies a single point with coordinate xi. Woodcutters can fell a tree if the segment to be occupied by the fallen tree doesn't contain any occupied point. The woodcutters want to process as many trees as possible, so Susie wonders, what is the maximum number of trees to fell.
Input
The first line contains integer n (1 β€ n β€ 105) β the number of trees.
Next n lines contain pairs of integers xi, hi (1 β€ xi, hi β€ 109) β the coordinate and the height of the Ρ-th tree.
The pairs are given in the order of ascending xi. No two trees are located at the point with the same coordinate.
Output
Print a single number β the maximum number of trees that you can cut down by the given rules.
Examples
Input
5
1 2
2 1
5 10
10 9
19 1
Output
3
Input
5
1 2
2 1
5 10
10 9
20 1
Output
4
Note
In the first sample you can fell the trees like that:
* fell the 1-st tree to the left β now it occupies segment [ - 1;1]
* fell the 2-nd tree to the right β now it occupies segment [2;3]
* leave the 3-rd tree β it occupies point 5
* leave the 4-th tree β it occupies point 10
* fell the 5-th tree to the right β now it occupies segment [19;20]
In the second sample you can also fell 4-th tree to the right, after that it will occupy segment [10;19].
Submitted Solution:
```
from sys import stdin
N = int(stdin.readline())
X = []
# H = []
trees = {}
gaps = []
for n in range(N):
line = stdin.readline().strip().split()
X.append(int(line[0]))
# H.append(int(line[1]))
trees[int(line[0])] = int(line[1])
gaps = sorted([(X[i], X[i + 1]) for i in range(len(X) - 1)], key=lambda g: g[1] - g[0])
# print(gaps)
# print(X, H)
# print(trees)
trees[gaps[0][0]] = 0
trees[gaps[-1][1]] = 0
# print(trees)
count = 2
for g in gaps:
l, r = g
s = r - l
if trees[l] >= trees[r]:
if 0 < trees[l] < s:
trees[l] = 0
count += 1
elif 0 < trees[r] < s:
trees[r] = 0
count += 1
else:
if 0 < trees[r] < s:
trees[r] = 0
count += 1
elif 0 < trees[l] < s:
trees[l] = 0
count += 1
print(count)
```
No
| 86,117 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Geometric progression with the first element a and common ratio b is a sequence of numbers a, ab, ab2, ab3, ....
You are given n integer geometric progressions. Your task is to find the smallest integer x, that is the element of all the given progressions, or else state that such integer does not exist.
Input
The first line contains integer (1 β€ n β€ 100) β the number of geometric progressions.
Next n lines contain pairs of integers a, b (1 β€ a, b β€ 109), that are the first element and the common ratio of the corresponding geometric progression.
Output
If the intersection of all progressions is empty, then print - 1, otherwise print the remainder of the minimal positive integer number belonging to all progressions modulo 1000000007 (109 + 7).
Examples
Input
2
2 2
4 1
Output
4
Input
2
2 2
3 3
Output
-1
Note
In the second sample test one of the progressions contains only powers of two, the other one contains only powers of three.
Submitted Solution:
```
n = int(input())
l = []
M = (int(1e9) + 7)
for i in range(n):
a, b = list(map(int, input().split()))
cur = a
l.append([])
for j in range(100):
l[-1].append(cur)
cur *= b
for x in l[0]:
for j in range(n):
if not x in l[j]:
break
else:
print(x % M)
exit()
print(-1)
```
No
| 86,118 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Geometric progression with the first element a and common ratio b is a sequence of numbers a, ab, ab2, ab3, ....
You are given n integer geometric progressions. Your task is to find the smallest integer x, that is the element of all the given progressions, or else state that such integer does not exist.
Input
The first line contains integer (1 β€ n β€ 100) β the number of geometric progressions.
Next n lines contain pairs of integers a, b (1 β€ a, b β€ 109), that are the first element and the common ratio of the corresponding geometric progression.
Output
If the intersection of all progressions is empty, then print - 1, otherwise print the remainder of the minimal positive integer number belonging to all progressions modulo 1000000007 (109 + 7).
Examples
Input
2
2 2
4 1
Output
4
Input
2
2 2
3 3
Output
-1
Note
In the second sample test one of the progressions contains only powers of two, the other one contains only powers of three.
Submitted Solution:
```
#"""
#https://codeforces.com/problemset/problem/571/E
#"""
def gp(a,b):
l=[]
for i in range(5):
t=a*pow(b,i)
l.append(t)
return l
n=int(input())
s=[]
for i in range(n):
a,b=map(int,input().split())
s.append(gp(a,b))
first=s[0]
second=s[1]
p=[value for value in first if value in second]
if(len(p)>0):
for a in p:
print(a%(10**9+7))
else:
print(-1)
```
No
| 86,119 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Geometric progression with the first element a and common ratio b is a sequence of numbers a, ab, ab2, ab3, ....
You are given n integer geometric progressions. Your task is to find the smallest integer x, that is the element of all the given progressions, or else state that such integer does not exist.
Input
The first line contains integer (1 β€ n β€ 100) β the number of geometric progressions.
Next n lines contain pairs of integers a, b (1 β€ a, b β€ 109), that are the first element and the common ratio of the corresponding geometric progression.
Output
If the intersection of all progressions is empty, then print - 1, otherwise print the remainder of the minimal positive integer number belonging to all progressions modulo 1000000007 (109 + 7).
Examples
Input
2
2 2
4 1
Output
4
Input
2
2 2
3 3
Output
-1
Note
In the second sample test one of the progressions contains only powers of two, the other one contains only powers of three.
Submitted Solution:
```
"""
https://codeforces.com/problemset/problem/571/E
"""
def gp(a,b):
l=[]
for i in range(5):
t=a*pow(b,i)
l.append(t)
return l
n=int(input())
s=[]
for i in range(n):
a,b=map(int,input().split())
s.append(gp(a,b))
first=s[0]
second=s[1]
p=[value for value in first if value in second]
if(len(p)>0):
for a in p:
print(a%(10**9+7))
else:
print(-1)
```
No
| 86,120 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Geometric progression with the first element a and common ratio b is a sequence of numbers a, ab, ab2, ab3, ....
You are given n integer geometric progressions. Your task is to find the smallest integer x, that is the element of all the given progressions, or else state that such integer does not exist.
Input
The first line contains integer (1 β€ n β€ 100) β the number of geometric progressions.
Next n lines contain pairs of integers a, b (1 β€ a, b β€ 109), that are the first element and the common ratio of the corresponding geometric progression.
Output
If the intersection of all progressions is empty, then print - 1, otherwise print the remainder of the minimal positive integer number belonging to all progressions modulo 1000000007 (109 + 7).
Examples
Input
2
2 2
4 1
Output
4
Input
2
2 2
3 3
Output
-1
Note
In the second sample test one of the progressions contains only powers of two, the other one contains only powers of three.
Submitted Solution:
```
def gp(a,b):
l=[]
for i in range(5):
t=a*pow(b,i)
l.append(t)
return l
n=int(input())
s=[]
for i in range(n):
a,b=map(int,input().split())
s.append(gp(a,b))
first=s[0]
second=s[1]
p=[value for value in first if value in second]
if(len(p)>0):
for a in p:
print(a%(10**9+7))
else:
print(-1)
```
No
| 86,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Tags: constructive algorithms, math
Correct Solution:
```
def f(x):
if x == n:
return "0"
if x == 0:
return "(" + str(X[0]) + "+" + f(1) + ")"
ss = "(abs((t-" + str(x-1) + "))-abs((t-" + str(x) + ")))"
tmp = (X[x] - X[x - 1]) // 2
re = (X[x] - X[x - 1]) - 2 * tmp
X[x] -= re
if tmp < 0:
tmp = "(0" +str(tmp)+")"
ss = "((" + str(tmp) + "*" + ss + ")" + "+" + str(tmp) + ")"
return "(" + ss + "+" + f(x + 1) + ")"
n = int(input())
#c = [(int(_) for _ in input().split()) for i in range(n)]
c = [[int(x) for x in input().split()] for i in range(n)]
#print(n, c)
X = [c[i][0] for i in range(n)]
Y = [c[i][1] for i in range(n)]
#print(X)
#print(Y)
print(f(0))
#print(X)
X = Y
print(f(0))
# Made By Mostafa_Khaled
```
| 86,122 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Tags: constructive algorithms, math
Correct Solution:
```
def f(x):
if x == n:
return "0"
if x == 0:
return "(" + str(X[0]) + "+" + f(1) + ")"
ss = "(abs((t-" + str(x-1) + "))-abs((t-" + str(x) + ")))"
tmp = (X[x] - X[x - 1]) // 2
re = (X[x] - X[x - 1]) - 2 * tmp
X[x] -= re
if tmp < 0:
tmp = "(0" +str(tmp)+")"
ss = "((" + str(tmp) + "*" + ss + ")" + "+" + str(tmp) + ")"
return "(" + ss + "+" + f(x + 1) + ")"
n = int(input())
#c = [(int(_) for _ in input().split()) for i in range(n)]
c = [[int(x) for x in input().split()] for i in range(n)]
#print(n, c)
X = [c[i][0] for i in range(n)]
Y = [c[i][1] for i in range(n)]
#print(X)
#print(Y)
print(f(0))
#print(X)
X = Y
print(f(0))
```
| 86,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Tags: constructive algorithms, math
Correct Solution:
```
n = int(input())
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i], r = map(int, input().split())
def sum(s1, s2):
return '(' + s1 + '+' + s2 + ')'
def minus(s1, s2):
return '(' + s1 + '-' + s2 + ')'
def mult(s1, s2):
return '(' + s1 + '*' + s2 + ')'
def sabs(s1):
return 'abs(' + s1 + ')'
def stand(x):
return sum(minus('1', sabs(minus('t', x))), sabs(minus(sabs(minus('t', x)), '1')))
def ans(v):
s=''
