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.
Oh, New Year. The time to gather all your friends and reflect on the heartwarming events of the past year...
n friends live in a city which can be represented as a number line. The i-th friend lives in a house with an integer coordinate x_i. The i-th friend can come celebrate the New Year to the house with coordinate x_i-1, x_i+1 or stay at x_i. Each friend is allowed to move no more than once.
For all friends 1 ≤ x_i ≤ n holds, however, they can come to houses with coordinates 0 and n+1 (if their houses are at 1 or n, respectively).
For example, let the initial positions be x = [1, 2, 4, 4]. The final ones then can be [1, 3, 3, 4], [0, 2, 3, 3], [2, 2, 5, 5], [2, 1, 3, 5] and so on. The number of occupied houses is the number of distinct positions among the final ones.
So all friends choose the moves they want to perform. After that the number of occupied houses is calculated. What is the minimum and the maximum number of occupied houses can there be?
Input
The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of friends.
The second line contains n integers x_1, x_2, ..., x_n (1 ≤ x_i ≤ n) — the coordinates of the houses of the friends.
Output
Print two integers — the minimum and the maximum possible number of occupied houses after all moves are performed.
Examples
Input
4
1 2 4 4
Output
2 4
Input
9
1 1 8 8 8 4 4 4 4
Output
3 8
Input
7
4 3 7 1 4 3 3
Output
3 6
Note
In the first example friends can go to [2, 2, 3, 3]. So friend 1 goes to x_1+1, friend 2 stays at his house x_2, friend 3 goes to x_3-1 and friend 4 goes to x_4-1. [1, 1, 3, 3], [2, 2, 3, 3] or [2, 2, 4, 4] are also all valid options to obtain 2 occupied houses.
For the maximum number of occupied houses friends can go to [1, 2, 3, 4] or to [0, 2, 4, 5], for example.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import sys
from collections import Counter
def input(): return sys.stdin.readline().strip()
def list2d(a, b, c): return [[c] * b for i in range(a)]
def list3d(a, b, c, d): return [[[d] * c for j in range(b)] for i in range(a)]
def list4d(a, b, c, d, e): return [[[[e] * d for j in range(c)] for j in range(b)] for i in range(a)]
def ceil(x, y=1): return int(-(-x // y))
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(N=None): return list(MAP()) if N is None else [INT() for i in range(N)]
def Yes(): print('Yes')
def No(): print('No')
def YES(): print('YES')
def NO(): print('NO')
INF = 10 ** 18
MOD = 10 ** 9 + 7
N = INT()
A = LIST()
C = Counter(A)
def check1(C):
C2 = [0] * (N+2)
i = 0
while i < N+2:
if C[i] > 0 and (C[i+1] > 0 or C[i+2] > 0):
C2[i+1] = C[i] + C[i+1] + C[i+2]
C[i] = C[i+1] = C[i+2] = 0
i += 3
else:
C2[i] = C[i]
C[i] = 0
i += 1
return C2
def check2(C):
C2 = [0] * (N+2)
i = 0
while i < N:
if C[i+1] > 0 and C2[i] == 0:
C2[i] += 1
C[i+1] -= 1
if C[i+1] >= 2:
C2[i+2] += 1
C[i+1] -= 1
C2[i+1] += C[i+1]
C[i+1] = 0
i += 1
return C2
res1 = check1(C.copy())
res2 = check2(C.copy())
ans1 = N + 2 - res1.count(0)
ans2 = N + 2 - res2.count(0)
print(ans1, ans2)
```
No
| 97,100 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
def erase(s):
c =0
if len(s) == 1:
return 0
ns = ""
if "1" in s:
ns = s[s.index("1")]
for i in range(s.index("1"),len(s)-1):
if ns[-1] == "1" and s[i] == "0" and "1" in s[i:]:
ns+="1"
c+=1
return c
tc = int(input())
for i in range(tc):
s = input()
print(erase(s))
```
| 97,101 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
l = []
c=0
for i in range(len(s)):
if s[i]=='1':
c+=1
l.append(i)
if len(l)>1:
d = l[-1]-l[0]+1
print(d-c)
else:
print('0')
```
| 97,102 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
t=int(input())
while t>0:
t-=1
l=input()
d=dict()
d['0']=[]
d['1']=[]
for i in range(len(l)):
d[l[i]].append(i)
if len(d['0'])==0 or len(d['1'])==0:
print(0)
else:
n=len(d['0'])-(d['1'])[0]
n-=len(l)-1-(d['1'])[len(d['1'])-1]
print(n)
```
| 97,103 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
t = int(input())
for _ in range(t):
s = input()
i = 0
while s[i] == '0' and i != len(s) - 1:
i += 1
j = -1
while s[j] == '0' and -j != len(s):
j -= 1
print(s[i:j].count('0'))
```
| 97,104 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
q = int(input())
for _ in range(q):
s = input()
t = s[::-1]
try:
id1 = s.index('1')
id2 = len(s) - t.index('1')
a = s[id1:id2]
b = a.replace('0','')
print(len(a)-len(b))
except:
print(0)
```
| 97,105 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
for _ in [0]*int(input()):
First1 = False
One = False
ans,scet0 = 0,0
for i in input():
if i =="1":
First1 = True
One ==True
ans += scet0
scet0 = 0
else:
if First1:
scet0+=1
print(ans)
```
| 97,106 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
for _ in range(int(input())):
s = input()
i = 0
l = len(s)
while i<l:
if s[i]=='1':
break
i+=1
j = l-1
while j>-1:
if s[j]=='1':
break
j-=1
s = s[i:j+1]
print(s.count('0'))
```
| 97,107 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Tags: implementation, strings
Correct Solution:
```
#A erasing zeroes
for i in range(int(input())):
arr = input()
a = []
sum = 0
for j in range(0,len(arr)):
if(arr[j]=='1'):
a.append(j)
for j in range(len(a)-1):
sum = sum + (a[j+1]-a[j]-1)
print(sum)
```
| 97,108 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
t = int(input())
for _ in range(t):
s = str(int(input()))
cnt = 0
hasOne = False
for i in reversed(s):
if i=="1":
hasOne = True
if i=="0" and hasOne:
cnt+=1
print(cnt)
```
Yes
| 97,109 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
n=int(input())
for i in range(n):
t=str(input())
h=list(t)
for j in range(len(h)):
if h[j]=='0':
h[j]='k'
else:
break
r=len(h)
for y in range(r-1,-1,-1):
if h[y]=='0':
h[y]='k'
else:
break
print(h.count('0'))
```
Yes
| 97,110 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
t = int(input())
for i in range(t):
s = input()
enter = False
endind = -1
counter = 0
for l in range(1, len(s) + 1):
if s[len(s) - l] == '1':
endind = len(s) - l
break
if endind != -1:
s1 = s[s.index('1'):endind]
for k in s1:
if k == '0':
counter += 1
print(counter)
```
Yes
| 97,111 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
# ------------------- fast io --------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------- fast io --------------------
for _ in range(int(input())):
s = input()
ans = 0
for i in range(len(s)):
if s[i] == "1":
for j in range(i+1, s.rfind("1")):
if s[j] == "0": ans += 1
break
print(ans)
```
Yes
| 97,112 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
T = int(input())
for i in range(1, T+1):
s = list(input())
if '1' not in s:
print(0)
else:
first = s.index('1')
for j in range(len(s)-1, -1, -1):
if s[j] == '1':
last = j
count = 0
for k in range(first, last):
if s[k] == '0':
count += 1
print(count)
```
No
| 97,113 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
for _ in range(int(input())):
s=input()
if '1' not in s:
print(0);continue
t=s.index('1');m=10000000000
for i in range(t+1,len(s)):
if s[i]=='1':
if m>=i-t and i-t>1:
m=i-t
t=i
print(m-1) if m<10000000000 else print(0)
```
No
| 97,114 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
t=int(input())
while t!=0:
s=input()
li=list(s)
if '1' not in li:
print(0)
else:
count=0
index=li.index('1')
for i in range(index,len(li)-1,1):
if li[i+1]=='0':
count+=1
elif li[i+1]=='1':
break
print(count)
t=t-1
```
No
| 97,115 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. Each character is either 0 or 1.
You want all 1's in the string to form a contiguous subsegment. For example, if the string is 0, 1, 00111 or 01111100, then all 1's form a contiguous subsegment, and if the string is 0101, 100001 or 11111111111101, then this condition is not met.
You may erase some (possibly none) 0's from the string. What is the minimum number of 0's that you have to erase?
Input
The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.
Then t lines follow, each representing a test case. Each line contains one string s (1 ≤ |s| ≤ 100); each character of s is either 0 or 1.
Output
Print t integers, where the i-th integer is the answer to the i-th testcase (the minimum number of 0's that you have to erase from s).
Example
Input
3
010011
0
1111000
Output
2
0
0
Note
In the first test case you have to delete the third and forth symbols from string 010011 (it turns into 0111).
