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.
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q1q2... qk. The algorithm consists of two steps:
1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2.
2. Rearrange the letters of the found subsequence randomly and go to step 1.
Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.
Sereja wants to test his algorithm. For that, he has string s = s1s2... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
Input
The first line contains non-empty string s, its length (n) doesn't exceed 105. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.
The second line contains integer m (1 ≤ m ≤ 105) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
Output
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
Examples
Input
zyxxxxxxyyz
5
5 5
1 3
1 11
1 4
3 6
Output
YES
YES
NO
YES
NO
Note
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
Submitted Solution:
```
s = input()
n = int(input())
tests = []
for _ in range(n):
a, b = map(int, input().split())
a -= 1
b -= 1
tests.append((a, b))
count_left = {}
count_left[-1] = (0, 0, 0)
x, y, z = 0, 0, 0
for i, c in enumerate(s):
if c == 'x':
x += 1
elif c == 'y':
y += 1
elif c == 'z':
z += 1
count_left[i] = (x, y, z)
def count(a, b):
x2, y2, z2 = count_left[b]
x1, y1, z1 = count_left[a - 1]
x = x2 - x1
y = y2 - y1
z = z2 - z1
return x,y,z
def is_valid(s, a, b):
x,y,z = count(a, b)
diffs = map(abs, [x-y, x-z, y-z])
return not any(d > 1 for d in diffs)
for a, b in tests:
print("YES" if is_valid(s, a, b) else "NO")
```
No
| 98,200 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q1q2... qk. The algorithm consists of two steps:
1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2.
2. Rearrange the letters of the found subsequence randomly and go to step 1.
Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.
Sereja wants to test his algorithm. For that, he has string s = s1s2... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
Input
The first line contains non-empty string s, its length (n) doesn't exceed 105. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.
The second line contains integer m (1 ≤ m ≤ 105) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
Output
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
Examples
Input
zyxxxxxxyyz
5
5 5
1 3
1 11
1 4
3 6
Output
YES
YES
NO
YES
NO
Note
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
Submitted Solution:
```
import os,sys
from io import BytesIO, IOBase
from collections import deque, Counter,defaultdict as dft
from heapq import heappop ,heappush
from math import log,sqrt,factorial,cos,tan,sin,radians,log2,ceil,floor
from bisect import bisect,bisect_left,bisect_right
from decimal import *
import sys,threading
from itertools import permutations, combinations
from copy import deepcopy
input = sys.stdin.readline
ii = lambda: int(input())
si = lambda: input().rstrip()
mp = lambda: map(int, input().split())
ms= lambda: map(str,input().strip().split(" "))
ml = lambda: list(mp())
mf = lambda: map(float, input().split())
alphs = "abcdefghijklmnopqrstuvwxyz"
# stuff you should look for
# int overflow, array bounds
# special cases (n=1?)
# do smth instead of nothing and stay organized
# WRITE STUFF DOWN
# DON'T GET STUCK ON ONE APPROACH
#
def solve():
s=input()
n=len(s)
dct={'x':0,'y':1,'z':2}
ls=[[0,0,0] for i in range(n+1)]
for i in range(n):
idx=dct[s[i]]
for j in range(3):
ls[i+1][j]+=ls[i][j]
ls[i+1][idx]+=1
#print(ls)
m=ii()
for _ in range(m):
l,r=mp()
if r-l+1<3:
print("YES")
continue
x=ls[r][0]-ls[l-1][0]
y=ls[r][1]-ls[l-1][1]
z=ls[r][2]-ls[l-1][2]
mx=max(x,y,z)
l=[mx-x,mx-y,mx-z]
l.sort()
#print(l)
if l==[0,1,1] or l==[0,0,0]:
print("YES")
else:print("NO")
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")
def print(*args, **kwargs):
"""Prints the values to a stream, or to sys.stdout by default."""
sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
at_start = True
for x in args:
if not at_start:
file.write(sep)
file.write(str(x))
at_start = False
file.write(kwargs.pop("end", "\n"))
if kwargs.pop("flush", False):
file.flush()
if sys.version_info[0] < 3:
sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
else:
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
# endregion
if __name__ == "__main__":
tc=1
#tc = ii()
for i in range(tc):
solve()
```
No
| 98,201 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Sereja loves all sorts of algorithms. He has recently come up with a new algorithm, which receives a string as an input. Let's represent the input string of the algorithm as q = q1q2... qk. The algorithm consists of two steps:
1. Find any continuous subsequence (substring) of three characters of string q, which doesn't equal to either string "zyx", "xzy", "yxz". If q doesn't contain any such subsequence, terminate the algorithm, otherwise go to step 2.
2. Rearrange the letters of the found subsequence randomly and go to step 1.
Sereja thinks that the algorithm works correctly on string q if there is a non-zero probability that the algorithm will be terminated. But if the algorithm anyway will work for infinitely long on a string, then we consider the algorithm to work incorrectly on this string.