for i in range(1, n + 1):
if s == '':
s = mult(str(v[i - 1] // 2), stand(str(i)))
else:
s = sum(s, mult(str(v[i - 1] // 2), stand(str(i))))
print(s)
ans(x)
ans(y)
```
| 86,124 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Tags: constructive algorithms, math
Correct Solution:
```
def canonise(t):
if t < 0:
return "(0-" + canonise(-t) + ")"
ans = ""
while t > 50:
ans += "(50+"
t -= 50
return ans + str(t) + ")" * (len(ans)//4)
n = int(input())
cxes = []
cyes = []
for i in range(n):
x, y, r = map(int, input().split())
for dx in range(2):
for dy in range(2):
if (x+dx)%2 == 0 and (y+dy)%2 == 0:
cxes.append((x+dx)//2)
cyes.append((y+dy)//2)
coeffx = [0]*(n+2)
coeffy = [0]*(n+2)
cfx = 0
cfy = 0
for i in range(n):
if i == 0:
cfx += cxes[i]
coeffx[i+1] -= cxes[i]
coeffx[i+2] += cxes[i]
cfy += cyes[i]
coeffy[i+1] -= cyes[i]
coeffy[i+2] += cyes[i]
elif i == n-1:
cfx += cxes[i]
coeffx[i] += cxes[i]
coeffx[i+1] -= cxes[i]
cfy += cyes[i]
coeffy[i] += cyes[i]
coeffy[i+1] -= cyes[i]
else:
coeffx[i] += cxes[i]
coeffx[i+1] -= 2*cxes[i]
coeffx[i+2] += cxes[i]
coeffy[i] += cyes[i]
coeffy[i+1] -= 2*cyes[i]
coeffy[i+2] += cyes[i]
rx = ""
ry = ""
for i in range(1, n+1):
s = f"abs((t-{i}))"
if i != n:
rx += f"(({s}*{canonise(coeffx[i])})+"
ry += f"(({s}*{canonise(coeffy[i])})+"
else:
rx += f"({s}*{canonise(coeffx[i])})" + ")"*(n-1)
ry += f"({s}*{canonise(coeffy[i])})" + ")"*(n-1)
print(f"({rx}+{canonise(cfx)})")
print(f"({ry}+{canonise(cfy)})")
```
| 86,125 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Tags: constructive algorithms, math
Correct Solution:
```
def ex(values):
e = None
for i, v in enumerate(values):
e_ = f'({v//2}*((1-abs((t-{i})))+abs((1-abs((t-{i}))))))'
if e is None:
e = e_
else:
e = f'({e}+{e_})'
return e
def solve(circles):
xs = [c[0] for c in circles]
ys = [c[1] for c in circles]
return ex(xs), ex(ys)
def pc(line):
t = tuple(map(int, line.split()))
assert len(t) == 3, f"Invalid circle: {line}"
return t
def main():
n = int(input())
circles = [pc(input()) for _ in range(n)]
f, g = solve(circles)
print(f)
print(g)
if __name__ == '__main__':
main()
```
| 86,126 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Submitted Solution:
```
def spec(i, targ):
return ("(((abs(((2 * t) - (2i - 1))) + ((2 * t) - (2i - 1))) * (abs(((2 * t) - (2i + 1))) - ((2 * t) - (2i + 1))))".replace("2i", str(2 * i)) + " * " + str((targ + 2) // 4) + ")").replace(" ", "")
def p(i):
x = eval(X.replace("t", str(i)))
y = eval(Y.replace("t", str(i)))
return (x, y)
n = int(input())
X = ["(0)"]
Y = ["(0)"]
for i in range(n):
x, y, r = map(int, input().split(" "))
X = ["("] + X + ["+"] + [spec(i, x)] + [")"]
Y = ["("] + Y + ["+"] + [spec(i, y)] + [")"]
X = "".join(X)
Y = "".join(Y)
print(X)
print(Y)
```
No
| 86,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Submitted Solution:
```
def f(x):
if x == n:
return "0"
if x == 0:
return "(" + str(X[0]) + "+" + f(1) + ")"
ss = "(abs(t-" + str(x-1) + ")-abs(t-" + str(x) + "))"
tmp = (X[x] - X[x - 1]) // 2
re = (X[x] - X[x - 1]) - 2 * tmp
X[x] -= re
ss = "((" + str(tmp) + "*" + ss + ")" + "+" + str(tmp) + ")"
return "(" + ss + "+" + f(x + 1) + ")"
n = int(input())
#c = [(int(_) for _ in input().split()) for i in range(n)]
c = [[int(x) for x in input().split()] for i in range(n)]
#print(n, c)
X = [c[i][0] for i in range(n)]
Y = [c[i][1] for i in range(n)]
#print(X)
#print(Y)
print(f(0))
#print(X)
X = Y
print(f(0))
```
No
| 86,128 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Submitted Solution:
```
def main():
n = int(input())
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i], _ = [int(i) for i in input().split()]
resx = []
resy = []
for i in range(n):
resx.append("({0}*((1-abs(t-{1}))+abs(1-abs(t-{1}))))".format(x[i] // 2, i))
resy.append("({0}*((1-abs(t-{1}))+abs(1-abs(t-{1}))))".format(y[i] // 2, i))
if i > 0:
resx[-1] += ')'
resy[-1] += ')'
print('(' * (n - 1) + '+'.join(resx))
print('(' * (n - 1) + '+'.join(resy))
main()
```
No
| 86,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Every day Ruslan tried to count sheep to fall asleep, but this didn't help. Now he has found a more interesting thing to do. First, he thinks of some set of circles on a plane, and then tries to choose a beautiful set of points, such that there is at least one point from the set inside or on the border of each of the imagined circles.
Yesterday Ruslan tried to solve this problem for the case when the set of points is considered beautiful if it is given as (xt = f(t), yt = g(t)), where argument t takes all integer values from 0 to 50. Moreover, f(t) and g(t) should be correct functions.
Assume that w(t) and h(t) are some correct functions, and c is an integer ranging from 0 to 50. The function s(t) is correct if it's obtained by one of the following rules:
1. s(t) = abs(w(t)), where abs(x) means taking the absolute value of a number x, i.e. |x|;
2. s(t) = (w(t) + h(t));
3. s(t) = (w(t) - h(t));
4. s(t) = (w(t) * h(t)), where * means multiplication, i.e. (w(t)Β·h(t));
5. s(t) = c;
6. s(t) = t;
Yesterday Ruslan thought on and on, but he could not cope with the task. Now he asks you to write a program that computes the appropriate f(t) and g(t) for any set of at most 50 circles.
In each of the functions f(t) and g(t) you are allowed to use no more than 50 multiplications. The length of any function should not exceed 100Β·n characters. The function should not contain spaces.
Ruslan can't keep big numbers in his memory, so you should choose f(t) and g(t), such that for all integer t from 0 to 50 value of f(t) and g(t) and all the intermediate calculations won't exceed 109 by their absolute value.
Input
The first line of the input contains number n (1 β€ n β€ 50) β the number of circles Ruslan thinks of. Next follow n lines, each of them containing three integers xi, yi and ri (0 β€ xi, yi β€ 50, 2 β€ ri β€ 50) β the coordinates of the center and the raduis of the i-th circle.
Output
In the first line print a correct function f(t). In the second line print a correct function g(t). The set of the points (xt = f(t), yt = g(t)) (0 β€ t β€ 50) must satisfy the condition, that there is at least one point inside or on the border of each of the circles, Ruslan thinks of at the beginning.
Examples
Input
3
0 10 4
10 0 4
20 10 4
Output
t
abs((t-10))
Note
Correct functions:
1. 10
2. (1+2)
3. ((t-3)+(t*4))
4. abs((t-10))
5. (abs((((23-t)*(t*t))+((45+12)*(t*t))))*((5*t)+((12*t)-13)))
6. abs((t-(abs((t*31))+14))))
Incorrect functions:
1. 3+5+7 (not enough brackets, it should be ((3+5)+7) or (3+(5+7)))
2. abs(t-3) (not enough brackets, it should be abs((t-3))
3. 2+(2-3 (one bracket too many)
4. 1(t+5) (no arithmetic operation between 1 and the bracket)
5. 5000*5000 (the number exceeds the maximum)
<image> The picture shows one of the possible solutions
Submitted Solution:
```
def main():
n = int(input())
x = [0] * n
y = [0] * n
for i in range(n):
x[i], y[i], _ = [int(i) for i in input().split()]
resx = []
resy = []
for i in range(n):
resx.append("{0}*((1-abs(t-{1}))+abs(1-abs(t-{1})))".format(x[i] // 2, i))
resy.append("{0}*((1-abs(t-{1}))+abs(1-abs(t-{1})))".format(y[i] // 2, i))
print('+'.join(resx))
print('+'.join(resy))
main()
```
No
| 86,130 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,copy,functools
sys.setrecursionlimit(10**7)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def LF(): return [float(x) for x in sys.stdin.readline().split()]
def LS(): return sys.stdin.readline().split()
def I(): return int(sys.stdin.readline())
def F(): return float(sys.stdin.readline())
def S(): return input()
def pf(s): return print(s, flush=True)
def main():
n,m = LI()
a = [LI() for _ in range(n)]
s = set()
for c in a:
s |= set(c[1:])
if len(s) == m:
return 'YES'
return 'NO'
print(main())
```
| 86,131 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
n,m=map(int,input().split())
yy=[]
for i in range(n):
x=list(map(int,input().split()))
x=x[1:]
yy.append(x)
main=[]
for i in yy:
for j in i:
main.append(j)
main=set(main)
gg={0}
for i in range(1,m+1):
gg.add(i)
gg.remove(0)
if(gg==main):
print("YES")
else:
print("NO")
```
| 86,132 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
n,m=map(int,input().split(" "))
L=[0]*(m)
for k in range(n):
i=0
for j in map(int,input().split(" ")):
if(i!=0):
L[j-1]=1
i+=1
if(sum(L)==m):
print("YES")
else:
print("NO")
```
| 86,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
if __name__ == '__main__':
n, m = map(int, input().split())
s = set()
for _ in range(n):
s.update(map(int, input().split()[1:]))
print('YES' if len(s) == m else 'NO')
```
| 86,134 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
a,b=map(int,input().split())
y=[]
k=0
for i in range(0,a):
x=list(map(int,input().split()))
y.extend(x[1:])
for i in range(1,b+1):
if(y.count(i)<1):
k=-1
print("NO")
break
if(k==0):
print("YES")
```
| 86,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
n , m = list(map(int,input().split()))
p = []
for i in range(n):
l = list(map(int,input().split()))
for k in range(1,len(l)):
p.append(l[k])
setp = set(p)
o = []
for i in range(1,m+1):
o.append(i)
seto = set(o)
if len(seto)==len(setp):
print("YES")
else:
print("NO")
```
| 86,136 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
def main():
n, m = map(int, input().split())
a = set({})
for i in range(n):
b = list(map(int, input().split()))
a.update(b[1:])
print("YES" if(len(a) == m) else "NO")
if __name__ == "__main__":
main()
```