Submitted Solution:
```
s = []
for i in range(int(input())):
s.append(input())
def working(pattern):
start = 0
end = 0
zeros_removed = 0
while start == 0:
start_one = pattern.find('1')
end_one = pattern.rfind('1')
start_zero = pattern.find('0')
if start_one < start_zero < end_one:
pattern = pattern[:start_zero] + pattern[start_zero + 1:]
zeros_removed += 1
else:
start = 1
while end == 0:
start_one = pattern.find('1')
end_one = pattern.rfind('1')
end_zero = pattern.rfind('0')
if start_one < end_zero < end_one:
pattern = pattern[:end_zero] + pattern[end_zero + 1:]
zeros_removed += 1
else:
end = 1
return zeros_removed
for i in s:
print(working(i))
```
No
| 97,116 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
def main():
for _ in range(int(input())):
n=int(input())
visited=[False]*n
q=[list(map(int,input().split())) for i in range(n)]
t1,x=0,0
for i in range(n):
t=-1
if(len(q[i])>1):
t=min(q[i][1:],key=lambda x: n+1 if(visited[x-1]) else x)
elif(t1==0 and x==0):
t1=-1
x=i
if(t!=-1 and not visited[t-1]):
visited[t-1]=True
elif(t1==0 and x==0):
t1=-1
x=i
if(t1==-1):
print("IMPROVE")
print(x+1,visited.index(False)+1)
else:
print("OPTIMAL")
main()
```
| 97,117 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
input = __import__('sys').stdin.readline
print = __import__('sys').stdout.write
for _ in range(int(input())):
n = int(input())
prince = set(range(1, n+1))
princess = set(range(1, n+1))
princess_list = [[] for _ in range(n+1)]
for i in range(n):
k = list(map(int, input().split()))[1:]
princess_list[i+1] = k
if k:
for j in k:
if j in prince:
prince.remove(j)
princess.remove(i+1)
break
# print(str(princess) + '\n')
# print(str(prince) + '\n')
# print(str(princess_list) +'\n')
# exit()Y
tmp = False
for pr1 in princess:
for pr2 in prince:
if pr2 in princess_list[pr1]:
continue
else:
print (f'IMPROVE\n')
print (f'{pr1} {pr2}\n')
tmp = True
break
if tmp:
break
else:
print ('OPTIMAL\n')
```
| 97,118 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
# daughters = [False for _ in range(n+1)]
# boys = [False for _ in range(n+1)]
boys = [False]*(n+1)
dul = -1
boys[0] = True
chk = True
for i in range(n):
x = list(map(int, input().split()))
for j in range(1, x[0]+1):
if boys[x[j]]==False:
boys[x[j]]=True
break
else:
if chk==True:
dul = i+1
chk = False
if dul!=-1:
print("IMPROVE")
print(dul, boys.index(False))
else:
print("OPTIMAL")
# print(boys)
# lst = [list(map(int, input().split())) for _ in range(n)]
```
| 97,119 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
dc ={}
for i in range(1,n+1):
dc[i] = []
prc = [False for i in range(n+1)]
pcc = [False for i in range(n+1)]
b = []
for i in range(1,n+1):
x = [int(o) for o in input().split()]
dc[i] = x[1:]
count = 0
for i in range(1,n+1):
if len(dc[i]) is 0:
continue
else:
temp = dc[i]
for boy in temp:
if prc[boy] is False:
pcc[i] = True
prc[boy] = True
count += 1
break
if count == n:
print("OPTIMAL")
else:
flag = 0
for girl in range(1,n+1):
if pcc[girl] is False:
for boy in range(1,n+1):
if prc[boy] is False and len(dc[girl]) is 0:
flag = 1
print("IMPROVE")
print(girl,boy)
break
elif prc[boy] is False and boy not in dc[girl]:
flag = 1
print("IMPROVE")
print(girl,boy)
break
if flag is 1:
break
if flag is 0:
print("OPTIMAL")
```
| 97,120 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
t = int(input())
for _ in range(t):
n = int(input())
used = [False for i in range(n)]
v = -1
for i in range(n):
l = [int(x) - 1 for x in input().split()][1:]
for j in l:
if not used[j]:
used[j] = True
break
else:
v = i
if v == -1:
print("OPTIMAL")
else:
u = used.index(False)
print("IMPROVE")
print((v + 1) ,(u + 1))
```
| 97,121 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
def main():
n = int(input())
prince = set()
princes = set()
for i in range(1, n + 1):
prince.add(i)
to_change = -1
for i in range(n):
lst = list(map(int, input().split()))
have = False
for k in lst[1:]:
if k in prince:
have = True
prince.remove(k)
princes.add(i)
break
if not have:
to_change = i
if to_change == -1:
print("OPTIMAL")
return
print("IMPROVE")
print(to_change + 1, min(prince))
if __name__ == "__main__":
t = int(input())
for i in range(t):
main()
```
| 97,122 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
t=int(input())
for _ in range(t):
n=int(input())
l1=[False]*n
l2=[False]*n
for i in range(n):
l=list(map(int,input().split()))[1:]
for j in l:
if not l2[j-1]:
l1[i]=True
l2[j-1]=True
break
f1=0
f2=0
s=" "
for i in range(n):
if f1==0 and not l1[i]:
f1=1
s=str(i+1)+s
if f2==0 and not l2[i]:
f2=1
s=s+str(i+1)
if f1==1 and f2==1:
break
if f1==1 and f2==1:
print("IMPROVE")
print(s)
else:
print("OPTIMAL")
```
| 97,123 |
Provide tags and a correct Python 3 solution for this coding contest problem.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Tags: brute force, graphs, greedy
Correct Solution:
```
import sys
input=sys.stdin.readline
from math import *
t=int(input())
while t>0:
t-=1
n=int(input())
f=[0 for i in range(n+1)]
l=[]
flag=0
for i in range(n):
a=[int(x) for x in input().split()]
a=a[1:]
a.sort()
p=0
for j in range(len(a)):
if f[a[j]]==0:
f[a[j]]=1
p=1
break
#print(f,p)
if p==0:
l.append(i)
#print(f)
if f[1:].count(0)>=1 and len(l)>0:
print("IMPROVE")
print(l[0]+1,f[1:].index(0)+1)
else:
print("OPTIMAL")
```
| 97,124 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
import bisect
import os
import io
from collections import Counter
from collections import defaultdict
import math
import random
import heapq as hq
from math import sqrt
import sys
from functools import reduce
from collections import deque
import threading
# sys.setrecursionlimit(2000000)
# input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
def input():
return sys.stdin.readline().strip()
def iinput():
return int(input())
def tinput():
return input().split()
def rinput():
return map(int, tinput())
def rlinput():
return list(rinput())
mod = int(1e9)+7
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
# ----------------------------------------------------
# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')
for _ in range(iinput()):
n = iinput()
val = []
for i in range(n):
a = rlinput()
a.pop(0)
val.append(a)
ans1 = -1
visited = [False]*(n+1)
for i in range(n):
flag = False
for j in val[i]:
if not visited[j]:
visited[j] = True
flag = True
break
if not flag:
ans1 = i+1
if ans1 == -1:
print('OPTIMAL')
else:
print('IMPROVE')
ans2 = -1
for i in range(1, n+1):
if not visited[i]:
ans2 = i
break
print(ans1, ans2)
```
Yes
| 97,125 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
t=int(input())
for i in range(t):
n=int(input())
arr = []
for j in range(n):
a1= input().split(" ")
arr.append([])
for c in range(len(a1)):
arr[j].append(int(a1[c]))
done = set()
remain = []
count=0
for k in range(len(arr)):
p = arr[k]
flag=0
for q in range(1,len(p)):
if(p[q] not in done):
done.add(p[q])
flag=1
count+=1
break
if(flag==0):
p.append(k)
remain.append(p)
if(count==n):
print("OPTIMAL")
continue
prince = list(done)
prince.sort()
num = -1
for w in range(len(prince)-1):
if(prince[w]!=prince[w+1]-1):
num = prince[w]+1
break
if(num==-1):
if(len(prince)>0 and prince[0]>=2):
num = 1
else:
num = len(prince)+1
add = -1
for z in range(len(remain)):
element = remain[z]
if(num not in element[1:len(element)-1]):
add = element[len(element)-1]
break
print("IMPROVE")
print(str(add+1),num)
```
Yes
| 97,126 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
t = int(input())
for j in range(t):
n = int(input())
taken_prince = [False for i in range(n)]
taken_princess = [False for i in range(n)]
for x in range(n):
wanted_prince_list = list(map(int, input().split()))
for i in wanted_prince_list[1:]:
if taken_prince[i-1] == False:
taken_prince[i-1] = True
taken_princess[x] = True
break
toprint = "OPTIMAL"
for i in range(n):
if taken_princess[i]==False:
u = taken_prince.index(False)
toprint = 'IMPROVE\n' + str(i+1) + ' ' + str(u+1)
break
print(toprint)
```
Yes
| 97,127 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
for _ in range(int(input())):
n=int(input())
dic={};m={};boo=True;ans=0
for i in range(n):
a=[int(j) for j in input().split()]
boo=True
for j in a[1:]:
if(j not in m.keys()):
m[j]=i+1
boo=False
break
if(boo):
ans=i+1
ans2=list(set(x for x in range(1,n+1))-set(m.keys()))
if(len(ans2)==0):
print('OPTIMAL')
else:
print('IMPROVE')
print(ans,ans2[0])
```
Yes
| 97,128 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
for _ in range(int(input())):
dlst = []
n = int(input())
prince = set(range(1, 1+n))
notf = None
for i in range(1, n+1):
dlst.append([int(x) for x in input().split()])
for val in dlst[-1]:
if val in prince:
prince.remove(val)
break
else:
notf = i
if len(prince) == 0:
ans = "OPTIMAL"
else:
ans = "IMPROVE\n"
ans += str(notf) + ' ' + str(prince.pop())
print(ans)
```
No
| 97,129 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
import sys
for _ in range(int(sys.stdin.readline())):
a=[]
b=int(sys.stdin.readline())
d=0
for i in range(b):
c=sys.stdin.readline()
if len(c)==1:
a.append(list())
else:
a.append(list(map(int,c.split())))
c=[i for i in range(1,b+1)]
for i in range(b):
for k in range(1,a[i][0]+1):
if a[i][k] in c:
c.remove(a[i][k])
break
elif k == a[i][0]:
d=i
if len(c)==0:
sys.stdout.write('OPTIMAL\n')
else:
sys.stdout.write('IMPROVE\n')
sys.stdout.write('{} {}\n'.format(d+1,c[0]))
```
No
| 97,130 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
t = int(input())
for i in range(t):
n = int(input())
pr = [0 for j in range(n+1)]
dr = [0 for j in range(n+1)]
if_good = 0
for j in range(1, n+1):
if_good = 0
flag = 0
dr[j] = list(map(int, input().split()))
if dr[j][0] == 0:
print(dr)
print('IMPROVE')
for p in range(1, n+1):
if pr[p]==0:
print(j, p)
flag = 1
break
if flag == 1:
break
else:
for p in dr[j][1:]:
if pr[p]==0:
pr[p]=1
flag = 1
if_good = 1
break
if (flag == 0):
for p in range(1, n+1):
if pr[p]==0:
print('IMPROVE')
print(j, p)
flag = 1
break
if if_good== 1:
print('OPTIMAL')
```
No
| 97,131 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
The King of Berland Polycarp LXXXIV has n daughters. To establish his power to the neighbouring kingdoms he wants to marry his daughters to the princes of these kingdoms. As a lucky coincidence there are n other kingdoms as well.
So Polycarp LXXXIV has enumerated his daughters from 1 to n and the kingdoms from 1 to n. For each daughter he has compiled a list of kingdoms princes of which she wanted to marry.
Polycarp LXXXIV is very busy, so he finds a couple for his daughters greedily one after another.
For the first daughter he takes the kingdom with the lowest number from her list and marries the daughter to their prince. For the second daughter he takes the kingdom with the lowest number from her list, prince of which hasn't been taken already. If there are no free princes in the list then the daughter marries nobody and Polycarp LXXXIV proceeds to the next daughter. The process ends after the n-th daughter.
For example, let there be 4 daughters and kingdoms, the lists daughters have are [2, 3], [1, 2], [3, 4], [3], respectively.
<image>
In that case daughter 1 marries the prince of kingdom 2, daughter 2 marries the prince of kingdom 1, daughter 3 marries the prince of kingdom 3, leaving daughter 4 nobody to marry to.
Actually, before starting the marriage process Polycarp LXXXIV has the time to convince one of his daughters that some prince is also worth marrying to. Effectively, that means that he can add exactly one kingdom to exactly one of his daughter's list. Note that this kingdom should not be present in the daughter's list.
Polycarp LXXXIV wants to increase the number of married couples.
Unfortunately, what he doesn't have the time for is determining what entry to add. If there is no way to increase the total number of married couples then output that the marriages are already optimal. Otherwise, find such an entry that the total number of married couples increases if Polycarp LXXXIV adds it.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
For your and our convenience you are asked to answer t independent test cases.
Input
The first line contains a single integer t (1 ≤ t ≤ 10^5) — the number of test cases.
Then t test cases follow.
The first line of each test case contains a single integer n (1 ≤ n ≤ 10^5) — the number of daughters and the number of kingdoms.
Each of the next n lines contains the description of each daughter's list. The first integer k (0 ≤ k ≤ n) is the number of entries in the i-th daughter's list. After that k distinct integers follow g_i[1], g_i[2], ..., g_i[k] (1 ≤ g_i[j] ≤ n) — the indices of the kingdoms in the list in the increasing order (g_i[1] < g_i[2] < ... < g_i[k]).
It's guaranteed that the total number of daughters over all test cases does not exceed 10^5.
It's also guaranteed that the total number of kingdoms in lists over all test cases does not exceed 10^5.
Output
For each test case print the answer to it.
Print "IMPROVE" in the first line if Polycarp LXXXIV can add some kingdom to some of his daughter's list so that the total number of married couples increases. The second line then should contain two integers — the index of the daughter and the index of the kingdom Polycarp LXXXIV should add to that daughter's list.
If there are multiple ways to add an entry so that the total number of married couples increases then print any of them.
Otherwise the only line should contain one word "OPTIMAL".
Example
Input
5
4
2 2 3
2 1 2
2 3 4
1 3
2
0
0
3
3 1 2 3
3 1 2 3
3 1 2 3
1
1 1
4
1 1
1 2
1 3
1 4
Output
IMPROVE
4 4
IMPROVE
1 1
OPTIMAL
OPTIMAL
OPTIMAL
Note
The first test case is depicted in the statement. Adding the fourth kingdom to the list of the fourth daughter makes her marry the prince of the fourth kingdom.
In the second test case any new entry will increase the number of marriages from 0 to 1.
In the third and the fourth test cases there is no way to add an entry.
In the fifth test case there is no way to change the marriages by adding any entry.