Sereja wants to test his algorithm. For that, he has string s = s1s2... sn, consisting of n characters. The boy conducts a series of m tests. As the i-th test, he sends substring slisli + 1... sri (1 ≤ li ≤ ri ≤ n) to the algorithm input. Unfortunately, the implementation of his algorithm works too long, so Sereja asked you to help. For each test (li, ri) determine if the algorithm works correctly on this test or not.
Input
The first line contains non-empty string s, its length (n) doesn't exceed 105. It is guaranteed that string s only contains characters: 'x', 'y', 'z'.
The second line contains integer m (1 ≤ m ≤ 105) — the number of tests. Next m lines contain the tests. The i-th line contains a pair of integers li, ri (1 ≤ li ≤ ri ≤ n).
Output
For each test, print "YES" (without the quotes) if the algorithm works correctly on the corresponding test and "NO" (without the quotes) otherwise.
Examples
Input
zyxxxxxxyyz
5
5 5
1 3
1 11
1 4
3 6
Output
YES
YES
NO
YES
NO
Note
In the first example, in test one and two the algorithm will always be terminated in one step. In the fourth test you can get string "xzyx" on which the algorithm will terminate. In all other tests the algorithm doesn't work correctly.
Submitted Solution:
```
def f(x, y, z): return abs(y - x) > 1 or abs(z - x) > 1
t = input()
n, p = len(t), {'x': 0, 'y': 1, 'z': 2}
s = [[0] * (n + 1) for i in range(3)]
for i, c in enumerate(t, 1): s[p[c]][i] = 1
for i in range(3):
for j in range(1, n): s[i][j + 1] += s[i][j]
a, b, c = s
q = [map(int, input().split()) for i in range(int(input()))]
d = ['YES'] * len(q)
for i, (l, r) in enumerate(q):
if r - l > 1 and f(a[r] - a[l - 1], b[r] - b[l - 1], c[r] - c[l - 1]): d[i] = 'NO'
print('\n'.join(d))
```
No
| 98,202 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Tags: dp, sortings
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n=int(input())
d=dict()
b=[]
for j in range(n):
x,y=map(int,input().split())
d[x]=y
b.append(x)
b.sort()
dp=[[float("inf"),float("inf")] for i in range(n)]
dp[0][1]=d[b[0]]
for i in range(1,n):
dp[i][1]=min(dp[i][1],min(dp[i-1])+d[b[i]])
j=i-1
s=b[i]
p=1
while(j>=0):
dp[i][0]=min(dp[i][0],dp[j][1]+s-p*b[j])
s+=b[j]
j+=-1
p+=1
print(min(dp[n-1]))
```
| 98,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Tags: dp, sortings
Correct Solution:
```
n=int(input() )
a=sorted([list(map(int, input().split())) for _ in range(n)])
a.append( [ a[-1][0]+1, 0] )
res=[10**15]*(n+1)
res[0]=a[0][1]
ind=a[0][0]
for i in range(n+1):
acc=0
for j in range(i+1,n+1):
res[j]=min(res[j], res[i]+acc+a[j][1])
acc=acc+a[j][0]-a[i][0]
print(res[-1])
```
| 98,204 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Tags: dp, sortings
Correct Solution:
```
from typing import Tuple, List
def compute(n: int, m: List[Tuple[int, int]]) -> int:
m = sorted(m)
state = [0] * (n + 1)
state[n - 1] = m[n - 1][1]
for i in range(n - 2, -1, -1):
min_cost = state[i + 1]
acc = 0
for j in range(i + 2, n + 1):
min_cost = min(min_cost, state[j] + abs(m[i][0] - m [j - 1][0]) + acc)
acc += abs(m[i][0] - m [j - 1][0])
state[i] = min_cost + m[i][1]
return state[0]
if __name__ == '__main__':
N = int(input())
M = [input().split() for _ in range(N)]
M = [(int(x), int(c)) for x, c in M]
print(compute(N, M))
```
| 98,205 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Tags: dp, sortings
Correct Solution:
```
import math
R = lambda: map(int, input().split())
n = int(input())
arr = sorted(list(R()) for i in range(n))
dp = [[math.inf, math.inf] for i in range(n + 1)]
dp[0][0] = arr[0][1]
for i in range(1, n):
dp[i][0] = min(dp[i - 1]) + arr[i][1]
sm = arr[i][0]
for j in range(i - 1, -1, -1):
dp[i][1] = min(dp[i][1], dp[j][0] + sm - (i - j) * arr[j][0])
sm += arr[j][0]
print(min(dp[n - 1]))
```
| 98,206 |
Provide tags and a correct Python 3 solution for this coding contest problem.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Tags: dp, sortings
Correct Solution:
```
n = int(input())
arr = []
for i in range(n):
p, c = map(int, input().split())
arr += [[p, c]]
arr.sort()
dp = [[10**100, 10**100] for i in range(n)]
dp[0][0] = 10**100
dp[0][1] = arr[0][1]
for i in range(1, n):
dp[i][1] = min(dp[i-1]) + arr[i][1]
tot = arr[i][0]
count = 1
for j in range(i-1, -1, -1):
temp = dp[j][1] + tot - count*arr[j][0]
dp[i][0] = min(dp[i][0], temp)
tot += arr[j][0]
count += 1
print(min(dp[-1]))
```
| 98,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Submitted Solution:
```
# http://codeforces.com/problemset/problem/38/E
# let's go rolling
#input = raw_input
def go_rolling(a):
"""
a: [[x0, c0], ..., [xn, cn]]
"""
n = len(a)
w0 = [False for e in range(n)]
w0[0] = True
w1 = [0 for e in range(n)]
w1[0] = a[0][1]
for d in range(1, n):
k = 0
for p in range((d - 1), -1, -1):
if w0[p]:
k = w1[d - 1] + (a[d][0] - a[p][0])
break
if k >= w1[d - 1] + a[d][1]:
w0[d] = True
k = w1[d - 1] + a[d][1]
w1[d] = k
return w1[n - 1]
def fun():
n = int(input())
a = [[0, 0] for c in range (n)]
for c in range(n):
a[c][0], a[c][1] = map(int, input().split())
a.sort()
print(go_rolling(a))
fun()
```
No
| 98,208 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Submitted Solution:
```
inf = 10**(10)
n = int(input())
p = n
x = []
c = []
ls = []
while(p>0):
p=p-1
a = input()
A = list(map(int,list(a.split())))
ls.append(A)
ls.sort()
for i in range(n):
x.append(ls[i][0])
c.append(ls[i][1])
dp = [[0 for i in range(n)] for j in range(2)]
f = [0]*(n)
dp[0][0] = inf
dp[1][0] = c[0]
if n>1:
dp[0][1] = abs(x[1]-x[0])+c[0]
dp[1][1] = c[0]+c[1]
f[1] = 0
for i in range(2,n):
if dp[0][i-1]+abs(x[i]-x[f[i-1]])<dp[1][i-1]+abs(x[i]-x[i-1]):
dp[0][i] = dp[0][i-1]+abs(x[i]-x[f[i-1]])
f[i] = f[i-1]
else:
dp[0][i] =dp[1][i-1]+abs(x[i]-x[i-1])
f[i] = i-1
dp[1][i] = min(dp[0][i-1],dp[1][i-1])+c[i]
print(min(dp[0][n-1],dp[1][n-1]))
```
No
| 98,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Submitted Solution:
```
li=[]
for _ in range(int(input())):
x,y= map(int, input().split())
li.append([x,y])
li.sort(key = lambda x: x[0])
s=li[0][1]
pre=0
for i in range(1, len(li)):
val1= li[i][1]
val2= abs( abs(li[pre][0]) -abs(li[i][0]) )
if val1 <= val2:
pre= i
s= s+ min( val1, val2)
print(s)
```
No
| 98,210 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
On a number axis directed from the left rightwards, n marbles with coordinates x1, x2, ..., xn are situated. Let's assume that the sizes of the marbles are infinitely small, that is in this task each of them is assumed to be a material point. You can stick pins in some of them and the cost of sticking in the marble number i is equal to ci, number ci may be negative. After you choose and stick the pins you need, the marbles will start to roll left according to the rule: if a marble has a pin stuck in it, then the marble doesn't move, otherwise the marble rolls all the way up to the next marble which has a pin stuck in it and stops moving there. If there is no pinned marble on the left to the given unpinned one, it is concluded that the marble rolls to the left to infinity and you will pay an infinitely large fine for it. If no marble rolled infinitely to the left, then the fine will consist of two summands:
* the sum of the costs of stuck pins;
* the sum of the lengths of the paths of each of the marbles, that is the sum of absolute values of differences between their initial and final positions.
Your task is to choose and pin some marbles in the way that will make the fine for you to pay as little as possible.
Input
The first input line contains an integer n (1 ≤ n ≤ 3000) which is the number of marbles. The next n lines contain the descriptions of the marbles in pairs of integers xi, ci ( - 109 ≤ xi, ci ≤ 109). The numbers are space-separated. Each description is given on a separate line. No two marbles have identical initial positions.
Output
Output the single number — the least fine you will have to pay.