| 86,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Tags: implementation
Correct Solution:
```
(n, m) = map(int, input().split())
a = []
for i in range(n):
a += list(map(int, input().split()))[1:]
print( 'YES' if len(set(a)) == m else 'NO')
```
| 86,138 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n, m = [int(s) for s in input().split(' ')]
bulbs = set()
for i in range(n):
for bulb in input().split(' ')[1:]:
bulbs.add(int(bulb))
if len(bulbs) < m:
print('NO')
else:
print('YES')
```
Yes
| 86,139 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n,m=input().split();
n=int(n);
m=int(m);
a=[1]*(m+1);
for i in range(n):
k=input().split();
w=int(k[0]);
for j in range(w):
x=int(k[j+1]);
m-=a[x];
a[x]=0;
if m==0 :print('YES');
else :print('NO');
```
Yes
| 86,140 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
'''
Amirhossein Alimirzaei
Telegram : @HajLorenzo
Instagram : amirhossein_alimirzaei
University of Bojnourd
'''
n=list(map(int,input().split()))
blb=[]
for _ in range(n[0]):
tmp=(list(map(int,input().split())))
for __ in tmp[1:]:
if __ not in blb:
blb.append(__)
print("YES" if len(blb)==n[1] else "NO")
```
Yes
| 86,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
n,m = map(int,input().split())
ls=[]
x = []
no=[]
B=[]
for i in range(1,m+1):
B.append(i)
while(n):
n-=1
ls = list(map(int,input().split()))
x.append(ls[0])
for i in ls[1:]:
no.append(i)
no = set(no)
no=list(no)
if(no==B):
print('YES')
else:
print('NO')
```
Yes
| 86,142 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 3 13:36:06 2020
@author: zuoxichen
"""
s1=input()
l1=s1.split()
n=int(l1[0])
m=int(l1[1])
i=0
l4=list()
while i<n:
s2=input()
l2=s2.split()
l3=list(int(l) for l in l2)
l3.pop(0)
for k in l3:
l4.append(k)
i+=1
else:
for i in range(1,m):
if i not in l4:
print('NO')
break
else:
print('YES')
```
No
| 86,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
my_input = input()
buttons,bulbs = my_input.split()
bulbs = int(bulbs)
temp = []
for x in range(int(buttons)):
inp = input()
my_list = inp.split()
for item in my_list:
if item not in temp:
temp.append(item)
if len(temp) == bulbs:
print("YES")
else:
print("NO")
```
No
| 86,144 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
import sys
si = True
i=0
j=0
n=[]
no=[]
m=None
for line in sys.stdin:
l = line.split()
if(len(l)!=0):
#print(l)
if(i==0):
m=int(l[1])
for x in range(m):
n.append(x+1)
#print("n=", n)
else:
for x in range(int(l[0])):
#print("'", l[x+1], "'")
if(int(l[x+1]) in n and not int(l[x+1]) in n):
#print("true")
no.append(int(l[x+1]))
j+=1
else: si = False
i+=1
#print(i)
else: break
#print("j= ", j)
if(si and j==len(n)): print("YES")
else: print("NO")
```
No
| 86,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya wants to turn on Christmas lights consisting of m bulbs. Initially, all bulbs are turned off. There are n buttons, each of them is connected to some set of bulbs. Vasya can press any of these buttons. When the button is pressed, it turns on all the bulbs it's connected to. Can Vasya light up all the bulbs?
If Vasya presses the button such that some bulbs connected to it are already turned on, they do not change their state, i.e. remain turned on.
Input
The first line of the input contains integers n and m (1 β€ n, m β€ 100) β the number of buttons and the number of bulbs respectively.
Each of the next n lines contains xi (0 β€ xi β€ m) β the number of bulbs that are turned on by the i-th button, and then xi numbers yij (1 β€ yij β€ m) β the numbers of these bulbs.
Output
If it's possible to turn on all m bulbs print "YES", otherwise print "NO".
Examples
Input
3 4
2 1 4
3 1 3 1
1 2
Output
YES
Input
3 3
1 1
1 2
1 1
Output
NO
Note
In the first sample you can press each button once and turn on all the bulbs. In the 2 sample it is impossible to turn on the 3-rd lamp.
Submitted Solution:
```
num_but, num_lmp = map(int,input().split(" "))
switched = list("-" * num_lmp)
for i in range(num_but):
print("inputting button {}".format(i))
inp = input()
inp = inp.split(" ")
inp = inp[1:]
for lamp in inp:
print("switching on lamp {}".format(lamp))
switched[int(lamp) - 1] = "+"
for lamp in range(len(switched)):
if switched[lamp] == "-":
print("NO")
break
else:
print("YES")
```
No
| 86,146 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Tags: data structures
Correct Solution:
```
from functools import reduce
class SegmentTree():
def __init__(self, L, function = lambda x,y: x+y, initilizer = None):
self.function = function
self.initilizer = initilizer
N = self.size = len(L)
M = 1 << N.bit_length()
self.margin = 2*M - N
self.L = [None for i in range(self.margin)] + L
for i in range(M-1, 0, -1):
x, y = self.L[i<<1], self.L[i<<1|1]
self.L[i] = None if x is None or y is None else function(x, y)
def modify(self, pos, value):
p = pos + self.margin
self.L[p] = value
while p > 1:
x, y = self.L[p], self.L[p^1]
if p&1: x, y = y, x
self.L[p>>1] = None if x is None or y is None else self.function(x, y)
p>>=1
def query(self, left, right):
l, r = left + self.margin, right + self.margin
stack = []
void = True
if self.initilizer is not None:
void = False
result = self.initilizer
while l < r:
if l&1:
if void:
result = self.L[l]
void = False
else:
result = self.function(result, self.L[l])
l+=1
if r&1:
r-=1
stack.append(self.L[r])
l>>=1
r>>=1
init = stack.pop() if void else result
return reduce(self.function, reversed(stack), init)
import sys
n, k, a, b, q = [int(x) for x in input().split()]
orders = [0]*(n+2)
a_tree, b_tree = SegmentTree(orders, initilizer = 0), SegmentTree(orders, initilizer = 0)
for line in sys.stdin:
s = [int(x) for x in line.split()]
if s[0] == 1:
orders[s[1]] += s[2]
a_tree.modify(s[1], min(a, orders[s[1]]))
b_tree.modify(s[1], min(b, orders[s[1]]))
else:
query = b_tree.query(0, s[1]) + a_tree.query(s[1]+k, n+1)
print(query)
```
| 86,147 |
Provide tags and a correct Python 3 solution for this coding contest problem.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Tags: data structures
Correct Solution:
```
import sys
from math import *
def minp():
return sys.stdin.readline().strip()
def mint():
return int(minp())
def mints():
return map(int, minp().split())
def add(a,x,v):
while x<len(a):
a[x] += v
x |= x+1
def get(a,x):
r = 0
while x>=0:
r += a[x]
x = (x&(x+1))-1
return r
n, k, a, b, q = mints()
h1 = [0]*n
h2 = [0]*n
z = [0]*n
for i in range(q):
t = tuple(mints())
if t[0] == 1:
p = z[t[1]-1]
pp = p + t[2]
add(h1, t[1]-1, min(a,pp)-min(a,p))
add(h2, t[1]-1, min(b,pp)-min(b,p))
z[t[1]-1] = pp
else:
print(get(h2,t[1]-2)+get(h1,n-1)-get(h1,t[1]+k-2))
```
| 86,148 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Submitted Solution:
```
#!/usr/bin/env python3
n, k, a, b, q = [int(x) for x in input().split()]
tree = [[0] * n, [0] * n]
orders = [[0] * n, [0] * n]
total = [0, 0]
def add(d_i, a_i, b_i):
total[0] += a_i
# total[1] += b_i
i = d_i
while i < n:
tree[0][i] += a_i
# tree[1][i] += b_i
i |= i + 1
def ask(p):
# [1; p) - normal
# [p; p + k) - maintanance
# [p + k; n) - normal
first_interval = [0, 0]
i = p - 1
while i >= 0:
# first_interval[0] += tree[0][i]
first_interval[1] += tree[1][i]
i &= i + 1
i -= 1
second_interval = [0, 0]
i = p + k - 1
while i >= 0:
second_interval[0] += tree[0][i]
# second_interval[1] += tree[1][i]
i &= i + 1
i -= 1
# ans = first_interval[0] + \
# second_interval[1] - first_interval[1] + \
# total[0] - second_interval[0]
ans = first_interval[1] + \
total[0] - second_interval[0]
# ans = second_interval[1] - first_interval[1]
print(ans)
for i in range(0, q):
l = [int(x) for x in input().split()]
if l[0] == 1:
d_i, a_i, b_i = l[1] - 1, l[2], l[2]
if orders[0][d_i] + a_i > a:
a_i = a - orders[0][d_i]
orders[0][d_i] = a
else:
orders[0][d_i] += a_i
if orders[1][d_i] + b_i > b:
b_i = b - orders[1][d_i]
orders[1][d_i] = b
else:
orders[1][d_i] += b_i
add(d_i, a_i, b_i)
else:
# assert l[0] == 2
p_i = l[1] - 1
ask(p_i)
```
No
| 86,149 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Submitted Solution:
```
n, k, a, b, q = map(int, input().split())
cnt = [0] * (n + 1)
days = [0] * (n+1)
for i in range(q):
c = list(map(int, input().split()))
if len(c) == 3:
cnt[c[1] - 1] += cnt[c[1] - 2] + min(b, c[2])
days[c[1] - 1] += c[2]
else:
cop = cnt[::]
cop[c[1]:c[1]+k] = [0]*k
c[1] += k
cop[c[1] - 1] = cnt[c[1] - k - 2] + min(a, days[c[1]-1])
cop[c[1]:] = [0]*(n - c[1]+1)
for j in range(c[1], n+1):
cop[j] = cop[j - 1] + min(a, days[j])
print(cop[-1])
```
No
| 86,150 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Submitted Solution:
```
n, k, a, b, q = map(int, input().split())
g1 = [0] * (n + 1)
g2 = [0] * (n + 1)
p = []
for i in range(q):
s = input().split()
if len(s) == 3:
h, d = int(s[1]), int(s[2])
g1[h] += d
g2[h] += d
else:
pos = int(s[1])
sum = 0
for i in range(n + 1):
sum += min(g1[i], b)
g1[i] = sum
sum = 0
for i in range(n + 1):
sum += min(g2[i], a)
g2[i] = sum
sum1 = g1[max(pos - 1, 0)] - g1[0]
sum2 = g2[-1] - g2[min(pos + k - 1, n)]
print(sum1 + sum2)
```
No
| 86,151 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
A factory produces thimbles in bulk. Typically, it can produce up to a thimbles a day. However, some of the machinery is defective, so it can currently only produce b thimbles each day. The factory intends to choose a k-day period to do maintenance and construction; it cannot produce any thimbles during this time, but will be restored to its full production of a thimbles per day after the k days are complete.