Submitted Solution:
```
t=int(input())
while t>0:
t-=1
n=int(input())
li=[]
for i in range(n):
x=map(int,input().split())
li.append(x)
a=[]
b=[]
for i in range(n):
for j in li[i]:
if j not in a and j!=0:
a.append(j)
b.append(i+1)
break
#print(a)
#print(b)
z=0
for i in range(1,n+1):
if i not in b:
z=i
break
for i in range(1,n+1):
if i not in a:
print('IMPROVE')
print(z,i)
break
elif i==n and i in a:
print('OPTIMAL')
```
No
| 97,132 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
k=int(input())
out=''
t=list('codeforces')
d=[1]*10
m=1
i=0
while m<k:
#print(i,m)
d[i]+=1
i=(i+1)%10
m=1
for j in range(10):
m*=d[j]
out=''
for i in range(10):
out+=t[i]*d[i]
print(out)
```
| 97,133 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
#!/usr/bin/env python3
import io
import os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
def get_str():
return input().decode().strip()
def rint():
return map(int, input().split())
def oint():
return int(input())
n = oint()
s = 'codeforces'
l = len(s)
r = 1
while r**10 < n:
r += 1
if r**10 == n:
print(''.join([s[i]*r for i in range(l)]))
exit()
r -= 1
mul = r**l
cnt = 0
while True:
cnt += 1
mul = mul*(r+1)//r
if mul >= n:
break
ans = []
ans += [s[i]*(r+1) for i in range(cnt)]
ans += [s[i]*r for i in range(cnt, l)]
print(''.join(ans))
```
| 97,134 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
import math
n=int(input())
p=int(math.floor(math.pow(n,0.1)))
a=0
x=p**10
while n>x:
a+=1
x=((p+1)**a)*(p**(10-a))
s='codeforces'
ans=''
for i in range(10):
if i<a:
ans+=s[i]*(p+1)
else:
ans+=s[i]*p
print(ans)
```
| 97,135 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
from math import log, floor, ceil
k = int(input())
base = 1
while base ** 10 < k:
base += 1
counts_base = 0
count_1 = 10
while (base ** counts_base) * ((base-1) ** count_1) < k:
counts_base += 1
count_1 -= 1
# print("counts_base", counts_base)
# print("count_1", count_1)
codeforces = "codeforces"
ans = ""
for i in range(counts_base):
ans += codeforces[i] * base
for i in range(count_1):
ans += codeforces[counts_base+i] * (base-1)
print(ans)
```
| 97,136 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
n = int(input())
if n == 1:
print("codeforces")
exit()
num = 1
while num ** 10 < n:
num += 1
s = list("codeforces")
temp = 10
for i in reversed(range(11)):
if ((num ** i) * ((num - 1) ** (10 - i))) < n:
temp = i + 1
break
ans = ""
for i in range(10):
ans += s[i] * num
if i + 1 == temp:
num -= 1
print(ans)
```
| 97,137 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
def tc():
k = int(input())
base = 1
while pow(base, 10) < k:
base += 1
less = 0
while pow(base, 10 - less) * pow(base - 1, less) > k:
less += 1
if pow(base, 10 - less) * pow(base - 1, less) < k:
less -= 1
ans = ''
s = 'codeforces'
for ch in s[:less]:
ans += ch * (base - 1)
for ch in s[less:]:
ans += ch * base
print(ans)
################################
# T = int(input())
# for _ in range(T):
# tc()
tc()
```
| 97,138 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
def prod(arr):
ans=1
for i in arr:
ans *= i
return ans
def find_opt(arr,n):
if(prod(arr)>=n):
return arr
m = min(arr)
min_idx = arr.index(m)
arr[min_idx]+=1
return find_opt(arr,n)
n = int(input())
arr = find_opt([1]*10,n)
ans = ""
brr =['c','o','d','e','f','o','r','c','e','s']
for i in range(10):
ans += brr[i]*arr[i]
print(ans)
```
| 97,139 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Tags: brute force, constructive algorithms, greedy, math, strings
Correct Solution:
```
a=int(input())
l=[0,0,0,0,0,0,0,0,0,0]
id=0
g=1
while(1):
g=1
if id==10:
id=0
g=1
for j in range(10):
g=g*l[j]
if g>=a:
break
l[id]=l[id]+1
id=id+1
for i in range(l[0]):
print('c',end='')
for i in range(l[1]):
print('o',end='')
for i in range(l[2]):
print('d',end='')
for i in range(l[3]):
print('e',end='')
for i in range(l[4]):
print('f',end='')
for i in range(l[5]):
print('o',end='')
for i in range(l[6]):
print('r',end='')
for i in range(l[7]):
print('c',end='')
for i in range(l[8]):
print('e',end='')
for i in range(l[9]):
print('s',end='')
```
| 97,140 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
n=int(input())
l=[1]*10
p=1
j=0
while(p<n):
k=l[j]
l[j]+=1
p=(p//k)*l[j]
j=(j+1)%10
ss='codeforces'
m = [ ss[i]*l[i] for i in range(10)]
print(''.join(m))
```
Yes
| 97,141 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
k = int(input())
def product(s):
res = 1
for e in s:
res *= e
return res
string = "codeforces"
nums = [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
index = 0
while True:
p = product(nums)
if p < k:
nums[index] += 1
index = (index + 1) % len(nums)
else:
break
result = ""
for i, e in enumerate(nums):
result += e * string[i]
print(result)
```
Yes
| 97,142 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
k = int(input())
def getperm(occ):
val = 1
for v in occ:
val *= v
return val
letters = ['c', 'o', 'd', 'e', 'f', 'o', 'r', 'c', 'e', 's']
occurences = [1 for x in range(len(letters))]
permutations = 1
location = 0
while getperm(occurences) < k:
occurences[location] += 1
location += 1
location %= len(occurences)
for i, v in enumerate(occurences):
print(letters[i]*v, end='')
print()
```
Yes
| 97,143 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
# -*- coding: utf-8 -*-
"""
Created on Wed Jun 24 09:24:46 2020
@author: Dark Soul
"""
def mult(x):
res=1
for i in range(len(x)):
x[i]=int(x[i])
res=res*x[i]
return res
import math
n=input('')
n=int(n)
k=0
sign=0
cnt=[1,1,1,1,1,1,1,1,1,1]
str=['c','o','d','e','f','o','r','c','e','s']
str_final=''
i=0
j=0
while mult(cnt)<=n:
if mult(cnt)==n:
break;
cnt[i]=cnt[i]+1
i=i+1
if i>9:
i=0
for i in range(10):
str_final=str_final+str[i]*cnt[i]
print(str_final)
```
Yes
| 97,144 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
import math
k = int(input())
new = int(math.ceil(math.log2(k)) / 10)
# print(new)
new2 = math.ceil(math.log2(k)) % 10
# print(new2)
orig = list('codeforces')
ans = ''
if new > 0:
for i in range(10):
ans += orig[i] * ((1 + new + (i < new2)) * new)
else:
for i in range(10):
ans += orig[i] * (1 + new + (i < new2))
print(ans)
```
No
| 97,145 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
from math import inf as inf
from math import *
from collections import *
import sys
from itertools import permutations
input=sys.stdin.readline
t=1
while(t):
t-=1
n=int(input())
if(n==1):
print("codeforces")
continue
l=[[1,'c'],[1,'o'],[1,'d'],[1,'e'],[1,'f'],[1,'o'],[1,'r'],[1,'c'],[1,'e'],[1,'s']]
p=1
c1=0
while(p<n):
p=p*2
c1+=1
for i in range(c1):
l[i%10][0]*=2
for i in l:
print(i[1]*i[0],end="")
print()
```
No
| 97,146 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
import math
def solve():
t = int(input())
current = 0
letters = ['c', 'o', 'd', 'e', 'f', 'o', 'r', 'c', 'e', 's']
arr = [1] * len(letters)
product = 1
power = 2
while product < t:
arr[current] += 1
product *= power
current += 1
if current == len(letters):
current = 0
for i in range(len(letters)):
print(letters[i] * arr[i], end="")
print()
solve()
```
No
| 97,147 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Karl likes Codeforces and subsequences. He wants to find a string of lowercase English letters that contains at least k subsequences codeforces. Out of all possible strings, Karl wants to find a shortest one.
Formally, a codeforces subsequence of a string s is a subset of ten characters of s that read codeforces from left to right. For example, codeforces contains codeforces a single time, while codeforcesisawesome contains codeforces four times: codeforcesisawesome, codeforcesisawesome, codeforcesisawesome, codeforcesisawesome.
Help Karl find any shortest string that contains at least k codeforces subsequences.
Input
The only line contains a single integer k (1 ≤ k ≤ 10^{16}).
Output
Print a shortest string of lowercase English letters that contains at least k codeforces subsequences. If there are several such strings, print any of them.
Examples
Input
1
Output
codeforces
Input
3
Output
codeforcesss
Submitted Solution:
```
k = int(input())
primes = []
a = 2
while k >= a*a:
while k % a == 0:
primes.append(a)
k //= a
a += 1
if k != 1:
primes.append(k)
primes = list(reversed(sorted(primes)))
K = [1] * 10
for p in primes:
idx = K.index(min(K))
K[idx] *= p
s = [c*K[i] for i, c in enumerate('codeforces')]
print(''.join(s))
```
No
| 97,148 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
l1,r1=map(int,input().split())
l2,r2=map(int,input().split())
if l1>l2:
l1,r1,l2,r2=l2,r2,l1,r1
if l2>r1:
ans=float("inf")
for i in range(1,n+1):
temp=i*(l2-r1)
limit=(r2-l1)*i
if k<=limit:
temp+=k
ans=min(ans,temp)
else:
temp+=limit
K=k-limit
temp+=2*K
ans=min(ans,temp)
print(ans)
else:
count=n*(min(r1,r2)-l2)
res=0
if count>k:
print(res)
continue
b=max(r2,r1)-min(l2,l1)-(min(r1,r2)-l2)
for i in range(n):
if k-count<=b:
res+=k-count
break
else:
res+=b
count+=b
else:
res+=(k-count)*2
print(res)
```
| 97,149 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
t = int(input())
for _ in range(t):
n,k = list(map(int,input().split()))
l1,r1 = list(map(int,input().split()))
l2,r2 = list(map(int,input().split()))
#make it same
if r1>r2:
l1, r1, l2, r2 = l2, r2, l1, r1
#reduce k by overlap
overlap = max(0, r1-max(l1,l2))
k -= overlap * n
#check border condition
if k<=0:
print(0)
continue
touch_cost = max(0, max(l1,l2)-min(r1,r2))
overlap_cost = max(r2,r1) - min(l1,l2) - overlap
overlap_gain = max(r1,r2) - min(l1,l2)
#print(touch_cost, overlap_cost, overlap_gain)
#make one touch and double extend
ans = 2*k + touch_cost
#print(ans)
for i in range(1, n+1):
poss = touch_cost * i
if i * overlap_cost>=k:
poss += k
else:
poss += 2*k - overlap_cost*i
#print(i, poss)
ans = min(poss, ans)
print(ans)
```
| 97,150 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for _ in range(int(input())):
n,k=map(int,input().split())
l1,r1=map(int,input().split())
l2,r2=map(int,input().split())
over=min(r1,r2)-max(l1,l2)
moves=0
#Case1: Already overlapped
if (l1,r1)==(l2,r2):
if (r1-l1)*n>=k:
print(0)
else:
k-=(r1-l1)*n
print(2*k)
elif over>0:
over*=n
if over>=k:
print(0)
else:
o=max(r1,r2)-min(l1,l2)-over//n
if over+o*n>=k:
print(k-over)
else:
moves+=o*n
over+=o*n
print(moves+(k-over)*2)