Examples
Input
3
2 3
3 4
1 2
Output
5
Input
4
1 7
3 1
5 10
6 1
Output
11
Submitted Solution:
```
mod=10**9+7
inf= "a"
dp=[[inf for i in range(3001)] for j in range(3001)]
li=[]
N= int(input())
for _ in range(N):
x,y= map(int, input().split())
li.append([x,y])
li.sort(key = lambda x: x[0])
def sol( n , pre):
if n >= N:
return 0
if dp[n][pre]!=inf :
return dp[n][pre]
x= li[n][1] + sol(n+1, n)
x=x%mod
y= abs( li[n][0]-li[pre][0] ) + sol(n+1, pre)
y=y%mod
dp[n][pre] = (min(x,y))%mod
return dp[n][pre]
print( sol(1, 0)+ li[0][1] )
```
No
| 98,211 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
from fractions import Fraction
import sys
sys.setrecursionlimit(1000*100)
A=int(input())
p=[]
c=1
for _ in range(600):
p.append(c)
c*=12
r=[]
for i in range(600):
for j in range(i+1):
if p[j]+p[i-j]==A:
r.append(i+1)
break
s=set()
for i in r:
for j in range(i):
v=p[j]+p[i-1-j]
if v!=A:
s.add(v)
if len(r)==0:
print('NO')
else:
print('YES')
print(min(len(r),1000))
print(' '.join(map(str,r[:1000])))
print(min(len(s),1000))
if len(s)>0:
print(' '.join(map(str,sorted(s)[:1000])))
#c=int(input())
#a,b=tuple(map(int,input().split()))
#edges=dict((i,[]) for i in range(1,c+1))
#children=filter(lambda x: x != p, edges[r])
#cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
#if dp[r] is not None:
#chr(ord('a')+1)
#''.join(['a','b','c'])
#sys.exit()
# Made By Mostafa_Khaled
```
| 98,212 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**302
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if a <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if a <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
```
| 98,213 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
from fractions import Fraction
import sys
sys.setrecursionlimit(1000*100)
A=int(input())
p=[]
c=1
for _ in range(600):
p.append(c)
c*=12
r=[]
for i in range(600):
for j in range(i+1):
if p[j]+p[i-j]==A:
r.append(i+1)
break
s=set()
for i in r:
for j in range(i):
v=p[j]+p[i-1-j]
if v!=A:
s.add(v)
if len(r)==0:
print('NO')
else:
print('YES')
print(min(len(r),1000))
print(' '.join(map(str,r[:1000])))
print(min(len(s),1000))
if len(s)>0:
print(' '.join(map(str,sorted(s)[:1000])))
#c=int(input())
#a,b=tuple(map(int,input().split()))
#edges=dict((i,[]) for i in range(1,c+1))
#children=filter(lambda x: x != p, edges[r])
#cs.sort(key=lambda x:Fraction(x[0],x[1]),reverse=True)
#if dp[r] is not None:
#chr(ord('a')+1)
#''.join(['a','b','c'])
#sys.exit()
```
| 98,214 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
A = int(input())
T = A
x = 0
while (A % 12 == 0):
A = A // 12
x = x + 1
b = [0] * 1000
b[1] = 2
b[2] = 13
for i in range(3, 606):
b[i] = 13 * b[i - 1] - 12 * b[i - 2]
y = 1
while (b[y] < A):
y = y + 1
if (b[y] != A):
print("NO")
else:
print("YES\n1")
t = 2 * x + y
print(t)
c = [0] * 1000
d = 0
for i in range((t - 1) // 2 + 1):
if (i != x and t - 2 * i != 0):
c[d] = b[t - 2 * i] * 12 ** i
d = d + 1
print(d)
for i in range(d):
print(c[d - i - 1])
```
| 98,215 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
A=int(input())
res=[]
for i in range(300):
for j in range(300):
if(pow(12,i)+pow(12,j)==A):
res.append([i,j])
if(len(res)==0):
print("NO")
quit()
print("YES")
print(1)
sm=sum(res[0])
print(sm+1)
print(sm//2)
for i in range(sm//2,-1,-1):
V=pow(12,i)+pow(12,sm-i)
if(V!=A):
print(V)
```
| 98,216 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Tags: math
Correct Solution:
```
a = int(input())
ans = set()
def work(i, z, y):
z.add(y)
# print(">>", i, y)
if y == a:
ans.add(i)
if len(z) > 1000:
z.remove(max(z))
pos = [set(), set([2])]
for i in range(2):
for x in pos[i]:
if x == a:
ans.add(i)
def dfs(i, last, cur):
if i > 988:
return
while len(pos) - 1 < i:
pos.append(set())
if len(pos[i]) == 0 and cur > a:
return
if cur in pos[i]:
return
work(i, pos[i], cur)
dfs(i + 1, cur, last * 12)
dfs(i + 1, cur, cur * 13 - last * 12)
dfs(2, 2, 13)
if len(ans) == 0:
print("NO")
else:
print("YES")
count = 0
print(min(1000, len(ans)))
z = set()
for i in sorted(ans):
if count < 1000:
print(i)
count += 1
for y in pos[i]:
if y != a:
z.add(y)
if len(z) > 1000:
z.remove(max(z))
count = 0
print(min(1000, len(z)))
for i in sorted(z):
print(i)
count += 1
if count >= 1000:
break
```
| 98,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
A = int(input())
T = A
x = 0
while (A % 12 == 0):
A = A / 12
x = x + 1
b = [0] * 1000
b[1] = 2
b[2] = 13
for i in range(3, 603):
b[i] = 13 * b[i - 1] - 12 * b[i - 2]
y = 1
while (b[y] < A):
y = y + 1
if (b[y] != A):