Initially, no orders are pending. The factory receives updates of the form di, ai, indicating that ai new orders have been placed for the di-th day. Each order requires a single thimble to be produced on precisely the specified day. The factory may opt to fill as many or as few of the orders in a single batch as it likes.
As orders come in, the factory owner would like to know the maximum number of orders he will be able to fill if he starts repairs on a given day pi. Help the owner answer his questions.
Input
The first line contains five integers n, k, a, b, and q (1 β€ k β€ n β€ 200 000, 1 β€ b < a β€ 10 000, 1 β€ q β€ 200 000) β the number of days, the length of the repair time, the production rates of the factory, and the number of updates, respectively.
The next q lines contain the descriptions of the queries. Each query is of one of the following two forms:
* 1 di ai (1 β€ di β€ n, 1 β€ ai β€ 10 000), representing an update of ai orders on day di, or
* 2 pi (1 β€ pi β€ n - k + 1), representing a question: at the moment, how many orders could be filled if the factory decided to commence repairs on day pi?
It's guaranteed that the input will contain at least one query of the second type.
Output
For each query of the second type, print a line containing a single integer β the maximum number of orders that the factory can fill over all n days.
Examples
Input
5 2 2 1 8
1 1 2
1 5 3
1 2 1
2 2
1 4 2
1 3 2
2 1
2 3
Output
3
6
4
Input
5 4 10 1 6
1 1 5
1 5 5
1 3 2
1 5 2
2 1
2 2
Output
7
1
Note
Consider the first sample.
We produce up to 1 thimble a day currently and will produce up to 2 thimbles a day after repairs. Repairs take 2 days.
For the first question, we are able to fill 1 order on day 1, no orders on days 2 and 3 since we are repairing, no orders on day 4 since no thimbles have been ordered for that day, and 2 orders for day 5 since we are limited to our production capacity, for a total of 3 orders filled.
For the third question, we are able to fill 1 order on day 1, 1 order on day 2, and 2 orders on day 5, for a total of 4 orders.
Submitted Solution:
```
n, k, a, b, q = map(int, input().split())
g1 = [0] * (n + 1)
g2 = [0] * (n + 1)
p = []
for i in range(q):
s = input().split()
if len(s) == 3:
h, d = int(s[1]), int(s[2])
g1[h] += d
g2[h] += d
else:
p.append(int(s[1]))
sum = 0
for i in range(n + 1):
sum += min(g1[i], b)
g1[i] = sum
sum = 0
for i in range(n + 1):
sum += min(g2[i], a)
g2[i] = sum
for i in range(len(p)):
sum1 = g1[p[i] - 1] - g1[0]
sum2 = g2[-1] - g2[p[i] + k - 1]
print(sum1 + sum2)
```
No
| 86,152 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
import sys
input=sys.stdin.readline
s=input().rstrip()
l=list(s.split(" "))
p=1
m=0
for ss in l:
if ss=="+":
p+=1
if ss=="-":
m+=1
n=int(l[-1])
if n*p-m<n or p-m*n>n:
print("Impossible")
exit()
arr_p=[1]*p
arr_m=[1]*m
r=p-m
for i in range(p):
xx=min(n-1,max(n-r,0))
arr_p[i]+=xx
r+=xx
for i in range(m):
xx=min(n-1,max(r-n,0))
arr_m[i]+=xx
r-=xx
ans=[]
pre=1
for i in range(len(l)):
if l[i]=="?":
if pre:
ans.append(str(arr_p.pop()))
else:
ans.append(str(arr_m.pop()))
else:
if l[i]=="+":
pre=1
ans.append(l[i])
elif l[i]=="-":
pre=0
ans.append(l[i])
else:
ans.append(str(l[i]))
print("Possible")
print(" ".join(ans))
```
| 86,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
def binp(a,b):
st=a[0]
fin=a[1]
b=[-b[1],-b[0]]
while True:
s=(st+fin)//2
if s>=b[0]:
if s<=b[1]:
return(s)
else:
fin=s-1
else:
st=s+1
s=input()
for i in range(len(s)):
if s[i]=='=':
y=i
break
y+=1
n=int(s[y:])
now='+'
max_=0
min_=0
ctr=0
pls=0
minus=0
g=[[0,0]]
for i in s:
ctr+=1
if i=='?':
if now=='+':
max_+=n
min_+=1
pls+=1
else:
max_-=1
min_-=n
minus+=1
g.append([min_,max_])
elif i=='-' or i=='+':
now=i
elif i=='=':
break
v=n
if v<min_ or v>max_:
print('Impossible')
else:
print('Possible')
a=[pls*1,pls*v]
b=[minus*(-v)-n,minus*(-1)-n]
#print(a,b)
ctr=0
ctr=binp(a,b)
fir=ctr
j=[0]*pls
if pls!=0:
j=[fir//pls]*pls
for u in range(fir%pls):
j[u]+=1
sec=n-ctr
k=[0]*minus
if minus!=0:
k=[((-sec)//minus)]*minus
#print(k)
#print((-sec)%minus)
for u in range((-sec)%minus):
k[u]+=1
p=0
m=0
now='+'
for i in s:
if i=='?':
if now=='+':
print(j[p],end='')
p+=1
else:
print(k[m],end='')
m+=1
elif i=='-' or i=='+':
now=i
print(i,end='')
else:
print(i,end='')
```
| 86,154 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s = ('+ ' + input()).strip().split()
plus = 0
minus = 0
for c in s:
if c == '+':
plus += 1
elif c == '-':
minus += 1
n = int(s[-1])
maxn = plus * n - minus
minn = plus - minus * n
if not (maxn >= n >= minn):
print("Impossible")
exit()
ans = []
for i in range(1, len(s)):
need = min(maxn - n, n - 1)
if s[i-1] == '+':
ans.append(str(n - need))
maxn -= need
elif s[i-1] == '-':
ans.append(str(1 + need))
maxn -= need
else:
ans.append(s[i])
print("Possible")
print(' '.join(ans))
```
| 86,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
data = input()
data = data.split()
data.reverse()
n = int(data[0])
data.reverse()
pos = 1
neg = 0
for x in data:
if x == '-':
neg += 1
if x == '+':
pos += 1
ans = pos - neg
if ans <= n:
tmp = n - ans
if tmp > pos * (n - 1):
print('Impossible')
exit(0)
else:
print('Possible')
fin = ''
for x in range(len(data)):
if data[x] != '?':
fin += data[x] + ' '
continue
if x == 0 or data[x - 1] == '+':
t = min(n - 1, tmp)
tmp -= t
fin += str(1 + t) + ' '
else:
fin += '1 '
print(fin)
else:
tmp = ans - n
if tmp > neg * (n - 1):
print('Impossible')
exit(0)
else:
print('Possible')
fin = ''
for x in range(len(data)):
if data[x] != '?':
fin += data[x] + ' '
continue
if x != 0 and data[x - 1] == '-':
t = min(n - 1, tmp)
tmp -= t
fin += str(1 + t) + ' '
else:
fin += '1 '
print(fin)
```
| 86,156 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
def max(a, b):
if(a>=b):
return a
else:
return b
####################################
s=input().split()
length=len(s)
n=int(s[length-1])
plus=1
minus=0
for i in s:
if(i== '+'):
plus+=1
if(i== '-'):
minus+=1
if(plus*n - minus < n or plus - n*minus > n):
print('Impossible')
exit()
else:
print('Possible')
for i in range(0, length-1, 2): #initializing all placeholders with 1
s[i]='1'
# if(minus==0):
# s[0]= repr(n-plus+1)
# else:
# diff=plus-1-minus
# if(diff>0):
# s[0]=repr(n-diff)
# for i in range(2, length-1, 2):
# s[i]='1'
# flag=0
# if(diff<=0):
# s[0]=repr(n)
# for i in range(2, length-1, 2):
# s[i]='1'
# if(flag==0 and s[i-1] == '+'):
# flag=1
# s[i]=repr(1-diff)
res=n-plus+minus
for i in range(0, length-1, 2):
if((i==0 or s[i-1]=='+' ) and res>0):
val=int(s[i])
if(res>n-val):
res-=(n-val)
s[i]=repr(n)
else:
val+=res
s[i]=repr(val)
res=0
elif(s[i-1]=='-' and res<0):
val=int(s[i])
if(res<val-n):
res+=(n-val)
s[i]=repr(n)
else:
val-=res
s[i]=repr(val)
res=0
print (' '.join(s))
```
| 86,157 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
s=input()
a=s.split()
n=int(a[-1])
pos=1
neg=0
for item in a:
if item=='+':
pos+=1
if item=='-':
neg+=1
if n<pos-neg*n or n>pos*n-neg:
print("Impossible")
else:
print("Possible")
parr=[]
narr=[]
if pos>neg:
for i in range(neg):
narr.append(n)
x=int(n*(neg+1)//pos)
for i in range(pos):
if(i<n*(neg+1)%pos):
parr.append(x+1)
else:
parr.append(x)
else:
for i in range(pos):
parr.append(n)
x=int(n*(pos-1)//neg)
for i in range(neg):
if(i<n*(pos-1)%neg):
narr.append(x+1)
else:
narr.append(x)
sgn=1
for c in s:
if (c=='?'):
if (sgn==1):
print(parr[0],end="")
del parr[0]
else:
print(narr[0],end="")
del narr[0]
continue
elif (c=='+'):
sgn=1
elif c=='-':
sgn=-1
print(c,end="")
```
| 86,158 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
a = list(input().split())
n = int(a[-1])
R = int(a[-1])
one, two = 0, 0
for i in range(1,len(a),2):
if a[i] == '=':
break
if a[i] == '+':
R -= 1
one += 1
else:
R += 1
two += 1
R -= 1
one += 1
if R >= 0:
if one * (n - 1) >= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i == 0 or a[i - 1] == '+':
print(min(n - 1, R) + 1, a[i + 1], end = ' ')
R -= min(n - 1, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
else:
if two * (1 - n) <= R:
print('Possible')
for i in range(0, len(a), 2):
if i > 0 and a[i - 1] == '=':
print(a[i])
else:
if i > 0 and a[i - 1] == '-':
print(-(max(1 - n, R) - 1), a[i + 1], end = ' ')
R -= max(1 - n, R)
else:
print(1, a[i + 1], end = ' ')
else:
print('Impossible')
```
| 86,159 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Tags: constructive algorithms, expression parsing, greedy, math
Correct Solution:
```
from sys import stdin
s=stdin.readline().strip().split()
def smas(x,mas):
for i in range(len(s)):
if s[i]==1 and (i==0 or s[i-1]=="+"):
if x>0:
y=min(n-1,x)
s[i]+=y
x-=y
def smen(x,men):
for i in range(len(s)):
if s[i]==1 and i>0 and s[i-1]=="-":
if x>0:
y=min(n-1,x)
s[i]+=y
x-=y
n=int(s[-1])
ans=0
y=0
if s[0]=="-":
y=1
ind=-1
men=0
mas=0
for i in range(y,len(s)-1,2):
if i!=0 and s[i-1]=="-" :
ans-=1
men+=1
else:
mas+=1
if ind==-1:
ind=i
else:
ans+=1
s[i]=1
l=[-(n*men),-men]
t=True
for i in range(mas,n*mas+1):
if i>=n:
x=n-i
else:
continue
if x>=l[0] and x<=l[1]:
t=False
smas(i-mas,mas)
smen((-x)-men,men)
break
if t:
print("Impossible")
else:
print("Possible")
print(*s)
```
| 86,160 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
#!/bin/python3
a = input()
while a:
expr, n = a.split('=')
n = int(n.strip())
pos = 1
neg = 0
sg = [1]
for c in expr:
if c == '+':
pos += 1
sg.append(1)
elif c == '-':
neg += 1
sg.append(0)
csum = pos - neg
rez = []
for i in sg:
if csum < n:
if i > 0:
v = min(n-csum, n-1)
csum += v
rez.append(1+v)
else:
rez.append(-1)
else:
if i > 0:
rez.append(1)
else:
v = min(csum - n, n-1)
csum -= v
rez.append(-1-v)
if csum == n:
print("Possible")
ans = str(rez[0])
for j in rez[1:]:
ans += " " + ("+" if j > 0 else "-") + " "
ans += str(abs(j))
ans += " = " + str(n)
print(ans)
else:
print("Impossible")
break
a = input()
```
Yes
| 86,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
#!/usr/bin/env python3
# 664B_rebus.py - Codeforces.com/problemset/problem/664/B by Sergey 2016
import unittest
import sys
import re
###############################################################################
# Rebus Class (Main Program)
###############################################################################
class Rebus:
""" Rebus representation """
def __init__(self, test_inputs=None):
""" Default constructor """
it = iter(test_inputs.split("\n")) if test_inputs else None
def uinput():
return next(it) if it else sys.stdin.readline().rstrip()
# Reading single elements
self.str = uinput()
# End value
self.n = 0
m = re.search("(\d+)", self.str)
if m:
self.n = int(m.group(1))
# Signs
self.signs = ["+"] + re.findall("\? ([+-])", self.str)
def summ(self, nums, signs):
result = 0
for i in range(len(self.signs)):
if self.signs[i] == "+":
result += nums[i]
else:
result -= nums[i]
return result
def calculate(self):
""" Main calcualtion function of the class """
nums = [0] * len(self.signs)
for i in range(len(self.signs)):
nums[i] = 1
sum = self.summ(nums, self.signs)
for i in range(len(self.signs)):
if sum != self.n:
if self.signs[i] == "+" and sum < self.n:
nums[i] = min(self.n - sum + 1, self.n)
sum -= 1
sum += nums[i]
if self.signs[i] == "-" and sum > self.n:
nums[i] = min(sum + 1 - self.n, self.n)
sum += 1
sum -= nums[i]
if sum == self.n:
result = "Possible\n"
for i in range(len(self.signs)):
if i != 0:
result += self.signs[i] + " "
result += str(nums[i]) + " "
result += "= " + str(self.n)
else:
result = "Impossible"
return str(result)
###############################################################################
# Unit Tests
###############################################################################
class unitTests(unittest.TestCase):
def test_single_test(self):
""" Rebus class testing """
# Constructor test
test = "? + ? - ? + ? + ? = 42"
d = Rebus(test)
self.assertEqual(d.str, "? + ? - ? + ? + ? = 42")
self.assertEqual(d.n, 42)
self.assertEqual(d.signs, ["+", "+", "-", "+", "+"])
# Sample test
self.assertEqual(Rebus(test).calculate(), "Possible\n40 + 1 - 1 + 1 + 1 = 42")
# Sample test
test = "? - ? = 1"
self.assertEqual(Rebus(test).calculate(), "Impossible")
# Sample test
test = "? = 1000000"
self.assertEqual(Rebus(test).calculate(), "Possible\n1000000 = 1000000")
test = "? + ? + ? + ? - ? = 2"
self.assertEqual(Rebus(test).calculate(), "Possible\n1 + 1 + 1 + 1 - 2 = 2")
# My tests
test = ""
# self.assertEqual(Rebus(test).calculate(), "0")
# Time limit test
# self.time_limit_test(5000)
def time_limit_test(self, nmax):
""" Timelimit testing """
import random
import timeit
# Random inputs
test = str(nmax) + " " + str(nmax) + "\n"
numnums = [str(i) + " " + str(i+1) for i in range(nmax)]
test += "\n".join(numnums) + "\n"
nums = [random.randint(1, 10000) for i in range(nmax)]
test += " ".join(map(str, nums)) + "\n"
# Run the test
start = timeit.default_timer()
d = Rebus(test)
calc = timeit.default_timer()
d.calculate()
stop = timeit.default_timer()
print("\nTimelimit Test: " +
"{0:.3f}s (init {1:.3f}s calc {2:.3f}s)".
format(stop-start, calc-start, stop-calc))
if __name__ == "__main__":
# Avoiding recursion limitaions
sys.setrecursionlimit(100000)
if sys.argv[-1] == "-ut":
unittest.main(argv=[" "])
# Print the result string
sys.stdout.write(Rebus().calculate())
```
Yes
| 86,162 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
#!/usr/bin/env python3
import re
try:
while True:
s = input()
n = int(s[s.rfind(' '):])
pos = s.count('+') + 1
neg = s.count('-')
if n * pos - neg < n or pos - n * neg > n:
print("Impossible")
else:
print("Possible")
need = n - (pos - neg)
prev = '+'
first = True
for m in re.finditer(r"[+-]", s):
if first:
first = False
else:
print(prev, end=' ')
if prev == '+' and need > 0:
x = min(need + 1, n)
need -= x - 1
elif prev == '-' and need < 0:
x = min(-need + 1, n)
need += x - 1
else:
x = 1
print(x, end=' ')
prev = m.group()
if not first:
print(prev, end=' ')
if prev == '+' and need > 0:
x = min(need + 1, n)
need -= x - 1
elif prev == '-' and need < 0:
x = min(-need + 1, n)
need += x - 1
else:
x = 1
print(x, '=', n)
except EOFError:
pass
```
Yes
| 86,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
ar, ans = input().split('=')
ans = int(ans)
pl = ar.count('+') + 1
mn = ar.count('-')
if pl*ans - mn*1 < ans or pl*1 - mn*ans > ans:
print("Impossible")
exit(0)
print("Possible")
mat = [1]*(pl+mn)
sig = ('+' + ar).replace(' ', '').split('?')
del sig[-1]
while True:
i = 0
s = ""
for ch in ar:
if ch == '?':
s += str(mat[i])
i += 1
else:
s += ch
t = eval(s)
if t == ans:
print("{0}= {1}".format(s, ans))
exit(0)
d = ans - t
for i in range(len(mat)):
if sig[i] == '+' and d > 0:
mat[i] = min(1 + d, ans)
d -= mat[i] - 1
elif sig[i] == '-' and d < 0:
mat[i] = min(1 + abs(d), ans)
d += mat[i] - 1
```
Yes
| 86,164 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
expr = input().split()
target_number = int(expr[-1])
result_expr = None
def spread(number_sum, quantity):
spreaded = []
acc = number_sum
remaining = quantity - 1
for i in range(quantity):
next_number = min(target_number, acc - remaining)
spreaded.append(next_number)
acc -= next_number
remaining -= 1
return spreaded
plus_count = 1 + expr.count('+')
minus_count = expr.count('-')
if minus_count == 0:
if plus_count <= target_number <= plus_count*target_number:
spreaded = spread(target_number, plus_count)
result_expr = "{0} = {1}".format(" + ".join(map(str, spreaded)), target_number)
else:
negative = minus_count
if negative <= minus_count*target_number:
positive = negative + target_number
if plus_count <= positive <= plus_count*target_number:
spreaded_positive = spread(positive, plus_count)
spreaded_negative = spread(negative, minus_count)
result_expr = "{0} ".format(spreaded_positive.pop())
for i in range(1, len(expr) - 2):
if expr[i] == '+':
result_expr += '+ {0} '.format(spreaded_positive.pop())
elif expr[i] == '-':
result_expr += '- {0} '.format(spreaded_negative.pop())
result_expr += "= {0}".format(target_number)
if result_expr:
print("Possible")
print(result_expr)
else:
print("Impossible")
```
No
| 86,165 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
from sys import stdin,stdout
input=lambda:stdin.readline().strip()
lst=input().split()
nq=0
nm=0
for i in lst:
if i=='?':
nq+=1
elif i=='-':
nm+=1
np=nq-nm
n=int(lst[-1])
sum1=n+nm-np
lst1=[1]*np
if sum1<0:
print('Impossible')
else:
i=0
while i<np and sum1>0:
if sum1<=(n-1):
lst1[i]=lst1[i]+sum1
sum1=0
else:
lst1[i]=lst1[i]+(n-1)
sum1=sum1-(n-1)
i+=1
if sum1>0:
print('Impossible')
else:
print('Possible')
nn=len(lst)
j=0
if lst[0]=='?':
lst[0]=lst1[0]
j+=1
i=0
while i<nn:
if lst[i]=='+':
lst[i+1]=lst1[j]
j+=1
i+=1
elif lst[i]=='-':
lst[i+1]=1
i+=1
i+=1
print(*lst)
```
No
| 86,166 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
s=input().split()
f=False
p=0
c=0
for num,i in enumerate(s):
if i=='=': break
if i=='?' and (not f):
p=num
f=True
continue
if i=='?':
if(i==0 or s[num-1]=='+'):
c+=1
s[num]='1'
elif (s[num-1]=='-'):
c-=1
s[num]='1'
if c < 0 or c >= int(s[-1]):
print('Impossible')
exit(0)
else:
print('Possible')
s[p]=int(s[-1])-c
print(*s)
```
No
| 86,167 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a rebus of form ? + ? - ? + ? = n, consisting of only question marks, separated by arithmetic operation '+' and '-', equality and positive integer n. The goal is to replace each question mark with some positive integer from 1 to n, such that equality holds.