#Case 2: No overlap(Here the fun begins)
else:
d=-1*over
o=max(r1,r2)-min(l1,l2)
x=k//o
k=k%o
if n>x:
moves+=(x)*(d+o)
if d<k or x==0:
moves+=d+k
else:
moves+=2*k
else:
moves+=n*(d+o)+((x-n)*o+k)*2
print(moves)
```
| 97,151 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
n, k = map(int, input().split())
l1, r1 = map(int, input().split())
l2, r2 = map(int, input().split())
if l1 > l2:
l1, r1, l2, r2 = l2, r2, l1, r1
ans = 10**18
if r1 < l2:
shared = 0
initial_cost = l2 - r1
max_len = r2 - l1
else:
shared = min(r1, r2) - l2
initial_cost = 0
max_len = max(r1, r2) - l1 - shared
k = max(0, k - shared * n)
if k == 0:
ans = 0
# Fill with i-area
for i in range(n + 1):
if i == 0 and initial_cost > 0:
continue
# Connect i
cost = i * initial_cost
rest = max(0, k - i * max_len)
ans = min(ans, cost + (k - rest) + rest * 2)
print(ans)
```
| 97,152 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
for tt in range(int(input())):
n,z = input().split(' ')
Al,Ar = input().split(' ')
Bl,Br = input().split(' ')
n,z,Al,Ar,Bl,Br = int(n),int(z),int(Al),int(Ar),int(Bl),int(Br)
if Al>Bl:
Al,Ar,Bl,Br = Bl,Br,Al,Ar
if Ar<Bl:
blank = Bl-Ar
get = Br-Al
need = blank+get
if z<get:
print(blank+z)
elif z<=n*get:
print(int(z/get)*need+min(z%get,blank)+z%get)
else:
print(n*need+(z-n*get)*2)
else:
get = max(Br,Ar) - Al
init = n*(min(Br,Ar)-Bl)
if z<=init:
print(0)
elif z<=n*get:
print(z-init)
else:
print(n*get-init + 2*(z-n*get))
```
| 97,153 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
import sys
input=sys.stdin.buffer.readline
t=int(input())
for _ in range(t):
n,k=[int(x) for x in input().split()]
l1,r1=[int(x) for x in input().split()]
l2,r2=[int(x) for x in input().split()]
its=min(r1,r2)-max(l1,l2) #intersect
if its>=0: #intersect
k-=n*its
if k<=0:ans=0
else:
aoc=max(r1,r2)-min(l1,l2) #add-one-capacity
if n*(aoc-its)>=k:
ans=k
else:
ans=n*(aoc-its)
k-=n*(aoc-its)
ans+=k*2 #every k requires 2 steps
print(ans)
else: #no intersection
gap=-its
aoc=max(r1,r2)-min(l1,l2) #add-one-capacity, only after gap is filled up
minn=float('inf')
for i in range(1,n+1): #fill for i segments
total=gap*i
if aoc*i>=k:total+=k
else:
total+=aoc*i+(k-aoc*i)*2
minn=min(minn,total)
print(minn)
```
| 97,154 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
from sys import stdin
tt = int(stdin.readline())
for loop in range(tt):
#don't
n,k = map(int,stdin.readline().split() )
l1,r1 = map(int,stdin.readline().split())
l2,r2 = map(int,stdin.readline().split())
#don't
if l1 >= r2 or l2 >= r1:
ans = float("inf")
for usenum in range(1,n+1):
if max(abs(l1-r2),abs(l2-r1))*usenum >= k:
ans = min(ans , min(abs(l1-r2),abs(l2-r1))*usenum + k)
else:
rem = k - max(abs(l1-r2),abs(l2-r1))*usenum
ans = min(ans , min(abs(l1-r2),abs(l2-r1))*usenum + max(abs(l1-r2),abs(l2-r1))*usenum + rem*2)
print (ans)
else:
ans = float("inf")
over = min(r1-l1 , r2-l2 , r1-l2 , r2-l1)
usenum = n
if over * usenum >= k:
print (0)
continue
k -= over * usenum
useable = max(r1-l1 , r2-l2 , r1-l2 , r2-l1) - over
if useable * usenum >= k:
print (k)
continue
else:
rem = k - useable * usenum
print (useable * usenum + rem*2)
```
| 97,155 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
# e053d9db8efe450d25f052c2a1e0f08131c83762143ed09d442439e0eaa5681d
import sys
input = sys.stdin.buffer.readline
def print(val):
sys.stdout.write(str(val) + '\n')
def prog():
for _ in range(int(input())):
n,k = map(int,input().split())
l1,r1 = map(int,input().split())
l2,r2 = map(int,input().split())
if(l1 > l2):
p1 = l1
p2 = r1
l1 = l2
r1 = r2
l2 = p1
r2 = p2
total = 0
moves = 0
if (r1 >= l2):
intersection = min(r1,r2) - max(l1,l2)
l3 = min(l1,l2)
r3 = max(r1,r2)
leftover = r3 - l3 - intersection
k -= intersection*n
if (k <= 0):
print(0)
elif(k - leftover*n <= 0):
print(k)
else:
k -= leftover*n
moves += leftover*n
moves += k*2
k = 0
print(moves)
else:
l3 = l1
r3 = r2
a = l2 - r1
if (r3 - l3)*n >= k:
if (k <= r3 - l3):
print(a + k)
continue
else:
k -= r3 - l3
moves += a + r3 - l3
if (r3 -l3 + a >= 2*(r3-l3)):
moves += 2*k
else:
moves += (r3-l3+a)*(k//(r3-l3))
moves += min(k %(r3-l3) + a , 2*(k % (r3-l3)))
print(moves)
else:
k -= (r3 - l3)*n
moves += (r3-l3 + a)*n
moves += k*2
print(moves)
prog()
```
| 97,156 |
Provide tags and a correct Python 2 solution for this coding contest problem.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Tags: brute force, greedy, implementation, math
Correct Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
import heapq
raw_input = stdin.readline
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return (map(int,stdin.read().split()))
range = xrange # not for python 3.0+
for t in range(ni()):
n,k=li()
l1,r1=li()
l2,r2=li()
df=0
if not ( (l1<=r2 and l1>=l2) or (r1>=l2 and r1<=r2) or (l2<=r1 and l2>=l1) or (r2>=l1 and r2<=r1)):
df=min(int(abs(r2-l1)),int(abs(l2-r1)))
else:
x=[l1,r1,l2,r2]
x.sort()
ln=x[2]-x[1]
ans=min(k,ln*n)
k-=ans
ans=0
mx=max(l1,l2,r1,r2)-min(l1,l2,r1,r2)-ln
ans+=min(k,mx*n)
k-=min(k,mx*n)
pn(ans+(2*k))
continue
mx=max(l1,l2,r1,r2)-min(l1,l2,r1,r2)
ans=df
if k<mx:
print ans+k
continue
ans+=mx
k-=mx
if mx:
cst=(df+mx)/float(mx)
else:
cst=10**12
if cst<2:
for i in range(n-1):
if k<mx:
ans+=min(2*k,k+df)
k=0
else:
ans+=mx+df
k-=mx
ans+=2*k
pn(ans)
```
| 97,157 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
import sys
import math
read = sys.stdin.buffer.read
# input = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
t = int(input())
for case in range(t):
n, k = map(int, input().split())
l1, r1 = map(int, input().split())
l2, r2 = map(int, input().split())
origin_inter = max(0, min(r1, r2) - max(l1, l2))
ans = 0
k -= origin_inter * n
if k <= 0:
ans = 0
elif origin_inter > 0:
# まず伸ばせるまで伸ばす。その後、2ターンで1伸ばす
if k <= n * ((r1 - l1) + (r2 - l2) - origin_inter * 2):
ans += k
else:
ans = 2 * k - n * ((r1 - l1) + (r2 - l2) - origin_inter * 2)
else:
margin = max(l1, l2) - min(r1, r2)
len_inter = max(r1, r2) - min(l1, l2)
# 何本を使うかを決め打つ
ans = 10 ** 10
for i in range(1, n + 1):
tmp_i = margin * i
if k > len_inter * i:
tmp_i += len_inter * i + (k - len_inter * i) * 2
else:
tmp_i += k
# print(i, tmp_i, margin, len_inter)
ans = min(tmp_i, ans)
"""bit = 0
while bit == 0 or k > dif:
ans += dif
if max(r1, r2) - min(l1, l2) >= k: # kの方が小さい dif を足して +1でいくべき
ans += k
k = 0
break
else:
ans += max(r1, r2) - min(l1, l2)
k -= max(r1, r2) - min(l1, l2)
bit = 1
ans += 2 * k"""
print(ans)
```
Yes
| 97,158 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
import sys
readline = sys.stdin.readline
def solve():
N, K = map(int, readline().split())
L1, R1 = map(int, readline().split())
L2, R2 = map(int, readline().split())
if L1 > L2:
L1, L2 = L2, L1
R1, R2 = R2, R1
if R1 < L2:
gap = L2 - R1
union = R2 - L1
ans = gap
if K <= union:
ans += K
print(ans)
return
ans += union
rem = K - union
if N == 1:
print(ans + 2 * rem)
return
t = min(rem // union, N - 1)
ans += t * (gap + union)
rem -= t * union
if t == N - 1:
print(ans + 2 * rem)
return
if 2*rem < gap + rem:
print(ans + 2 * rem)
return
else:
print(ans + gap + rem)
return
else:
ovl = min(R1, R2) - L2
if ovl * N >= K:
print(0)
return
union = max(R1, R2) - L1
if union * N >= K:
print(K - ovl * N)
return
else:
ans = union * N - ovl * N
ans += 2 * (K - union * N)
print(ans)
return
T = int(readline())
for i in range(T):
solve()
```
Yes
| 97,159 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
# import sys
import heapq
import copy
import math
import decimal
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li)
#
#heapq.heappush(li,4)
#
#heapq.heappop(li)
#
# & Bitwise AND Operator 10 & 7 = 2
# | Bitwise OR Operator 10 | 7 = 15
# ^ Bitwise XOR Operator 10 ^ 7 = 13
# << Bitwise Left Shift operator 10<<2 = 40
# >> Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
def inp():
return(int(input()))
def inlt():
return(list(map(int,input().split())))
def insr():
s = input()
return(list(s[:len(s) - 1]))
def insr2():
s = input()
return((s[:len(s) - 1]))
def invr():
return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####
def pr_list(a):
print( *a , sep=" ")
def main():
tests = inp()
# tests = 1
mod = 1000000007
limit = 10**18
ans = 0
stack = []
hashm = {}
arr = []
heapq.heapify(arr)
for test in range(tests):
[n,k] = inlt()
[a1,a2] = inlt()
[b1,b2] = inlt()
# if test == 39:
# print(n,k,[a1,a2],[b1,b2])
# if (a1<b1)and
def swap(a,b):
temp = a
a= b
b =temp
return [a,b]
if a2 <b1 or b2<a1:
if a2 > b1:
[a1,b1] = swap(a1,b1)
[a2,b2] = swap(a2,b2)
mini = b1 - a2
maxi = b2 - a1
if k <= maxi:
print(mini + k)
else:
case1 = (k- maxi)*2 + mini + maxi
a = min ((k//maxi) , n)
if k//maxi >=n:
a = n
case2 = a * (mini + maxi) + 2*(k- a*maxi)
else:
a = k//maxi
case2 = a * (mini + maxi) + min(k-a*maxi + mini, 2*(k- a*maxi))
print(min(case1,case2))
else:
if a2 > b2:
[a1,b1] = swap(a1,b1)
[a2,b2] = swap(a2,b2)
curr = (a2 - max(b1,a1))*n
if k<=curr:
print(0)
else:
maxi = (b2 - min(a1,b1))*n
if k<=maxi:
print(k - curr)
else:
print(2*(k-maxi) + maxi - curr)
if __name__== "__main__":
main()
```
Yes
| 97,160 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
# ---------------------------iye ha aam zindegi---------------------------------------------
import math
import heapq, bisect
import sys
from collections import deque, defaultdict
from fractions import Fraction
mod = 10 ** 9 + 7
mod1 = 998244353
# ------------------------------warmup----------------------------
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# -------------------game starts now----------------------------------------------------import math
class TreeNode:
def __init__(self, k, v):
self.key = k
self.value = v
self.left = None
self.right = None
self.parent = None
self.height = 1
self.num_left = 1
self.num_total = 1
class AvlTree:
def __init__(self):
self._tree = None
def add(self, k, v):
if not self._tree:
self._tree = TreeNode(k, v)
return
node = self._add(k, v)
if node:
self._rebalance(node)
def _add(self, k, v):
node = self._tree
while node:
if k < node.key:
if node.left:
node = node.left
else:
node.left = TreeNode(k, v)
node.left.parent = node
return node.left
elif node.key < k:
if node.right:
node = node.right
else:
node.right = TreeNode(k, v)
node.right.parent = node
return node.right
else:
node.value = v
return
@staticmethod
def get_height(x):
return x.height if x else 0
@staticmethod
def get_num_total(x):
return x.num_total if x else 0
def _rebalance(self, node):
n = node
while n:
lh = self.get_height(n.left)
rh = self.get_height(n.right)
n.height = max(lh, rh) + 1
balance_factor = lh - rh
n.num_total = 1 + self.get_num_total(n.left) + self.get_num_total(n.right)
n.num_left = 1 + self.get_num_total(n.left)
if balance_factor > 1:
if self.get_height(n.left.left) < self.get_height(n.left.right):
self._rotate_left(n.left)
self._rotate_right(n)
elif balance_factor < -1:
if self.get_height(n.right.right) < self.get_height(n.right.left):
self._rotate_right(n.right)
self._rotate_left(n)
else:
n = n.parent
def _remove_one(self, node):
"""
Side effect!!! Changes node. Node should have exactly one child
"""
replacement = node.left or node.right
if node.parent:
if AvlTree._is_left(node):
node.parent.left = replacement
else:
node.parent.right = replacement
replacement.parent = node.parent
node.parent = None
else:
self._tree = replacement
replacement.parent = None
node.left = None
node.right = None
node.parent = None
self._rebalance(replacement)
def _remove_leaf(self, node):
if node.parent:
if AvlTree._is_left(node):
node.parent.left = None
else:
node.parent.right = None
self._rebalance(node.parent)
else:
self._tree = None
node.parent = None
node.left = None
node.right = None
def remove(self, k):
node = self._get_node(k)
if not node:
return
if AvlTree._is_leaf(node):
self._remove_leaf(node)
return
if node.left and node.right:
nxt = AvlTree._get_next(node)
node.key = nxt.key
node.value = nxt.value
if self._is_leaf(nxt):
self._remove_leaf(nxt)
else:
self._remove_one(nxt)
self._rebalance(node)
else:
self._remove_one(node)
def get(self, k):
node = self._get_node(k)
return node.value if node else -1
def _get_node(self, k):
if not self._tree:
return None
node = self._tree
while node:
if k < node.key:
node = node.left
elif node.key < k:
node = node.right
else:
return node
return None
def get_at(self, pos):
x = pos + 1
node = self._tree
while node:
if x < node.num_left:
node = node.left
elif node.num_left < x:
x -= node.num_left
node = node.right
else:
return (node.key, node.value)
raise IndexError("Out of ranges")
@staticmethod
def _is_left(node):
return node.parent.left and node.parent.left == node
@staticmethod
def _is_leaf(node):
return node.left is None and node.right is None
def _rotate_right(self, node):
if not node.parent:
self._tree = node.left
node.left.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.left
node.left.parent = node.parent
else:
node.parent.right = node.left
node.left.parent = node.parent
bk = node.left.right
node.left.right = node
node.parent = node.left
node.left = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
def _rotate_left(self, node):
if not node.parent:
self._tree = node.right
node.right.parent = None
elif AvlTree._is_left(node):
node.parent.left = node.right
node.right.parent = node.parent
else:
node.parent.right = node.right
node.right.parent = node.parent
bk = node.right.left
node.right.left = node
node.parent = node.right
node.right = bk
if bk:
bk.parent = node
node.height = max(self.get_height(node.left), self.get_height(node.right)) + 1
node.num_total = 1 + self.get_num_total(node.left) + self.get_num_total(node.right)
node.num_left = 1 + self.get_num_total(node.left)
@staticmethod
def _get_next(node):
if not node.right:
return node.parent
n = node.right
while n.left:
n = n.left
return n
# -----------------------------------------------binary seacrh tree---------------------------------------
class SegmentTree1:
def __init__(self, data, default='z', func=lambda a, b: min(a, b)):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------game starts now----------------------------------------------------import math
class SegmentTree:
def __init__(self, data, default=0, func=lambda a, b: a + b):
"""initialize the segment tree with data"""
self._default = default
self._func = func
self._len = len(data)
self._size = _size = 1 << (self._len - 1).bit_length()
self.data = [default] * (2 * _size)
self.data[_size:_size + self._len] = data
for i in reversed(range(_size)):
self.data[i] = func(self.data[i + i], self.data[i + i + 1])
def __delitem__(self, idx):
self[idx] = self._default
def __getitem__(self, idx):
return self.data[idx + self._size]
def __setitem__(self, idx, value):
idx += self._size
self.data[idx] = value
idx >>= 1
while idx:
self.data[idx] = self._func(self.data[2 * idx], self.data[2 * idx + 1])
idx >>= 1
def __len__(self):
return self._len
def query(self, start, stop):
if start == stop:
return self.__getitem__(start)
stop += 1
start += self._size
stop += self._size
res = self._default
while start < stop:
if start & 1:
res = self._func(res, self.data[start])
start += 1
if stop & 1:
stop -= 1
res = self._func(res, self.data[stop])
start >>= 1
stop >>= 1
return res
def __repr__(self):
return "SegmentTree({0})".format(self.data)
# -------------------------------iye ha chutiya zindegi-------------------------------------
class Factorial:
def __init__(self, MOD):
self.MOD = MOD
self.factorials = [1, 1]
self.invModulos = [0, 1]
self.invFactorial_ = [1, 1]
def calc(self, n):
if n <= -1:
print("Invalid argument to calculate n!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.factorials):
return self.factorials[n]
nextArr = [0] * (n + 1 - len(self.factorials))
initialI = len(self.factorials)
prev = self.factorials[-1]
m = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = prev * i % m
self.factorials += nextArr
return self.factorials[n]
def inv(self, n):
if n <= -1:
print("Invalid argument to calculate n^(-1)")
print("n must be non-negative value. But the argument was " + str(n))
exit()
p = self.MOD
pi = n % p
if pi < len(self.invModulos):
return self.invModulos[pi]
nextArr = [0] * (n + 1 - len(self.invModulos))
initialI = len(self.invModulos)
for i in range(initialI, min(p, n + 1)):
next = -self.invModulos[p % i] * (p // i) % p
self.invModulos.append(next)
return self.invModulos[pi]
def invFactorial(self, n):
if n <= -1:
print("Invalid argument to calculate (n^(-1))!")
print("n must be non-negative value. But the argument was " + str(n))
exit()
if n < len(self.invFactorial_):
return self.invFactorial_[n]
self.inv(n) # To make sure already calculated n^-1
nextArr = [0] * (n + 1 - len(self.invFactorial_))
initialI = len(self.invFactorial_)
prev = self.invFactorial_[-1]
p = self.MOD
for i in range(initialI, n + 1):
prev = nextArr[i - initialI] = (prev * self.invModulos[i % p]) % p
self.invFactorial_ += nextArr
return self.invFactorial_[n]
class Combination:
def __init__(self, MOD):
self.MOD = MOD
self.factorial = Factorial(MOD)
def ncr(self, n, k):
if k < 0 or n < k:
return 0
k = min(k, n - k)
f = self.factorial
return f.calc(n) * f.invFactorial(max(n - k, k)) * f.invFactorial(min(k, n - k)) % self.MOD
# --------------------------------------iye ha combinations ka zindegi---------------------------------
def powm(a, n, m):
if a == 1 or n == 0:
return 1
if n % 2 == 0:
s = powm(a, n // 2, m)
return s * s % m
else:
return a * powm(a, n - 1, m) % m
# --------------------------------------iye ha power ka zindegi---------------------------------
def sort_list(list1, list2):
zipped_pairs = zip(list2, list1)
z = [x for _, x in sorted(zipped_pairs)]
return z
# --------------------------------------------------product----------------------------------------
def product(l):
por = 1
for i in range(len(l)):
por *= l[i]
return por
# --------------------------------------------------binary----------------------------------------
def binarySearchCount(arr, n, key):
left = 0
right = n - 1
count = 0
while (left <= right):
mid = int((right + left) / 2)
# Check if middle element is
# less than or equal to key
if (arr[mid] <= key):
count = mid + 1
left = mid + 1
# If key is smaller, ignore right half
else:
right = mid - 1
return count
# --------------------------------------------------binary----------------------------------------
def countdig(n):
c = 0
while (n > 0):
n //= 10
c += 1
return c
def countGreater(arr, n, k):
l = 0
r = n - 1
# Stores the index of the left most element
# from the array which is greater than k
leftGreater = n
# Finds number of elements greater than k
while (l <= r):
m = int(l + (r - l) / 2)
if (arr[m] >= k):
leftGreater = m
r = m - 1
# If mid element is less than
# or equal to k update l
else:
l = m + 1
# Return the count of elements
# greater than k
return (n - leftGreater)
# --------------------------------------------------binary------------------------------------
for ik in range(int(input())):
n,k=map(int,input().split())
l1,r1=map(int,input().split())
l2,r2=map(int,input().split())
gap=max(0,-min(r1,r2)+max(l1,l2))
lap=max(0,min(r1,r2)-max(l1,l2))
total=max(r1,r2)-min(l1,l2)
#print(gap,lap,total)
if lap*n>=k:
print(0)
continue
lap1=lap
lap*=n
ans=0
t=min(k-lap,total-lap1)
ans+=gap+t
n-=1
lap+=t
#print(ans,lap,total,k)
while(lap<k):
t=min(k-lap,total-lap1)
#print(t)
if n==0:
break
elif gap<=t:
lap+=t
ans+=t+gap
n-=1
else:
ans+=2*t
lap+=t
#print(ans,lap)
if lap<k:
ans+=2*(k-lap)
print(ans)
```
Yes
| 97,161 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
import sys
readline = sys.stdin.buffer.readline
def even(n): return 1 if n%2==0 else 0
def main():
n,k = map(int,readline().split())
a = list(map(int,readline().split()))
b = list(map(int,readline().split()))
#left側が小さい方をaとする
if a[0] > b[0]:
a,b = b,a
score = 0
cost = 0
#初期状態でのスコア
score_init = max(0,a[1]-b[0])*n
if score_init >= k:
print(0)
return
#case 1
if a[0] <= b[0] and b[1] <= a[1]:
res = b[0] - a[0] + a[1] - b[1]
score += res*n
cost += res*n
if k <= score+score_init:
print(k-score_init)
return
#case 2-1
elif a[1] > b[0]:
res = b[0]-a[0] + b[1]-a[1]
score += res*n
cost += res*n
if k <= score+score_init:
print(k-score_init)
return
#case 3
else:
one_term_score = b[1]-a[0]
one_term_cost = a[1]-a[0]+b[1]-b[0]+(b[0]-a[1])*2
if k < one_term_score*n: #途中で終了条件に達しそうならば
#何回まで確定で回すか
let = k//one_term_score
score += one_term_score*let
cost += one_term_cost*let
#余りはコスト2払ったほうがお得な場合もあるのでまた場合分け
if k-score < b[0]-a[1]:
pass
else:
cost += b[0]-a[1]+(k-score)
score = k
else: #そうでないなら全部回しちゃう
score += one_term_score*n
cost += one_term_cost*n
#あとは全部コスト2払って1伸ばす
if score + score_init < k:
cost += 2*(k-score - score_init)
print(cost)
if __name__ == "__main__":
n = int(readline())
for i in range(n):
main()
```
No
| 97,162 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
from bisect import *
from collections import *
from math import gcd,ceil,sqrt,floor,inf
from heapq import *
from itertools import *
from operator import add,mul,sub,xor,truediv,floordiv
from functools import *
#------------------------------------------------------------------------
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#------------------------------------------------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
#------------------------------------------------------------------------
from types import GeneratorType
def bootstrap(f, stack=[]):
def wrappedfunc(*args, **kwargs):
if stack:
return f(*args, **kwargs)
else:
to = f(*args, **kwargs)
while True:
if type(to) is GeneratorType:
stack.append(to)
to = next(to)
else:
stack.pop()
if not stack:
break
to = stack[-1].send(to)
return to
return wrappedfunc
farr=[1]
ifa=[]
def fact(x,mod=0):
if mod:
while x>=len(farr):
farr.append(farr[-1]*len(farr)%mod)
else:
while x>=len(farr):
farr.append(farr[-1]*len(farr))
return farr[x]
def ifact(x,mod):
global ifa
ifa.append(pow(farr[-1],mod-2,mod))
for i in range(x,0,-1):
ifa.append(ifa[-1]*i%mod)
ifa=ifa[::-1]
def per(i,j,mod=0):
if i<j: return 0
if not mod:
return fact(i)//fact(i-j)
return farr[i]*ifa[i-j]%mod
def com(i,j,mod=0):
if i<j: return 0
if not mod:
return per(i,j)//fact(j)
return per(i,j,mod)*ifa[j]%mod
def catalan(n):
return com(2*n,n)//(n+1)
def linc(f,t,l,r):
while l<r:
mid=(l+r)//2
if t>f(mid):
l=mid+1
else:
r=mid
return l
def rinc(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t<f(mid):
r=mid-1
else:
l=mid
return l
def ldec(f,t,l,r):
while l<r:
mid=(l+r)//2
if t<f(mid):
l=mid+1
else:
r=mid
return l
def rdec(f,t,l,r):
while l<r:
mid=(l+r+1)//2
if t>f(mid):
r=mid-1
else:
l=mid
return l
def isprime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True
def binfun(x):
c=0
for w in arr:
c+=ceil(w/x)
return c
def lowbit(n):
return n&-n
def inverse(a,m):
a%=m
if a<=1: return a
return ((1-inverse(m,a)*m)//a)%m
class BIT:
def __init__(self,arr):
self.arr=arr
self.n=len(arr)-1
def update(self,x,v):
while x<=self.n:
self.arr[x]+=v
x+=x&-x
def query(self,x):
ans=0
while x:
ans+=self.arr[x]
x&=x-1
return ans
'''
class SMT:
def __init__(self,arr):
self.n=len(arr)-1
self.arr=[0]*(self.n<<2)
self.lazy=[0]*(self.n<<2)
def Build(l,r,rt):
if l==r:
self.arr[rt]=arr[l]
return
m=(l+r)>>1
Build(l,m,rt<<1)
Build(m+1,r,rt<<1|1)
self.pushup(rt)
Build(1,self.n,1)
def pushup(self,rt):
self.arr[rt]=self.arr[rt<<1]+self.arr[rt<<1|1]
def pushdown(self,rt,ln,rn):#lr,rn表区间数字数
if self.lazy[rt]:
self.lazy[rt<<1]+=self.lazy[rt]
self.lazy[rt<<1|1]=self.lazy[rt]
self.arr[rt<<1]+=self.lazy[rt]*ln
self.arr[rt<<1|1]+=self.lazy[rt]*rn
self.lazy[rt]=0
def update(self,L,R,c,l=1,r=None,rt=1):#L,R表示操作区间
if r==None: r=self.n
if L<=l and r<=R:
self.arr[rt]+=c*(r-l+1)
self.lazy[rt]+=c
return
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
if L<=m: self.update(L,R,c,l,m,rt<<1)
if R>m: self.update(L,R,c,m+1,r,rt<<1|1)
self.pushup(rt)
def query(self,L,R,l=1,r=None,rt=1):
if r==None: r=self.n
#print(L,R,l,r,rt)
if L<=l and R>=r:
return self.arr[rt]
m=(l+r)>>1
self.pushdown(rt,m-l+1,r-m)
ans=0
if L<=m: ans+=self.query(L,R,l,m,rt<<1)
if R>m: ans+=self.query(L,R,m+1,r,rt<<1|1)
return ans
'''
class DSU:#容量+路径压缩
def __init__(self,n):
self.c=[-1]*n
def same(self,x,y):
return self.find(x)==self.find(y)
def find(self,x):
if self.c[x]<0:
return x
self.c[x]=self.find(self.c[x])
return self.c[x]
def union(self,u,v):
u,v=self.find(u),self.find(v)
if u==v:
return False
if self.c[u]<self.c[v]:
u,v=v,u
self.c[u]+=self.c[v]
self.c[v]=u
return True
def size(self,x): return -self.c[self.find(x)]
class UFS:#秩+路径
def __init__(self,n):
self.parent=[i for i in range(n)]
self.ranks=[0]*n
def find(self,x):
if x!=self.parent[x]:
self.parent[x]=self.find(self.parent[x])
return self.parent[x]
def union(self,u,v):
pu,pv=self.find(u),self.find(v)
if pu==pv:
return False
if self.ranks[pu]>=self.ranks[pv]:
self.parent[pv]=pu
if self.ranks[pv]==self.ranks[pu]:
self.ranks[pu]+=1
else:
self.parent[pu]=pv
def Prime(n):
c=0
prime=[]
flag=[0]*(n+1)
for i in range(2,n+1):
if not flag[i]:
prime.append(i)
c+=1
for j in range(c):
if i*prime[j]>n: break
flag[i*prime[j]]=prime[j]
if i%prime[j]==0: break
return prime
def dij(s,graph):
d={}
d[s]=0
heap=[(0,s)]
seen=set()
while heap:
dis,u=heappop(heap)
if u in seen:
continue
for v in graph[u]:
if v not in d or d[v]>d[u]+graph[u][v]:
d[v]=d[u]+graph[u][v]
heappush(heap,(d[v],v))
return d
def GP(it): return [(ch,len(list(g))) for ch,g in groupby(it)]
class DLN:
def __init__(self,val):
self.val=val
self.pre=None
self.next=None
t=N()
for i in range(t):
n,k=RL()
l1,r1=RL()
l2,r2=RL()
if l1>l2:
l1,r1,l2,r2=l2,r2,l1,r1
if r1<l2:
#print((r2-l1),(r2-r1+l2-l1))
if n==1 or 2*(r2-l1)<=(r2-r1+l2-l1):
if k<=r2-r1:
ans=l2-r1+k
else:
ans=r2-r1+l2-l1+(k-r2+l1)*2
else:
d,m=divmod(k,r2-l1)
if d+1<=n:
ans=d*(r2-r1+l2-l1)+min(l2-r1+m,m*2)
else:
m=k-(r2-l1)*n
ans=n*(r2-r1+l2-l1)+m*2
else:
a=min(r1,r2)-l2
p=max(r1,r2)-l1
if n*a>=k:
ans=0
elif n*p>=k:
ans=k-n*a
else:
ans=n*(p-a)+(k-n*p)*2
print(ans)
'''
sys.setrecursionlimit(200000)
import threading
threading.stack_size(10**8)
t=threading.Thread(target=main)
t.start()
t.join()
'''
```
No
| 97,163 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# ------------------------------
def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
# import heapq as hq
# import bisect as bs
# from collections import deque as dq
from collections import defaultdict as dc
# from math import ceil,floor,sqrt
# from collections import Counter
for _ in range(N()):
n,k = RL()
l1,r1 = RL()
l2,r2 = RL()
l1,l2,r1,r2 = min(l1,l2),max(l1,l2),min(r1,r2),max(r1,r2)
dic = max(0,l2-r1)
inc = max(0,r1-l2)
al = r2-l1
now = inc*n
if now>=k:
print(0)
else:
res = 0
tt = dic+(al-inc)
tmp = k-now-(al-inc)
if tmp<=0:
print(dic+k-now)
else:
r1 = tt+tmp*2
key = max((k-now)//(al-inc),n)
r3 = tt*key+2*((k-now)-key*(al-inc))
r2 = tt*key
if (k-now)%(al-inc)>0 and key<n:
r2+=dic+((k-now)%(al-inc))
else:
r2 = r3
res = min(r1,r2,r3)
print(res)
```
No
| 97,164 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
import time,math as mt,bisect,sys
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
mx=10**7
spf=[mx]*(mx+1)
def SPF():
spf[1]=1
for i in range(2,mx+1):
if spf[i]==mx:
spf[i]=i
for j in range(i*i,mx+1,i):
if i<spf[j]:
spf[j]=i
return
#####################################################################################
mod=10**9+7
def solve():
n,k=IP()
l1,r1=IP()
l2,r2=IP()
a=[[l1,r1] for i in range(n)]
b=[[l2,r2] for i in range(n)]
x,y=max(l1,l2),min(r1,r2)
inter=max(0,y-x)
if (y>x):
k-=(y-x)*n
steps=0
for i in range(n):
if k<=0:
break
reqd=min(k,max(r1,r2)-min(l1,l2)-inter)
cost=max(0,(x-y))+reqd
if i>=2:
cost=min(cost,reqd*2)
k-=reqd
steps+=cost
if k>0:
steps+=k*2
P(steps)
return
t=II()
for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
```
No
| 97,165 |
Evaluate the correctness of the submitted Python 2 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given two lists of segments [al_1, ar_1], [al_2, ar_2], ..., [al_n, ar_n] and [bl_1, br_1], [bl_2, br_2], ..., [bl_n, br_n].
Initially, all segments [al_i, ar_i] are equal to [l_1, r_1] and all segments [bl_i, br_i] are equal to [l_2, r_2].
In one step, you can choose one segment (either from the first or from the second list) and extend it by 1. In other words, suppose you've chosen segment [x, y] then you can transform it either into [x - 1, y] or into [x, y + 1].
Let's define a total intersection I as the sum of lengths of intersections of the corresponding pairs of segments, i.e. ∑_{i=1}^{n}{intersection_length([al_i, ar_i], [bl_i, br_i])}. Empty intersection has length 0 and length of a segment [x, y] is equal to y - x.
What is the minimum number of steps you need to make I greater or equal to k?
Input
The first line contains the single integer t (1 ≤ t ≤ 1000) — the number of test cases.
The first line of each test case contains two integers n and k (1 ≤ n ≤ 2 ⋅ 10^5; 1 ≤ k ≤ 10^9) — the length of lists and the minimum required total intersection.
The second line of each test case contains two integers l_1 and r_1 (1 ≤ l_1 ≤ r_1 ≤ 10^9) — the segment all [al_i, ar_i] are equal to initially.
The third line of each test case contains two integers l_2 and r_2 (1 ≤ l_2 ≤ r_2 ≤ 10^9) — the segment all [bl_i, br_i] are equal to initially.
It's guaranteed that the sum of n doesn't exceed 2 ⋅ 10^5.
Output
Print t integers — one per test case. For each test case, print the minimum number of step you need to make I greater or equal to k.
Example
Input
3
3 5
1 2
3 4
2 1000000000
1 1
999999999 999999999
10 3
5 10
7 8
Output
7
2000000000
0
Note
In the first test case, we can achieve total intersection 5, for example, using next strategy:
* make [al_1, ar_1] from [1, 2] to [1, 4] in 2 steps;
* make [al_2, ar_2] from [1, 2] to [1, 3] in 1 step;
* make [bl_1, br_1] from [3, 4] to [1, 4] in 2 steps;
* make [bl_2, br_2] from [3, 4] to [1, 4] in 2 steps.
In result, I = intersection_length([al_1, ar_1], [bl_1, br_1]) + intersection_length([al_2, ar_2], [bl_2, br_2]) + \\\ + intersection_length([al_3, ar_3], [bl_3, br_3]) = 3 + 2 + 0 = 5
In the second test case, we can make [al_1, ar_1] = [0, 1000000000] in 1000000000 steps and [bl_1, br_1] = [0, 1000000000] in 1000000000 steps.
In the third test case, the total intersection I is already equal to 10 > 3, so we don't need to do any steps.
Submitted Solution:
```
from sys import stdin, stdout
from collections import Counter, defaultdict
pr=stdout.write
import heapq
raw_input = stdin.readline
def ni():
return int(raw_input())
def li():
return list(map(int,raw_input().split()))
def pn(n):
stdout.write(str(n)+'\n')
def pa(arr):
pr(' '.join(map(str,arr))+'\n')
# fast read function for total integer input
def inp():