print("NO")
else:
print("YES\n1")
t = 2 * x + y
print(t)
c = [0] * 1000
d = 0
for i in range((t - 1) // 2 + 1):
if (i != x and t - 2 * i != 0):
c[d] = b[t - 2 * i] * 12 ** i
d = d + 1
print(d)
for i in range(d):
print(c[d - i - 1])
```
No
| 98,218 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**302
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if b <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if b <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
```
No
| 98,219 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 2000):
d[j] = set()
big = 10**333
for i in range(1, 2000):
ok = 0
for a, b in d[i]:
c = 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
break
n = int(input())
g = set()
for i in range(1, 2000):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
```
No
| 98,220 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Berland scientists noticed long ago that the world around them depends on Berland population. Due to persistent research in this area the scientists managed to find out that the Berland chronology starts from the moment when the first two people came to that land (it is considered to have happened in the first year). After one Berland year after the start of the chronology the population had already equaled 13 people (the second year). However, tracing the population number during the following years was an ultimately difficult task, still it was found out that if di — the number of people in Berland in the year of i, then either di = 12di - 2, or di = 13di - 1 - 12di - 2. Of course no one knows how many people are living in Berland at the moment, but now we can tell if there could possibly be a year in which the country population equaled A. That's what we ask you to determine. Also, if possible, you have to find out in which years it could be (from the beginning of Berland chronology). Let's suppose that it could be in the years of a1, a2, ..., ak. Then you have to define how many residents could be in the country during those years apart from the A variant. Look at the examples for further explanation.
Input
The first line contains integer A (1 ≤ A < 10300). It is guaranteed that the number doesn't contain leading zeros.
Output
On the first output line print YES, if there could be a year in which the total population of the country equaled A, otherwise print NO.
If the answer is YES, then you also have to print number k — the number of years in which the population could equal A. On the next line you have to output precisely k space-separated numbers — a1, a2, ..., ak. Those numbers have to be output in the increasing order.
On the next line you should output number p — how many variants of the number of people could be in the years of a1, a2, ..., ak, apart from the A variant. On each of the next p lines you have to print one number — the sought number of residents. Those number also have to go in the increasing order.
If any number (or both of them) k or p exceeds 1000, then you have to print 1000 instead of it and only the first 1000 possible answers in the increasing order.
The numbers should have no leading zeros.
Examples
Input
2
Output
YES
1
1
0
Input
3
Output
NO
Input
13
Output
YES
1
2
0
Input
1729
Output
YES
1
4
1
156
Submitted Solution:
```
d = {1: {(2, 13)}}
for j in range(2, 600):
d[j] = set()
big = 10**302
h = -1
for i in range(1, 600):
ok = 0
for a, b in d[i]:
c = 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
c = 13*b - 12*a
if c <= big:
ok = 1
d[i+1].add((b, c))
if not ok:
h = i
break
n = int(input())
g = set()
for i in range(1, h):
for a, b in d[i]:
if a == n:
g.add(i)
if len(g) > 0:
print('YES')
print(len(g))
print(*list(sorted(g)))
c = []
for i in g:
for a, b in d[i]:
if a != n:
c += [a]
c = list(sorted(list(set(c))))
if len(c) > 1000:
c = c[:1000]
print(len(c))
print(*c)
else:
print('NO')
```
No
| 98,221 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Tags: constructive algorithms
Correct Solution:
```
# https://codeforces.com/problemset/problem/45/J
def solve(n, m):
if n*m == 1:
return [[1]]
if n*m <= 3 or (n == 2 and m == 2):
return -1
a = [[0] * m for _ in range(n)]
base1 = 1
base2 = (n*m) // 2 + 1
for i in range(n):
for j in range(m):
if (i+j) % 2 == 0:
a[i][j] = base2
base2+=1
else:
a[i][j] = base1
base1+=1
return a
def pr(a, n, m):
if a == -1:
print(a)
else:
for _ in range(n):
print(' '.join([str(x) for x in a[_]]))
n, m = map(int, input().split())
pr(solve(n, m), n, m)
```