Input
The only line of the input contains a rebus. It's guaranteed that it contains no more than 100 question marks, integer n is positive and doesn't exceed 1 000 000, all letters and integers are separated by spaces, arithmetic operations are located only between question marks.
Output
The first line of the output should contain "Possible" (without quotes) if rebus has a solution and "Impossible" (without quotes) otherwise.
If the answer exists, the second line should contain any valid rebus with question marks replaced by integers from 1 to n. Follow the format given in the samples.
Examples
Input
? + ? - ? + ? + ? = 42
Output
Possible
9 + 13 - 39 + 28 + 31 = 42
Input
? - ? = 1
Output
Impossible
Input
? = 1000000
Output
Possible
1000000 = 1000000
Submitted Solution:
```
s = input()
plus = 1
minus = 0
for i in s :
if i=='+':
plus+=1
if i=='-':
minus+=1
n= int (s[s.rfind(' '):])
if plus<=minus :
print('Impossible')
else:
if plus>minus:
ras=plus-minus
znak=0
if (ras>n):
print('Impossible')
else:
out = str(n-ras+1);
for i in s[1:s.rfind(' ')]:
if i=='?':
out+='1'
if i=='+':
out+='+'
if i=='-':
out+='-'
if i=='=':
out=out+'= '+str(n)
if i==' ':
out+=' '
print(out)
```
No
| 86,168 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Tags: dp
Correct Solution:
```
f = lambda: list(map(int, input().split()))
n = int(input())
a, b = f(), f()
d = [[None] * 10001 for i in range(n)]
def g(i, s):
if s <= 0: return (0, s)
if i == n: return (1e7, 0)
if not d[i][s]:
x, y = g(i + 1, s - b[i])
d[i][s] = min(g(i + 1, s), (x + 1, y + b[i] - a[i]))
return d[i][s]
x, y = g(0, sum(a))
print(x, y)
```
| 86,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Tags: dp
Correct Solution:
```
import math
from collections import defaultdict
import os
import sys
from io import BytesIO, IOBase
from types import GeneratorType
from collections import defaultdict
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")
sys.setrecursionlimit(10**5)
def calc(i,total):
if total <= 0:
return [0, total]
if i==-1:
return [float('inf'),float('inf')]
if dp[(i,total)][0]!=-1:
return dp[(i,total)]
poss1=calc(i-1,total)
poss2=calc(i-1,total-b[i])
poss2[0]+=1
poss2[1]+=(b[i]-a[i])
if poss1[0]<poss2[0]:
dp[(i,total)][0]=poss1[0]
dp[(i, total)][1] = poss1[1]
elif poss1[0]>poss2[0]:
dp[(i, total)][0] = poss2[0]
dp[(i, total)][1] = poss2[1]
else:
if poss1[1] < poss2[1]:
dp[(i, total)][0] = poss1[0]
dp[(i, total)][1] = poss1[1]
else:
dp[(i, total)][0] = poss2[0]
dp[(i, total)][1] = poss2[1]
return [dp[(i,total)][0],dp[(i,total)][1]]
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
s=sum(a)
dp=defaultdict(lambda:[-1,0])
val=calc(n-1,s)
print(val[0],val[1])
```
| 86,170 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Tags: dp
Correct Solution:
```
import sys
input = sys.stdin.buffer.readline
n = int(input())
soda = list(map(int, input().split()))
volume = list(map(int, input().split()))
dp = [[{} for j in range(n + 1)] for k in range(n + 1)]
dp[0][1][volume[0] - soda[0]] = 0
total = soda[0]
for i in range(0, n - 1):
total += soda[i + 1]
dp[i + 1][1][volume[i + 1] - total] = total - soda[i + 1]
for bottle in range(1, i + 2):
for volume_left in dp[i][bottle]:
if volume_left - soda[i + 1] not in dp[i + 1][bottle]:
dp[i + 1][bottle][volume_left - soda[i + 1]] = dp[i][bottle][volume_left] + soda[i + 1]
dp[i + 1][bottle][volume_left - soda[i + 1]] = min(dp[i][bottle][volume_left] + soda[i + 1],dp[i + 1][bottle][volume_left - soda[i + 1]])
if volume_left + volume[i + 1] - soda[i + 1] not in dp[i + 1][bottle + 1]:
dp[i + 1][bottle + 1][volume_left + volume[i + 1] - soda[i + 1]] = dp[i][bottle][volume_left]
dp[i + 1][bottle + 1][volume_left + volume[i + 1] - soda[i + 1]] = min(dp[i][bottle][volume_left],dp[i + 1][bottle + 1][volume_left + volume[i + 1] -soda[i + 1]])
flag = 0
mini = 99999999999
i = 0
for i in range(1, n + 1):
for j in dp[n - 1][i]:
if j >= 0:
mini = min(mini, dp[n - 1][i][j])
flag = 1
if flag == 1:
break
print(i, mini)
```
| 86,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Tags: dp
Correct Solution:
```
# minTime[bottle][bottlesUsed][volume]
n = int(input())
a = [int(x) for x in input().split()]
b = [int(x) for x in input().split()]
minTime = [[{} for j in range(101)] for k in range(101)]
minTime[1][1][b[0] - a[0]] = 0
minTime[1][0][-a[0]] = a[0]
validVolumes = {-a[0], b[0]-a[0]}
for bottle in range(1, n):
validVolumesNew = set()
for bottlesUsed in range(0, bottle+1):
for volumeLeft in validVolumes:
if volumeLeft in minTime[bottle][bottlesUsed]:
currTime = minTime[bottle][bottlesUsed][volumeLeft]
# pouredTuple = (bottle+1, bottlesUsed, volumeLeft - a[bottle])
pouredTime = currTime + a[bottle]
# includedTuple = (bottle+1, bottlesUsed+1, volumeLeft + b[bottle] - a[bottle])
includedTime = currTime
if volumeLeft - a[bottle] not in minTime[bottle+1][bottlesUsed]:
minTime[bottle + 1][bottlesUsed][volumeLeft - a[bottle]] = 100000000
minTime[bottle+1][bottlesUsed][volumeLeft - a[bottle]] = min(minTime[bottle+1][bottlesUsed][volumeLeft - a[bottle]], pouredTime)
if volumeLeft + b[bottle] - a[bottle] not in minTime[bottle+1][bottlesUsed+1]:
minTime[bottle + 1][bottlesUsed + 1][volumeLeft + b[bottle] - a[bottle]] = 100000000
minTime[bottle+1][bottlesUsed+1][volumeLeft + b[bottle] - a[bottle]] = min(minTime[bottle+1][bottlesUsed+1][volumeLeft + b[bottle] - a[bottle]], includedTime)
validVolumesNew.add(volumeLeft - a[bottle])
validVolumesNew.add(volumeLeft + b[bottle] - a[bottle])
validVolumes = validVolumesNew
validVolumesNew = set()
exitFlag = False
lowestTime = 1000000000
for bottlesUsed in range(0,n+1):
for volumeLeft in range(0, n*101):
if volumeLeft in minTime[n][bottlesUsed]:
exitFlag = True
lowestTime = min(lowestTime, minTime[n][bottlesUsed][volumeLeft])
# print(lowestTime, n, bottlesUsed, volumeLeft)
if exitFlag:
print(bottlesUsed, lowestTime)
exit()
```
| 86,172 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Submitted Solution:
```
n= int(input())
m = [int(x) for x in input().split()]
t = [int(x) for x in input().split()]
v = sum(m)
b = sorted(t)
c = 0
d = 0
for i in range(n-1,-1,-1):
if v <= 0:
break
v = v - b[i]
c+=1
for i in range(n):
for j in range(n-1):
if t[j]>t[j+1]:
t[j+1],t[j] = t[j],t[j+1]
m[j+1],m[j] = m[j],m[j+1]
for i in range(n-c):
d = d+m[i]
if d==276:
d = d-59
if d == 313:
d =d-3
if d == 762:
d = d -199
print(c,d)
```
No
| 86,173 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Submitted Solution:
```
n= int(input())
m = [int(x) for x in input().split()]
t = [int(x) for x in input().split()]
v = sum(m)
b = sorted(t)
c = 0
d = 0
for i in range(n-1,-1,-1):
if v <= 0:
break
v = v - b[i]
c+=1
for i in range(n):
for j in range(n-1):
if t[j]>t[j+1]:
t[j+1],t[j] = t[j],t[j+1]
m[j+1],m[j] = m[j],m[j+1]
for i in range(n-c):
d = d+m[i]
if d==276:
d = d-59
if d == 313:
d =d-3
print(c,d)
```
No
| 86,174 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Submitted Solution:
```
n=int(input())
s11 = list(map(int,input().split()))
s22 = list(map(int,input().split()))
l=[]
for i in range(n):
l.append([])
l[i].append(s22[i])
l[i].append(s11[i])
coun1 = 0;
step = 0;stept = 0;
for i in range(n):
coun1+=l[i][1]
# print(l)
for j in range(n):
for i in range(1,n):
if(l[i-1][0]<l[i][0]):
# print(l[i-1][0],l[i][0])
d=l[i-1];
l[i-1]=l[i]
l[i]=d
for i in range(n):
if(coun1>0):
# print(coun1,l[i][0],step)
coun1-=l[i][0]
step+=1;
else:
# print(stept,l[i][1])
stept+=l[i][1]
print(step,stept)
```
No
| 86,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai β€ bi).
Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.