# this function returns whole input of
# space/line seperated integers
# Use Ctrl+D to flush stdin.
return (map(int,stdin.read().split()))
range = xrange # not for python 3.0+
for t in range(ni()):
n,k=li()
l1,r1=li()
l2,r2=li()
df=0
if not ( (l2<=r1 and l2>=l1) or (r2>=l1 and r2<=r1)):
df=min(int(abs(r2-l1)),int(abs(l2-r1)))
else:
if (l2<=r1 and l2>=l1):
ln=min(r2,r1)-l2
else:
ln=r2-max(l1,l2)
ans=min(k,ln*n)
k-=ans
ans=0
mx=max(r2-l2,r1-l1)
ans+=min(k,mx*n)
k-=min(k,mx*n)
pn(ans+(2*k))
continue
mx=max(l1,l2,r1,r2)-min(l1,l2,r1,r2)
ans=df
if k<mx:
print ans+k
continue
ans+=mx
k-=mx
if mx:
cst=(df+mx)/float(mx)
else:
cst=10**12
if cst<2:
for i in range(n-1):
if k<mx:
ans+=min(2*k,k+df)
k-=k
else:
ans+=mx+df
k-=mx
ans+=2*k
pn(ans)
```
No
| 97,166 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
from math import asin, pi
n, R, r = map(int, input().split())
if n == 1 and r <= R:
print('YES')
elif 2 * r > R:
print('NO')
elif (pi / asin(r / (R - r))) + 10 ** -6 >= n:
print('YES')
else:
print('NO')
```
| 97,167 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
import math
def Rp(n, R, r):
if n == 1:
return r
else:
return r * (1 + 1 / math.sin(math.pi / n)) - 1e-6
n, R, r = map(int, input().split())
print("YES" if R >= Rp(n, R, r) else "NO")
```
| 97,168 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
import math
n, R, r = map(int, input().split())
if n == 1:
print("YES" if r <= R else "NO")
exit(0)
theta = math.pi / n
RR = r / math.sin(theta) + r
print("YES" if R >= RR else "NO")
```
| 97,169 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
"""
Author : Arif Ahmad
Date :
Algo :
Difficulty :
"""
def main():
n, R, r = map(int, input().split())
if n == 1:
if r <= R: print('YES')
else: print('NO')
else:
import math
# calculate side of inscribed n-gon
a = (R - r) * math.sin(math.pi / n)
#print(a)
if r < a+1e-7: print('YES')
else: print('NO')
if __name__ == '__main__':
main()
```
| 97,170 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
from math import sin, pi
def table(n, R, r):
if r > R:
return "NO"
if n == 1:
return "YES"
if 2 * r < (R - r) * sin(pi / n) * 2 + 1e-8:
return "YES"
return "NO"
n, R, r = [int(i) for i in input().split()]
print(table(n, R, r))
```
| 97,171 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
import math
n,R,r=map(int,input().split())
if n==1 :
if r>R :
print("NO")
else:
print("YES")
quit()
if n==2 :
if 2*r>R :
print("NO")
else:
print("YES")
quit()
Rv=2*r/(2*math.sin(360/2/n*math.pi/180))
t=r+Rv
if t>R :
print("NO")
else :
print("YES")
```
| 97,172 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
from math import pi, sin
n, R, r = map(int, input().split())
print('NO' if r > R or (n > 1 and ((R - r) * sin(pi / n) + 0.0000001) < r) else 'YES')
```
| 97,173 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Tags: geometry, math
Correct Solution:
```
import math
n,R,r=map(int,input().split())
if 2*r>R and n>1:
print('NO')
exit()
if n==1 and r<=R:
print('YES')
exit()
if n==1 and r>R:
print('NO')
exit()
if n>1 and r>=R:
print('NO')
exit()
c = 2*math.pi/(2*math.asin(r/(R-r)))
if c-n>-10**-6:
print('YES')
else:
print('NO')
```
| 97,174 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
from math import cos,pi
p,R,r = map(int,input().split())
if p==1:
if R>=r:
print("YES")
else:
print("NO")
exit()
angle = (p-2)*pi/(p*2)
side = r/cos(angle)
# print(side)
if side+r<=R:
print("YES")
else:
print("NO")
```
Yes
| 97,175 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
n,R,r = map(int,input().strip().split())
if r>R:
print("NO")
elif 2*r>R:
if n==1:
print("YES")
else:
print("NO")
else:
import math
cnt = math.pi/(math.asin(r/(R-r)))
if cnt>n or abs(cnt-n)<=1e-6:
print("YES")
else:
print("NO")
```
Yes
| 97,176 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
import math
n,R,r=map(int,input().split())
if n==1 and R>=r:
print("YES")
elif (R-r)*math.sin(math.pi/n)+1e-9<r:
print("NO")
else:
print("YES")
```
Yes
| 97,177 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
from math import (asin, pi)
def calc(r1, r2):
res = r2 / r1
if res > 1:
return 0
elif 0.5 < res <= 1:
return 1
else:
tmp = r2 / (r1 - r2)
theta = asin(tmp)
return pi / theta
if __name__ == '__main__':
n, R, r = map(int, input().split())
ans = calc(R, r)
if ans >= n or abs(ans - n) <= 1e-6:
print("YES")
else:
print("NO")
```
Yes
| 97,178 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
#Code by Sounak, IIESTS
#------------------------------warmup----------------------------
import os
import sys
import math
from io import BytesIO, IOBase
from fractions import Fraction
from collections import defaultdict
from itertools import permutations
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,R,r=map(int,input().split())
y=2*r/math.sin(math.pi/n)+r
if y>=R:
print("YES")
else:
print("NO")
```
No
| 97,179 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
from math import asin, pi
n, R, r = map(int, input().split())
if r > R:
print('NO')
elif 2 * r > R:
print('YES' if n == 1 else 'NO')
else:
a = 2 * asin(r / (R - r))
maxx = (2 * pi) / a + 1
print('YES' if maxx >= n else 'NO')
```
No
| 97,180 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
import math
n, R, r = map(int, input().split())
if r > R: print("NO")
elif n < 3: print("YES") if n * r <= R else print("NO")
else: print("YES") if n*r <= (R-r)*math.pi else print("NO")
```
No
| 97,181 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.
Input
The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.
Output
Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".
Remember, that each plate must touch the edge of the table.
Examples
Input
4 10 4
Output
YES
Input
5 10 4
Output
NO
Input
1 10 10
Output
YES
Note
The possible arrangement of the plates for the first sample is:
<image>
Submitted Solution:
```
# cook your dish here
import math
l = list(map(int,input().split()))
if l[1] == l[2]:
if l[0] == 1:
print("YES")
else:
print("NO")
else:
x = math.asin(l[2]/(l[1]-l[2]))
n = math.pi / x
print(n)
if l[0]-n < 1e-9 or l[0]<=n:
print("YES")
else:
print("NO")
```
No
| 97,182 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
def mergesort(l,n):
if n==1:
return l,0
else:
a,ans1=mergesort(l[0:n//2],n//2)
b,ans2=mergesort(l[n//2:],n-n//2)
c,ans3=combine(a,n//2,b,n-n//2)
#print (c,ans1,ans2,ans3)
return (c,ans1+ans2+ans3)
def combine(l1,n,l2,m):
i=0
j=0
ans=[]
inversions=0
while (i<n and j<m):
if l1[i]<=l2[j]:
ans.append(l1[i])
i+=1
else:
ans.append(l2[j])
inversions += n-i
j+=1
if i==n:
for k in range(j,m):
ans.append(l2[k])
elif j==m:
for k in range(i,n):
ans.append(l1[k])
return (ans,inversions)
n = int(input())
start = input()
s = start[::-1]
loc = {}
for i in range(n):
if s[i] in loc:
loc[s[i]].append(i)
else:
loc[s[i]] = [i]
new = {}
for i in loc:
new[i] = loc[i][::-1]
arr = []
for i in range(n):
arr.append(new[start[i]].pop())
# print (arr)
print (mergesort(arr, len(arr))[1])
```
| 97,183 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
import functools
import datetime
import math
import collections
import heapq
maxn = int(2e5 + 5)
n = int(input())
s = input()
t = s[::-1]
a = [[] for i in range(31)]
# for i in range(30):
# a.append([])
tree = [0] * maxn
num = [0] * maxn
g = [0] * 30
def lowbit(x):
return x & (-x)
def tadd(x, val):
while x <= maxn:
tree[x] += val
x += lowbit(x)
def tsum(x):
ans = 0
while x > 0:
ans += tree[x]
x -= lowbit(x)
return ans
for i in range(0, len(t)):
ch = ord(t[i]) - ord('a')
a[ch].append(i)
for i in range(0, len(s)):
ch = ord(s[i]) - ord('a')
num[i] = a[ch][g[ch]]
g[ch] += 1
ans = 0
for i in range(0, len(s)):
tadd(num[i] + 1, 1)
# print(i+1-tsum(num[i]+1))
ans += i + 1 - tsum(num[i] + 1)
print(ans)
```
| 97,184 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
N = int(input())
S = input()
dic = {}
for i, s in enumerate(S):
if s in dic:
dic[s].append(i)
else:
dic[s] = [i]
X = list(range(N-1, -1, -1))
for c, idcs in dic.items():
for p, q in zip(idcs, reversed(idcs)):
X[p] = N-1-q
class BIT():
def __init__(self, N):
self.N = N
self.T = [0]*(N+1)
def add(self, i, x):
i += 1
while i <= self.N:
self.T[i] += x
i += i & -i
def _sum(self, i): #[0,i)
ret = 0
while i > 0:
ret += self.T[i]
i ^= i & -i
return ret
def sum(self, l, r): #[l,r)
return self._sum(r)-self._sum(l)
def lower_bound(self, w):
if w <= 0:
return 0
x = 0
k = 1<<self.N.bit_length()
while k:
if x+k <= self.N and self.T[x+k] < w:
w -= self.t[x+k]
x += k
k >>= 1
return x+1
b = BIT(N)
ans = 0
for i, x in enumerate(X):
ans += i - b._sum(x+1)
b.add(x, 1)
print(ans)
```
| 97,185 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
# cook your dish here
import heapq
import collections
from math import log2
import itertools
from functools import lru_cache
from sys import setrecursionlimit as srl
srl(2*10**6)
N = 200001
class fenwick:
def __init__(self,n):
self.n = n
self.arr = [0]*(n+1)
def update(self,ind,x):
if ind <= 0:
return
while ind <= self.n:
self.arr[ind] += x
ind += (ind)&(-ind)
def query(self,ind):
s = 0
while ind > 0:
s += self.arr[ind]
ind -= (ind)&(-ind)
return s
def solve(n,s):
fen = fenwick(n+1)
inv = 0
t = s[::-1]
chars = collections.defaultdict(collections.deque)
for i in range(n):
chars[ord(s[i])-ord('a')].append(i)
for i in range(n):
v = chars[ord(t[i])-ord('a')].popleft()
fen.update(v+1,1)
v += (i+1)-fen.query(v+1)
inv += v-i
return inv
n = int(input())
s = input()
print(solve(n,s))
```
| 97,186 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
def mergeSortswap(arr):
if len(arr) >1:
mid = len(arr)//2
L = arr[:mid]
R = arr[mid:] # into 2 halves
x = mergeSortswap(L) # Sorting the first half
y = mergeSortswap(R)
i = j = k = 0
z = 0
while i < len(L) and j < len(R):
if L[i] < R[j]:
arr[k] = L[i]
i+= 1
else:
arr[k] = R[j]
z += (mid-i)
j+= 1
k+= 1
while i < len(L):
arr[k] = L[i]
i+= 1
k+= 1
while j < len(R):
arr[k] = R[j]
j+= 1
k+= 1
return z+x+y
else:
return 0
import collections,bisect
n = int(input())
s = input().strip(' ')
d = collections.defaultdict(collections.deque)
s_reverse = s[::-1]
if s == s_reverse:
print(0)
else:
for i in range(n):
d[s_reverse[i]].append(i)
to_sort = []
for i in range(n):
ind = d[s[i]].popleft()
to_sort.append(ind)
c = mergeSortswap(to_sort)
print(c)
```
| 97,187 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
n = int(input())
s = input()
SZ = 30
lim = n + 233
g = [[] for i in range(SZ)]
a = [0] * SZ
for i in range(n):
cur = ord(s[i]) - 97
g[cur].append(i)
bit = [0] * (lim+1)
def lowbit(x):
return x & (-x)
def add(pos, val):
while pos <= lim:
bit[pos] += val
pos += lowbit(pos)
def query(pos):
res = 0
while pos:
res += bit[pos]
pos -= lowbit(pos)
return res
ans = 0
for i in range(n-1, -1, -1):
cur = ord(s[i]) - 97
val = g[cur][a[cur]]
ans += val - query(val+1)
a[cur] += 1
add(val+1, 1)
print(ans)
```
| 97,188 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
import sys
input=sys.stdin.readline
def segment_tree(ar,n):
data=[0]*n+ar.copy()
for i in range(n-1,0,-1):
data[i]+=(data[i*2]+data[i*2+1])
return data
def update(data,idx,n,value):
idx+=n
data[idx]+=value
while(idx>1):
idx//=2
data[idx]+=value
def summation(data,l,r,n):
l+=n
r+=n
maxi=0
while(l<r):
if(l%2!=0):
maxi+=data[l]
l+=1
if(r%2!=0):
r-=1
maxi+=data[r]
l//=2
r//=2
return maxi
n=int(input())
st=list(input())[:-1]
ar=[]
dic={}
for i in range(n):
ar.append(1)
if(st[i] in dic):
dic[st[i]].append(i)
else:
dic[st[i]]=[i]
ar[0]=0
for i in dic:
dic[i]=dic[i][::-1]
rev=st[::-1]
data=segment_tree(ar, n)
ans=0
for i in range(n):
cur=rev[i]
ind=dic[cur].pop()
ind1=summation(data, 0, ind+1, n)
tem=ind1-i
ans+=tem
if(tem):
update(data, 0, n, 1)
update(data,ind,n,-ar[ind])
print(ans)
```