| 98,222 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Tags: constructive algorithms
Correct Solution:
```
n, m = map(int, input().split())
transpose = False
if n > m:
n, m = m, n
transpose = True
if n == 1 and m == 1:
print(1)
import sys; sys.exit()
if (n == 1 and m < 4) or (n == 2 and m < 3):
print(-1)
import sys; sys.exit()
if n == 1:
row = m * [None]
for c in range(1, m, 2):
row[c] = (c + 1) // 2
first = m // 2 + 1
for c in range(0, m, 2):
row[c] = first + c // 2
result = [row]
else:
result = n * [None]
for r in range(n):
result[r] = [ c for c in range(r * m + 1, (r + 1) * m + 1) ]
for r in range(1, n):
for c in range(r % 2, m, 2):
result[r - 1][c], result[r][c] = result[r][c], result[r - 1][c]
if transpose:
arr = m * [None]
for c in range(m):
arr[c] = [ result[r][c] for r in range(n) ]
result = arr
for row in result:
print(' '.join(map(str, row)))
```
| 98,223 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Tags: constructive algorithms
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def main():
n,m = map(int,input().split())
arr = [[0]*m for _ in range(n)]
if m >= 4:
arr[0] = list(range(m-1,0,-2))+list(range(m,0,-2))
for i in range(1,n):
for j in range(m):
arr[i][j] = arr[i-1][j]+m
elif n >= 4:
for ind,i in enumerate(list(range(n-1,0,-2))+list(range(n,0,-2))):
arr[ind][0] = i
for j in range(1,m):
for i in range(n):
arr[i][j] = arr[i][j-1]+n
else:
if n == 2 and m == 3:
arr = [[3,6,2],[5,1,4]]
elif n == 3 and m == 2:
arr = [[3,5],[6,1],[2,4]]
elif n == 3 and m == 3:
arr = [[1,7,4],[3,9,6],[5,2,8]]
elif n == 1 and m == 1:
arr = [[1]]
else:
print(-1)
return
for i in arr:
print(*i)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
```
| 98,224 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Tags: constructive algorithms
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
z=n*m
y,a=z-(z%2==0),[[] for _ in range(n)]
for i in range(n):
for j in range(m):
a[i].append(y)
y-=2
if y<0:
y=z-(z%2)
for i in range(n):
for j in range(m):
if i+1<n and abs(a[i][j]-a[i+1][j])<2:
if j>0 and abs(a[i][j-1]-a[i+1][j])>1:
a[i][j-1],a[i][j]=a[i][j],a[i][j-1]
elif j+1<m and abs(a[i][j+1]-a[i+1][j])>1:
a[i][j+1],a[i][j]=a[i][j], a[i][j+1]
for i in range(n):
for j in range(m):
if (j+1<m and abs(a[i][j]-a[i][j+1])<2) or (i+1<n and abs(a[i][j]-a[i+1][j])<2):
print(-1)
return
for i in a:
print(*i)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 98,225 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Tags: constructive algorithms
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
z,f=n*m,1
y,a=z-(z%2==0),[[] for _ in range(n)]
for i in range(n):
for j in range(m):
a[i].append(y)
y-=2
if y<0:
y=z-(z%2)
for i in range(n):
for j in range(m):
if i+1<n and abs(a[i][j]-a[i+1][j])<2:
if j>0 and abs(a[i][j-1]-a[i+1][j])>1:
a[i][j-1],a[i][j]=a[i][j],a[i][j-1]
elif j+1<m and abs(a[i][j+1]-a[i+1][j])>1:
a[i][j+1],a[i][j]=a[i][j], a[i][j+1]
for i in range(n):
for j in range(m):
if (j+1<m and abs(a[i][j]-a[i][j+1])<2) or (i+1<n and abs(a[i][j]-a[i+1][j])<2):
f=0
break
if not f:
break
if f:
for i in a:
print(*i)
else:
print(-1)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 98,226 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# https://codeforces.com/problemset/problem/45/J
def solve(n, m):
if n*m == 1:
return [1]
if n*m < 5:
return [-1]
if n == 2 and m == 3:
return [[1,5,3], [4,2,6]]
a = [[0] * m for _ in range(n)]
base1 = 1
base2 = (n*m+1) // 2 + 1
for i in range(n):
for j in range(m):
if (i+j) % 2 == 0:
a[i][j] = base1
base1+=1
else:
a[i][j] = base2
base2+=1
return a
def pr(a, n, m):
if len(a) == 1:
print(a[0])
else:
for _ in range(n):
print(' '.join([str(x) for x in a[_]]))
n, m = map(int, input().split())
pr(solve(n, m), n, m)
```
No
| 98,227 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
def some_random_function():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function5():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
import os,sys
from io import BytesIO,IOBase
def main():
n,m = map(int,input().split())
arr = [[0]*m for _ in range(n)]
if m >= 4:
arr[0] = list(range(m-1,0,-2))+list(range(m,0,-2))
for i in range(1,n):
for j in range(m):
arr[i][j] = arr[i-1][j]+m
elif n >= 4:
for i,ind in enumerate(list(range(m-1,0,-2))+list(range(m,0,-2))):
arr[ind][0] = i
for j in range(1,m):
for i in range(n):
arr[i][j] = arr[i][j-1]+n
else:
if n == 2 and m == 3:
arr = [[3,6,2],[5,1,4]]
elif n == 3 and m == 2:
arr = [[3,5],[6,1],[2,4]]
elif n == 3 and m == 3:
arr = [[1,7,4],[3,9,6],[5,2,8]]
elif n == 1 and m == 1:
arr = [[1]]
else:
print(-1)
return
for i in arr:
print(*i)
#Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
def some_random_function1():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function2():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function3():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function4():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function6():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function7():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
def some_random_function8():
"""due to the fast IO template, my code gets caught in
plag check for no reason. That is why, I am making
random functions"""
x = 10
x *= 100
i_dont_know = x
why_am_i_writing_this = x*x
print(i_dont_know)
print(why_am_i_writing_this)
if __name__ == '__main__':
main()
```
No
| 98,228 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
if n==1 and m==1:
print(1)
elif n<=2 and m<=2:
print(-1)
else:
a=[]
z=1
for i in range(n):
a.append([])
for j in range(m):
a[-1].append(z)
z+=1
for j in range(1,m,2):
for i in range(n-1):
a[i][j],a[i+1][j]=a[i+1][j],a[i][j]
for i in a:
print(*i)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,229 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Vasya is a Greencode wildlife preservation society proponent. One day he found an empty field nobody owned, divided it into n × m squares and decided to plant a forest there. Vasya will plant nm trees of all different heights from 1 to nm. For his forest to look more natural he wants any two trees growing in the side neighbouring squares to have the absolute value of difference in heights to be strictly more than 1. Help Vasya: make the plan of the forest planting for which this condition is fulfilled.
Input
The first line contains two space-separated integers n and m (1 ≤ n, m ≤ 100) — the number of rows and columns on Vasya's field
Output
If there's no solution, print -1. Otherwise, print n lines containing m numbers each — the trees' planting plan. In every square of the plan the height of a tree that should be planted on this square should be written. If there are several solutions to that problem, print any of them.
Examples
Input
2 3
Output
3 6 2
5 1 4
Input
2 1
Output
-1
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
import sys
from io import BytesIO, IOBase
def main():
n,m=map(int,input().split())
if n==1 and m==1:
print(1)
elif n<2 or m<2 or (n==2 and m==2):
print(-1)
else:
a=[]
z=1
for i in range(n):
a.append([])
for j in range(m):
a[-1].append(z)
z+=1
for j in range(1,m,2):
for i in range(n-1):
a[i][j],a[i+1][j]=a[i+1][j],a[i][j]
for i in a:
print(*i)
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,230 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Tags: constructive algorithms, data structures, trees
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
po=[1]
for i in range(30):
po.append(po[-1]*2)
n,m=map(int,input().split())
q=[]
b=[[0 for _ in range(30)] for _ in range(n+2)]
for i in range(m):
l,r,x=map(int,input().split())
q.append((l,r,x))
j=0
while x:
if x&1:
b[l][j]+=1
b[r+1][j]-=1
x=x>>1
j+=1
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
for i in range(1,n+1):
for j in range(30):
if b[i][j]>=2:
b[i][j]=1
b[i][j]+=b[i-1][j]
f=1
for i in q:
l,r,x=i
z=0
for j in range(30):
if b[r][j]-b[l-1][j]==(r-l+1):
z+=po[j]
if z!=x:
f=0
break
if f:
print("YES")
for i in range(1,n+1):
z=0
for j in range(30):
if b[i][j]-b[i-1][j]==1:
z+=po[j]
print(z,end=" ")
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 98,231 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Tags: constructive algorithms, data structures, trees
Correct Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
po=[1]
for i in range(30):
po.append(po[-1]*2)
n,m=map(int,input().split())
q=[]
b=[[0 for _ in range(30)] for _ in range(n+2)]
for i in range(m):
l,r,x=map(int,input().split())
q.append((l,r,x))
j=0
while x:
if x&1:
b[l][j]+=1
b[r+1][j]-=1
x=x>>1
j+=1
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
for i in range(1,n+1):
for j in range(30):
if b[i][j]>=2:
b[i][j]=1
b[i][j]+=b[i-1][j]
f=1
for i in q:
l,r,x=i
z=0
for j in range(30):
if b[r][j]-b[l-1][j]==(r-l+1):
z+=po[j]
if z!=x:
f=0
break
if f:
print("YES")
a=[]
for i in range(1,n+1):
z=0
for j in range(30):
if b[i][j]-b[i-1][j]==1:
z+=po[j]
a.append(z)
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 98,232 |