Nick asks you to help him to determine k β the minimal number of bottles to store all remaining soda and t β the minimal time to pour soda into k bottles. A bottle can't store more soda than its volume. All remaining soda should be saved.
Input
The first line contains positive integer n (1 β€ n β€ 100) β the number of bottles.
The second line contains n positive integers a1, a2, ..., an (1 β€ ai β€ 100), where ai is the amount of soda remaining in the i-th bottle.
The third line contains n positive integers b1, b2, ..., bn (1 β€ bi β€ 100), where bi is the volume of the i-th bottle.
It is guaranteed that ai β€ bi for any i.
Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.
Examples
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Note
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
import collections
from itertools import permutations
from collections import defaultdict
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")
#-------------------game starts now-----------------------------------------------------
n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
c=list()
d=list()
for i in range (n):
c.append([b[i]-a[i],a[i],i])
d.append([a[i],b[i]-a[i],i])
c.sort()
d.sort()
s=set()
ct=0
j=n-1
time=0
#print(d)
#print(c)
for i in range (n):
vol=d[i][0]
ch=1
s.add(d[i][2])
j=n-1
m=c
while vol>0:
if j<0:
ch=0
break
if c[j][2] in s:
j-=1
continue
if c[j][0]>=vol:
c[j][0]-=vol
vol=0
break
else:
vol-=c[j][0]
c[j][0]=0
j-=1
#print(i,vol)
#print(c)
if vol!=0:
s.remove(d[i][2])
c=m
continue
ct+=1
time+=d[i][0]
print(n-ct,time)
```
No
| 86,176 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
#-------------Program--------------
#----Kuzlyaev-Nikita-Codeforces----
#-------------Training-------------
#----------------------------------
n1,n2=map(str,input().split())
n=int(input())
for i in range(n):
v,f=map(str,input().split())
print(n1,n2)
if v==n1:
n1=f
else:
n2=f
print(n1,n2)
```
| 86,177 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
line=input()
print(line)
now = line.split()
n=int(input())
for i in range(n):
inp=input().strip()
line=inp.split()
for i in range(len(now)):
if now[i]==line[0]:
now[i]=line[1]
for x in now:
print(x,end=' ')
print()
```
| 86,178 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
import math, sys
def main():
first, second = input().split()
n = int(input())
for i in range(n):
print(first, second)
kill, zher = input().split()
if first == kill:
first = zher
if second == kill:
second = zher
print(first, second)
if __name__=="__main__":
main()
```
| 86,179 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
a, b = map(str, input().split())
print(a, b, sep=' ')
n = int(input())
for i in range(n):
c, d = map(str, input().split())
if c == a:
a = d
if c == b:
b = d
print(a, b, sep=' ')
# CodeForcesian
# β₯
# Ψ²Ψ¨Ω
```
| 86,180 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
def main():
n1, n2 = input().split()
print(n1, n2)
names = {n1, n2}
k = int(input())
for i in range(k):
killed, new = input().split()
names.remove(killed)
names.add(new)
for name in names:
print(name, end=' ')
print()
if __name__ == '__main__':
main()
```
| 86,181 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
s = input().split()
print(' '.join(s))
n = int(input())
for i in range(n):
n1, n2 = input().split()
s[s.index(n1)] = n2
print(' '.join(s))
```
| 86,182 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
(a, b) = input().split()
n = int(input())
print(a, b)
for i in range(n):
(a1, a2) = input().split()
if a == a1:
a = a2
elif b == a1:
b = a2
elif a == a2:
a = a1
elif b == a2:
b = a1
print(a, b)
```
| 86,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Tags: brute force, implementation, strings
Correct Solution:
```
potential_victim_1, potential_victim_2 = input().split()
print(potential_victim_1, potential_victim_2)
n = int(input())
for i in range(n):
victim, next_potential_victim = input().split()
if potential_victim_1 == victim:
print(potential_victim_2, next_potential_victim)
potential_victim_1 = next_potential_victim
else:
print(potential_victim_1, next_potential_victim)
potential_victim_2 = next_potential_victim
```
| 86,184 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
a, b = input().split()
n = int(input())
print(a, b)
for _ in range(n):
old, new = input().split()
if old == a:
a = new
else:
b = new
print(a, b)
```
Yes
| 86,185 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
ans=[list(input().split())]
n=int(input())
for _ in range(n):
x,y=input().split()
xo,yo=ans[-1][0],ans[-1][1]
if xo==x:
ans+=[[y,yo]]
else:
ans+=[[xo,y]]
for i in range(n+1):
ans[i]=' '.join(ans[i])
print('\n'.join(ans))
```
Yes
| 86,186 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
s1, s2 = list(map(str, input().split()))
l = []
l += [s1, s2]
n = int(input())
for i in range(n):
s3, s4 = list(map(str, input().split()))
if s1 == s3:
s1 = s4
else:
s2 = s4
l += [s1, s2]
for i in range(2 * n + 2):
if i % 2 == 0:
print(l[i], l[i + 1])
```
Yes
| 86,187 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
t,v = input().split(' ')
n=int(input())
i=0
l = []
l1= []
l1.append(t)
l1.append(v)
l.append(l1)
while(i<n):
k1,k2 = input().split(' ')
l1=[]
if(t==k1):
l1.append(k2)
l1.append(v)
l.append(l1)
t=v
v=k2
if(v==k1):
l1.append(k2)
l1.append(t)
l.append(l1)
t=t
v=k2
i=i+1
for i in l:
print(i[0],i[1])
```
Yes
| 86,188 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
s = input()
n = int(input())
print(s)
for i in range(n):
n1, n2 = input().split()
if n1 in s: s = s.replace(n1, n2)
else: s = s.replace(n2, n1)
print(s)
```
No
| 86,189 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
s=input()
l=[s]
n=int(input())
for i in range(n):
x,y=map(str,input().split())
if(x in s):
s=s.replace(x,y)
l.append(s)
for i in l:
print(i)
```
No
| 86,190 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
a=input().split()
n=int(input())
print(*a)
for _ in range(n):
b=input().split()
a.remove(b[0])
a.append(b[1])
print(*b)
```
No
| 86,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Our beloved detective, Sherlock is currently trying to catch a serial killer who kills a person each day. Using his powers of deduction, he came to know that the killer has a strategy for selecting his next victim.
The killer starts with two potential victims on his first day, selects one of these two, kills selected victim and replaces him with a new person. He repeats this procedure each day. This way, each day he has two potential victims to choose from. Sherlock knows the initial two potential victims. Also, he knows the murder that happened on a particular day and the new person who replaced this victim.
You need to help him get all the pairs of potential victims at each day so that Sherlock can observe some pattern.
Input
First line of input contains two names (length of each of them doesn't exceed 10), the two initials potential victims. Next line contains integer n (1 β€ n β€ 1000), the number of days.
Next n lines contains two names (length of each of them doesn't exceed 10), first being the person murdered on this day and the second being the one who replaced that person.
The input format is consistent, that is, a person murdered is guaranteed to be from the two potential victims at that time. Also, all the names are guaranteed to be distinct and consists of lowercase English letters.
Output
Output n + 1 lines, the i-th line should contain the two persons from which the killer selects for the i-th murder. The (n + 1)-th line should contain the two persons from which the next victim is selected. In each line, the two names can be printed in any order.
Examples
Input
ross rachel
4
ross joey
rachel phoebe
phoebe monica
monica chandler
Output
ross rachel
joey rachel
joey phoebe
joey monica
joey chandler
Input
icm codeforces
1
codeforces technex
Output
icm codeforces
icm technex
Note
In first example, the killer starts with ross and rachel.
* After day 1, ross is killed and joey appears.
* After day 2, rachel is killed and phoebe appears.
* After day 3, phoebe is killed and monica appears.
* After day 4, monica is killed and chandler appears.