| 97,189 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Tags: data structures, greedy, strings
Correct Solution:
```
from collections import defaultdict, Counter
class FenwickTree:
def __init__(self, n):
self.tree = [0] * (n + 1)
def _get(self, x):
res = 0
while x >= 1:
res += self.tree[x]
x -= x & -x
return res
def get(self, l, r):
return self._get(r) - self._get(l - 1)
def add(self, x, d):
while x < len(self.tree):
self.tree[x] += d
x += x & -x
def inversions(a):
n = len(a)
res = 0
count = FenwickTree(n)
for e in a:
res += count.get(e + 1, n)
count.add(e, 1)
return res
n = int(input())
s = input()
d = defaultdict(list)
for i, e in enumerate(s[::-1]):
d[e].append(i + 1)
c = Counter()
a = []
for e in s:
a.append(d[e][c[e]])
c[e] += 1
print(inversions(a))
```
| 97,190 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
from collections import defaultdict, deque
from heapq import heappush, heappop
from math import inf
ri = lambda : map(int, input().split())
ro = lambda : int(input())
class FW:
def __init__(self, N):
self.A = [0] * (N + 1)
def add(self, idx, val):
while idx < len(self.A):
self.A[idx] += val
idx += idx & -idx
def qry(self, idx):
sm = 0
while idx > 0:
sm += self.A[idx]
idx -= idx & -idx
return sm
def solve():
n = ro()
fw = FW(n)
s = input()
rev = s[::-1]
cnt = defaultdict(int)
indexes = defaultdict(list)
for i in range(n):
indexes[rev[i]].append(i)
inv = []
for i in range(n):
inv.append(indexes[s[i]][cnt[s[i]]])
cnt[s[i]] += 1
ans = 0
for i in range(n-1, -1, -1):
ans += fw.qry(inv[i]+1)
fw.add(inv[i]+1, 1)
print(ans)
t = 1
#t = int(input())
while t:
t -= 1
solve()
```
Yes
| 97,191 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
# aadiupadhyay
from heapq import heappop
import os.path
from math import gcd, floor, ceil
from collections import *
import sys
from heapq import *
mod = 1000000007
INF = float('inf')
def st(): return list(sys.stdin.readline().strip())
def li(): return list(map(int, sys.stdin.readline().split()))
def mp(): return map(int, sys.stdin.readline().split())
def inp(): return int(sys.stdin.readline())
def pr(n): return sys.stdout.write(str(n)+"\n")
def prl(n): return sys.stdout.write(str(n)+" ")
if os.path.exists('input.txt'):
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
def solve():
def update(ind, val):
while ind <= n:
BIT[ind] += val
ind += ind & -ind
def query(ind):
su = 0
while ind > 0:
su += BIT[ind]
ind -= ind & -ind
return su
n = inp()
s = ['&']+st()
d = defaultdict(list)
n += 1
for i in range(1, n):
d[s[i]].append(i)
BIT = [0]*(n+2)
ans = 0
p = [0]*(n+1)
for i in range(n-1, 0, -1):
last = d[s[n-i]].pop()
p[i] = last
for i in range(1, n):
c = query(n) - query(p[i])
ans += c
update(p[i], 1)
pr(ans)
for _ in range(1):
solve()
```
Yes
| 97,192 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
import os
import sys
from io import BytesIO, IOBase
_str = str
BUFSIZE = 8192
def str(x=b''):
return x if type(x) is bytes else _str(x).encode()
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = 'x' in file.mode or 'r' not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b'\n') + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode('ascii'))
self.read = lambda: self.buffer.read().decode('ascii')
self.readline = lambda: self.buffer.readline().decode('ascii')
def inp():
return sys.stdin.readline().rstrip()
def mpint():
return map(int, inp().split(' '))
def itg():
return int(inp())
# ############################## import
# from ACgenerator.Y_Testing import get_code
from collections import defaultdict
def discretization(iterable):
""" [1, 2, 4, 2, 5] -> [0, 1, 2, 1, 3] """
iterable = tuple(iterable)
restore = sorted(set(iterable))
return list(map(dict(zip(restore, range(len(restore)))).__getitem__, iterable)), restore
class Fenwick:
""" Simpler FenwickTree """
def __init__(self, x):
self.bit = [0] * x
def update(self, idx, x):
"""updates bit[idx] += x"""
while idx < len(self.bit):
self.bit[idx] += x
idx |= idx + 1
def query(self, end):
"""calc sum(bit[:end])"""
x = 0
while end:
x += self.bit[end - 1]
end &= end - 1
return x
def findkth(self, k):
"""
Find largest idx such that sum(bit[:idx]) < k
(!) different from pyrival (just removed the '=')
"""
idx = -1
for d in reversed(range(len(self.bit).bit_length())):
right_idx = idx + (1 << d)
if right_idx < len(self.bit) and k > self.bit[right_idx]:
idx = right_idx
k -= self.bit[idx]
return idx + 1
def inversion(iterable):
"""
the number of j s.t (a[i] > a[j]) and (i < j) for each i
Example:
inversion([2, 3, 4, 2, 1])
[1, 2, 2, 1, 0]
"""
iterable = discretization(iterable)[0]
index = len(iterable)
bit = Fenwick(index)
result = [0] * index
for item in reversed(iterable):
index -= 1
result[index] = bit.query(item)
bit.update(item, 1)
return result
def discrete_by_permutation(seq1, seq2):
""" 'teenager', 'generate' -> 61325074 """
seq1, seq2 = reversed(seq1), tuple(seq2)
d_stack = defaultdict(lambda: [])
for index, item in enumerate(seq2):
d_stack[item].append(index)
return list(reversed([d_stack[item].pop() for item in seq1]))
def bubble_step(s1, s2):
"""
:return the minimum step that have to perform
s1 to s2 with swapping the neighboring elements
"""
return sum(inversion(discrete_by_permutation(s1, s2)))
# ############################## main
def main():
# print("YES" if solve() else "NO")
# print("yes" if solve() else "no")
# solve()
print(solve())
# for _ in range(itg()):
# print(solve())
def solve():
_, s = itg(), inp()
return bubble_step(s, s[::-1])
DEBUG = 0
URL = 'https://codeforces.com/contest/1430/problem/E'
if __name__ == '__main__':
if DEBUG:
import requests # ImportError: cannot import name 'md5' from 'sys' (unknown location)
from ACgenerator.Y_Test_Case_Runner import TestCaseRunner
runner = TestCaseRunner(main, URL)
inp = runner.input_stream
print = runner.output_stream
runner.checking()
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
main()
# Please check!
```
Yes
| 97,193 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
#import modules here
import math,sys,os
#from itertools import permutations, combinations
from collections import defaultdict,deque,OrderedDict,Counter
import bisect as bi
#import heapq
from io import BytesIO, IOBase
mod=10**9+7
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
#input functions
def minp(): return map(int, sys.stdin.readline().rstrip().split())
def linp(): return list(map(int, sys.stdin.readline().rstrip().split()))
def inp(): return int(input())
def print_list(l):
print(' '.join(map(str,l)))
#functions
def BinarySearch(a,x):
i=bi.bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
else:
return -1
def gcd(a,b):
return math.gcd(a,b)
def is_prime(n):
"""returns True if n is prime else False"""
if n < 5 or n & 1 == 0 or n % 3 == 0:
return 2 <= n <= 3
s = ((n - 1) & (1 - n)).bit_length() - 1
d = n >> s
for a in [2, 325, 9375, 28178, 450775, 9780504, 1795265022]:
p = pow(a, d, n)
if p == 1 or p == n - 1 or a % n == 0:
continue
for _ in range(s):
p = (p * p) % n
if p == n - 1:
break
else:
return False
return True
def mulinverse(a):
return pow(a,mod-2,mod)
####################Let's Go Baby########################
from collections import defaultdict as dc
def reversePairs(nums):
if not nums:
return 0
res = 0
def merge(a,b):
nonlocal res
temp = []
i,j = 0,0
m = len(a)
n = len(b)
while i<m and j<n:
if a[i]<=b[j]:
temp.append(a[i])
i+=1
else:
temp.append(b[j])
j+=1
res+=m-i
if i<m:
temp+=a[i:]
else:
temp+=b[j:]
return temp
def mergesort(L):
length = len(L)
if length==1:
return L
k = length>>1
a = mergesort(L[:k])
b = mergesort(L[k:])
return merge(a,b)
mergesort(nums)
return res
n = inp()
s = input()
t = s[::-1]
res = 0
flag = [0]*(n)
dic = dc(int)
for i in range(n):
c = t[i]
key = s.find(c,dic[c])
dic[c] = key+1
flag[key] = i
#print(dic)
print(reversePairs(flag))
```
Yes
| 97,194 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
n=int(input())
s=input()
if s==s[::-1]:
print(0)
else:
print(n//2)
```
No
| 97,195 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
import sys
input=sys.stdin.readline
def update(inp,add,ar):
global n
while(inp<n):
ar[inp]+=add
inp+=(inp&(-inp))
def fun1(inp,ar):
global n
ans=0
while(inp):
ans+=ar[inp]
inp-=(inp&(-inp))
return ans
n=int(input())
st=list(input())
st=st[:-1]
li=[0]*(n+1)
rev=st[::-1]
dic={}
for i in range(n):
if(st[i] in dic):
dic[st[i]].append(i)
else:
dic[st[i]]=[i]
for i in dic:
dic[i]=dic[i][::-1]
ans=0
pas=0
j=0
for i in range(n):
while(st[j]=='-'):
j+=1
pas+=1
if(st[j]==rev[i]):
dic[st[j]].pop()
j+=1
else:
ind=dic[rev[i]][-1]+1
st[ind-1]='-'
dic[rev[i]].pop()
su=fun1(ind,li)
ans+=(ind-su-j-1+pas)
update(ind,1,li)
print(ans)
```
No
| 97,196 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
def update(inp,add,ar):
global n
while(inp<n):
ar[inp]+=add
inp+=(inp&(-inp))
def fun1(inp,ar):
global n
ans=0
while(inp):
ans+=ar[inp]
inp-=(inp&(-inp))
return ans
n=int(input())
st=list(input())
li=[0]*(n+1)
rev=st[::-1]
dic={}
for i in range(n):
if(st[i] in dic):
dic[st[i]].append(i)
else:
dic[st[i]]=[i]
for i in dic:
dic[i]=dic[i][::-1]
ans=0
pas=0
j=0
for i in range(n):
while(st[j]=='-'):
j+=1
pas+=1
if(st[j]==rev[i]):
dic[st[j]].pop()
j+=1
else:
ind=dic[rev[i]][-1]+1
st[ind-1]='-'
dic[rev[i]].pop()
su=fun1(ind,li)
ans+=(max(ind-su-j-1+pas,0))
update(ind,1,li)
print(ans)
```
No
| 97,197 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
You are given a string s. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last letter before the reversal — and so on. For example, if your goal is to reverse the string "abddea", you should get the string "aeddba". To accomplish your goal, you can swap the neighboring elements of the string.
Your task is to calculate the minimum number of swaps you have to perform to reverse the given string.
Input
The first line contains one integer n (2 ≤ n ≤ 200 000) — the length of s.
The second line contains s — a string consisting of n lowercase Latin letters.
Output
Print one integer — the minimum number of swaps of neighboring elements you have to perform to reverse the string.
Examples
Input
5
aaaza
Output
2
Input
6
cbaabc
Output
0
Input
9
icpcsguru
Output
30
Note
In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps.
Since the string in the second example is a palindrome, you don't have to do anything to reverse it.
Submitted Solution:
```
a=int(input())
b=str(input())
if a==len(b):
print(b[::-1])
else:
pass
```
No
| 97,198 |
Provide tags and a correct Python 3 solution for this coding contest problem.
You are given one integer n (n > 1).
Recall that a permutation of length n is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2, 3, 1, 5, 4] is a permutation of length 5, but [1, 2, 2] is not a permutation (2 appears twice in the array) and [1, 3, 4] is also not a permutation (n = 3 but there is 4 in the array).
Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
You have to answer t independent test cases.
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Input
The first line of the input contains one integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.
The only line of the test case contains one integer n (2 ≤ n ≤ 100) — the length of the permutation you have to find.
Output
For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).
If there are several answers, you can print any. It can be proven that the answer exists for each n > 1.
Example
Input
2
2
5
Output
2 1
2 1 5 3 4
Tags: constructive algorithms, probabilities
Correct Solution:
```
for _ in range(int(input())):
n = int(input())
res = [n]
for i in range(1,n):
res.append(i)
print(*res)
```
| 97,199 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.