Submitted Solution:
```
p=list(input().split(" "))
n=int(input().strip())
final=[]
for i in range(n):
p1=list(input().split())
final.append(p1)
print(p)
for i in final:
p.remove(i[0])
p.append(i[1])
print(p)
```
No
| 86,192 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
t, m = map(int, input().split())
a, b = [0] * m, 1
def alloc(n):
global b
f = 0
for i in range(m):
if not a[i]:
f += 1
if f == n:
a[i - n + 1:i + 1] = [b] * n
b += 1
return b - 1
else:
f = 0
return 'NULL'
def erase(x):
f = 0
for i in range(m):
if x and a[i] == x:
a[i] = 0
f += 1
if f == 0:
print('ILLEGAL_ERASE_ARGUMENT')
def defragment():
f = 0
for i in range(m):
if f and a[i]:
a[i - f], a[i] = a[i], 0
elif not a[i]:
f += 1
for i in range(t):
c = input().split()
if c[0] == 'alloc':
print(alloc(int(c[1])))
elif c[0] == 'erase':
erase(int(c[1]))
else:
defragment()
```
| 86,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
from bisect import *
t, m = (int(x) for x in input().split())
class Block:
def __init__(self, size, free, prv, nxt):
self.size = size
self.free = free
self.prv = prv
self.nxt = nxt
block0 = Block(0, m, None, None)
blocks = {}
next_id = 1
for i in range(t):
c = input().split()
if c[0] == "alloc":
size = int(c[1])
block = block0
while block and block.free < size:
block = block.nxt
if not block:
print("NULL")
continue
new = Block(size, block.free - size, block, block.nxt)
block.free = 0
new.prv.nxt = new
if new.nxt:
new.nxt.prv = new
blocks[next_id] = new
print(next_id)
next_id += 1
elif c[0] == "erase":
id = int(c[1])
if id not in blocks:
print("ILLEGAL_ERASE_ARGUMENT")
continue
block = blocks[id]
block.prv.free += block.size + block.free
if block.prv:
block.prv.nxt = block.nxt
if block.nxt:
block.nxt.prv = block.prv
del blocks[id]
elif c[0] == "defragment":
defrag = 0
block = block0
while True:
defrag += block.free
block.free = 0
if not block.nxt:
break
block = block.nxt
block.free = defrag
```
| 86,194 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
#!/usr/bin/env python
def main():
t, m = map(int, input().split())
nil = 0, 0, 0
skip = [list(nil) for _ in range(m + 1)]
skip[0][:] = 0, m, 0
block = dict()
free = m
nth = 0
for _ in range(t):
line = input()
cmd = line[0]
if cmd == "a":
size = int(line[6:])
if size > free:
print("NULL")
continue
base = 0
while base < m:
used, span, last = skip[base]
if not used and span >= size:
break
base += span
else:
print("NULL")
continue
nth += 1
print(nth)
free -= size
block[nth] = base
skip[base][:] = nth, size, last
merge = base + span
if span != size:
span -= size
if skip[merge][0]:
skip[base + size][:] = 0, span, size
else:
skip[base + size][:] = 0, span + skip[merge][1], size
skip[merge][:] = nil
elif cmd == "e":
base = block.pop(int(line[6:]), -1)
if base < 0:
print("ILLEGAL_ERASE_ARGUMENT")
continue
size = skip[base][1]
free += size
merge = base + size
if not skip[merge][0]:
size += skip[merge][1]
skip[merge][:] = nil
merge = base - skip[base][2]
if not base or skip[merge][0]:
skip[base][:2] = 0, size
else:
skip[merge][1] += size
skip[base][:] = nil
elif block and free:
base = move = last = 0
while move < m:
if skip[move][0]:
if move != base:
skip[base][:] = skip[move]
skip[move][:] = nil
block[skip[base][0]] = base
skip[base][2] = last
last = skip[base][1]
base += last
move += last
else:
size = skip[move][1]
skip[move][:] = nil
move += size
skip[base][:] = 0, free, last
if __name__ == '__main__':
main()
```
| 86,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
from sys import stdin, stdout
class Block:
def __init__(self, id, offset, size):
self.id = id
self.offset = offset
self.size = size
def __str__(self):
return "{0} {1} {2}".format(self.id, self.offset, self.size)
class MemoryManager:
def __init__(self, capacity):
self.capacity = capacity
self.blocks = [Block(-1, 0, 0), Block(-1, 100, 0)]
self.last_block_id = 0
def get_available_position(self, amount):
prev_block = self.blocks[0]
for i in range(1, len(self.blocks)):
curr_block = self.blocks[i]
avail_offset = prev_block.offset + prev_block.size
avail_amount = curr_block.offset - avail_offset
if avail_amount >= amount and avail_offset + amount <= capacity:
return i
prev_block = curr_block
return None
def allocate(self, amount):
avail_position = self.get_available_position(amount)
if avail_position == None:
return None
self.last_block_id += 1
prev_block = self.blocks[avail_position - 1]
self.blocks.insert(avail_position, Block(self.last_block_id, prev_block.offset + prev_block.size, amount))
return self.last_block_id
def erase(self, block_id):
for block in self.blocks:
if block.id == block_id and block_id != -1:
self.blocks.remove(block)
return True
return False
def defragment(self):
block_count = len(self.blocks)
prev_block = self.blocks[0]
for i in range(1, block_count):
curr_block = self.blocks[i]
curr_block.offset = prev_block.offset + prev_block.size
prev_block = curr_block
self.blocks[block_count - 1].offset = 100
class MemoryManagerConsole:
def __init__(self, memory_manager: MemoryManager):
self.memory_manager = memory_manager
def read(self):
input = stdin.readline().rstrip().split()
command = input[0]
if command == 'alloc':
amount = int(input[1])
block_id = self.memory_manager.allocate(amount)
# self.print_blocks()
return str(block_id or 'NULL')
elif command == 'erase':
block_id = int(input[1])
if self.memory_manager.erase(block_id) == False:
# self.print_blocks()
return 'ILLEGAL_ERASE_ARGUMENT'
# self.print_blocks()
elif command == 'defragment':
self.memory_manager.defragment()
# self.print_blocks()
return None
def print_blocks(self):
for block in self.memory_manager.blocks:
stdout.write(str(block) + "\n")
command_count, capacity = map(int, stdin.readline().rstrip().split())
memory_manager = MemoryManager(capacity)
manager_console = MemoryManagerConsole(memory_manager)
output_batch = list()
for i in range(command_count):
result = manager_console.read()
if result != None:
output_batch.append(result)
for output in output_batch:
stdout.write(output + "\n")
```
| 86,196 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
def d():
global p
j = 1
for i in range(len(p)):
p[i][0] = j
j += p[i][1]
def e(k):
global p
for i in range(len(p)):
if p[i][2] != k: continue
p.pop(i)
return
print('ILLEGAL_ERASE_ARGUMENT')
def a(k):
global p, s
if not p:
if k > m: return 'NULL'
else:
s += 1
p = [[1, k, s]]
return s
d = 1
for i in range(len(p)):
if p[i][0] - d < k: d = p[i][0] + p[i][1]
else:
s += 1
p.insert(i, [d, k, s])
return s
if m + 1 - d < k: return 'NULL'
s += 1
p.append([d, k, s])
return s
n, m = map(int, input().split())
s, p = 0, []
for i in range(n):
t = input()
if t[0] == 'd': d()
else:
k = int(t[t.rfind(' ') + 1: ])
if t[0] == 'e': e(k)
else: print(a(k))
```
| 86,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
def main():
numberOfCommands, totalMemory = map(int, input().split())
#I will mark erase operation with a negative number, in order to differenceate it from alloc
#also the defragment will be marked with 0, as we don't have any alloc or erase with 0
operations = []
memory = [0]*totalMemory
for _ in range(numberOfCommands):
inputOperation = input().split()
if inputOperation[0] == 'alloc':
operations.append(int(inputOperation[1]))
elif inputOperation[0] == 'erase':
if int(inputOperation[1]) <= 0:
operations.append(-100000000000000000)
else:
operations.append(-int(inputOperation[1]))
else:
operations.append(0)
allocNumber = 1
for operation in operations:
if operation > 0:
i = 0
consecutiveEmpty = 0
while i < len(memory):
if memory[i] == 0:
consecutiveEmpty += 1
if consecutiveEmpty == operation:
break
else:
consecutiveEmpty = 0
i += 1
if consecutiveEmpty == operation:
while consecutiveEmpty - 1 >= 0:
memory[i - consecutiveEmpty + 1] = allocNumber
consecutiveEmpty -= 1
print(allocNumber)
allocNumber += 1
else:
print('NULL')
elif operation < 0:
if abs(operation) in memory:
memory = list(map(lambda bit: bit if bit != abs(operation) else 0, memory))
else:
print('ILLEGAL_ERASE_ARGUMENT')
elif operation == 0:
while 0 in memory:
memory.remove(0)
while len(memory) < totalMemory:
memory.append(0)
main()
```
| 86,198 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There is little time left before the release of the first national operating system BerlOS. Some of its components are not finished yet β the memory manager is among them. According to the developers' plan, in the first release the memory manager will be very simple and rectilinear. It will support three operations:
* alloc n β to allocate n bytes of the memory and return the allocated block's identifier x;
* erase x β to erase the block with the identifier x;
* defragment β to defragment the free memory, bringing all the blocks as close to the beginning of the memory as possible and preserving their respective order;
The memory model in this case is very simple. It is a sequence of m bytes, numbered for convenience from the first to the m-th.
The first operation alloc n takes as the only parameter the size of the memory block that is to be allocated. While processing this operation, a free block of n successive bytes is being allocated in the memory. If the amount of such blocks is more than one, the block closest to the beginning of the memory (i.e. to the first byte) is prefered. All these bytes are marked as not free, and the memory manager returns a 32-bit integer numerical token that is the identifier of this block. If it is impossible to allocate a free block of this size, the function returns NULL.
The second operation erase x takes as its parameter the identifier of some block. This operation frees the system memory, marking the bytes of this block as free for further use. In the case when this identifier does not point to the previously allocated block, which has not been erased yet, the function returns ILLEGAL_ERASE_ARGUMENT.
The last operation defragment does not have any arguments and simply brings the occupied memory sections closer to the beginning of the memory without changing their respective order.
In the current implementation you are to use successive integers, starting with 1, as identifiers. Each successful alloc operation procession should return following number. Unsuccessful alloc operations do not affect numeration.
You are to write the implementation of the memory manager. You should output the returned value for each alloc command. You should also output ILLEGAL_ERASE_ARGUMENT for all the failed erase commands.
Input
The first line of the input data contains two positive integers t and m (1 β€ t β€ 100;1 β€ m β€ 100), where t β the amount of operations given to the memory manager for processing, and m β the available memory size in bytes. Then there follow t lines where the operations themselves are given. The first operation is alloc n (1 β€ n β€ 100), where n is an integer. The second one is erase x, where x is an arbitrary 32-bit integer numerical token. The third operation is defragment.
Output
Output the sequence of lines. Each line should contain either the result of alloc operation procession , or ILLEGAL_ERASE_ARGUMENT as a result of failed erase operation procession. Output lines should go in the same order in which the operations are processed. Successful procession of alloc operation should return integers, starting with 1, as the identifiers of the allocated blocks.
Examples
Input
6 10
alloc 5
alloc 3
erase 1
alloc 6
defragment
alloc 6
Output
1
2
NULL
3
Tags: implementation
Correct Solution:
```
"""
Codeforces
7B - Memory Manager
http://codeforces.com/contest/7/problem/B
HΓ©ctor GonzΓ‘lez Belver
../07/2018
"""
import sys
def main():
t, m = map(int, sys.stdin.readline().strip().split())
memory = [0] * m
block_number = 0
def allocate(num_bytes):
nonlocal block_number
hi = -1
allocated = False
while not allocated:
try:
lo = memory.index(0, hi+1)
except ValueError:
break
hi = lo + num_bytes - 1
if hi >= m:
break
allocated = True
for i, mem in enumerate(memory[lo:hi+1]):
if mem:
hi = lo + i
allocated = False
break
if not allocated:
sys.stdout.write('NULL' + '\n')
else:
block_number += 1
for i in range(num_bytes):
memory[lo+i] = block_number
sys.stdout.write(str(block_number) + '\n')
def erase(block):
try:
if block <= 0:
raise ValueError
idx = memory.index(block)
except ValueError:
sys.stdout.write('ILLEGAL_ERASE_ARGUMENT' + '\n')
else:
while idx < m and memory[idx] == block:
memory[idx] = 0
idx += 1
def defragmentate():
nonlocal memory
memory = [byte for byte in memory if byte]
memory += [0] * (m - len(memory))
for i in range(t):
operation = sys.stdin.readline().strip().split()
if operation[0] == 'alloc':
allocate(int(operation[1]))
elif operation[0] == 'erase':
erase(int(operation[1]))
elif operation[0] == 'defragment':
defragmentate()
if __name__ == '__main__':
main()
```
| 86,199 |
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