Provide tags and a correct Python 3 solution for this coding contest problem.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Tags: constructive algorithms, data structures, trees
Correct Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
dp = [[0]*30 for _ in range(n+2)]
op = []
for _ in range(m):
op.append(tuple(map(int,input().split())))
l,r,q = op[-1]
mask,cou = 1,29
while mask <= q:
if mask&q:
dp[l][cou] += 1
dp[r+1][cou] -= 1
cou -= 1
mask <<= 1
ans = [[0]*30 for _ in range(n)]
for i in range(30):
a = 0
for j in range(n):
a += dp[j+1][i]
dp[j+1][i] = dp[j][i]
if a:
ans[j][i] = 1
dp[j+1][i] += 1
for i in op:
l,r,q = i
mask = 1
for cou in range(29,-1,-1):
if not mask&q and dp[r][cou]-dp[l-1][cou] == r-l+1:
print('NO')
return
mask <<= 1
for i in range(n):
ans[i] = int(''.join(map(str,ans[i])),2)
print('YES')
print(*ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
| 98,233 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
po=[1]
for i in range(30):
po.append(po[-1]*2)
n,m=map(int,input().split())
q=[]
b=[[0 for _ in range(30)] for _ in range(n+2)]
for i in range(m):
l,r,x=map(int,input().split())
q.append((l,r,x))
j=0
while x:
if x&1:
b[l][j]+=1
b[r+1][j]-=1
x=x>>1
j+=1
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
for i in range(1,n+1):
for j in range(30):
b[i][j]+=b[i-1][j]
f=1
for i in q:
l,r,x=i
z=0
for j in range(30):
if b[r][j]-b[l-1][j]>=(r-l+1):
z+=po[j]
if z!=x:
f=0
break
if f:
print("YES")
a=[]
for i in range(1,n+1):
z=0
for j in range(30):
if b[i][j]-b[i-1][j]==1:
z+=po[j]
a.append(z)
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,234 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# by the authority of GOD author: manhar singh sachdev #
import os,sys
from io import BytesIO, IOBase
def main():
n,m = map(int,input().split())
dp = [[0]*30 for _ in range(n+2)]
op = []
for _ in range(m):
op.append(tuple(map(int,input().split())))
l,r,q = op[-1]
mask,cou = 1,0
while mask <= q:
if mask&q:
dp[l][cou] += 1
dp[r+1][cou] -= 1
cou += 1
mask <<= 1
ans = [[0]*30 for _ in range(n)]
for i in range(30):
a = 0
for j in range(n):
a += dp[j+1][i]
dp[j+1][i] = dp[j][i]
if a:
ans[j][i] = 1
dp[j+1][i] += 1
for i in op:
l,r,q = i
mask,cou = 1,0
while mask <= q:
if not mask&q and dp[r][cou]-dp[l-1][cou] == r-l+1:
print('NO')
return
mask <<= 1
cou += 1
for i in range(n):
ans[i] = int(''.join(map(str,ans[i][::-1])),2)
print('YES')
print(*ans)
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,235 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
n, m = map(int, input().split())
a, f, v = [-1] * n, 1, [-1] * n
for i in range(m):
l, r, x = map(int, input().split())
if f:
j = l - 1
while j < r:
if a[j] != -1:
if a[j] != x:
f = 0
break
else:
j=v[j] - 1
a[j]=x
v[j]=r
j += 1
if f:
print("YES")
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,236 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
We'll call an array of n non-negative integers a[1], a[2], ..., a[n] interesting, if it meets m constraints. The i-th of the m constraints consists of three integers li, ri, qi (1 ≤ li ≤ ri ≤ n) meaning that value <image> should be equal to qi.
Your task is to find any interesting array of n elements or state that such array doesn't exist.
Expression x&y means the bitwise AND of numbers x and y. In programming languages C++, Java and Python this operation is represented as "&", in Pascal — as "and".
Input
The first line contains two integers n, m (1 ≤ n ≤ 105, 1 ≤ m ≤ 105) — the number of elements in the array and the number of limits.
Each of the next m lines contains three integers li, ri, qi (1 ≤ li ≤ ri ≤ n, 0 ≤ qi < 230) describing the i-th limit.
Output
If the interesting array exists, in the first line print "YES" (without the quotes) and in the second line print n integers a[1], a[2], ..., a[n] (0 ≤ a[i] < 230) decribing the interesting array. If there are multiple answers, print any of them.
If the interesting array doesn't exist, print "NO" (without the quotes) in the single line.
Examples
Input
3 1
1 3 3
Output
YES
3 3 3
Input
3 2
1 3 3
1 3 2
Output
NO
Submitted Solution:
```
# Legends Always Come Up with Solution
# Author: Manvir Singh
import os
from io import BytesIO, IOBase
import sys
from collections import defaultdict, deque, Counter
from math import sqrt, pi, ceil, log, inf, gcd, floor
from itertools import combinations, permutations
from bisect import *
from fractions import Fraction
from heapq import *
from random import randint
def main():
n, m = map(int, input().split())
a, f, v = [-1] * n, 1, [-1] * n
b=defaultdict(list)
for i in range(m):
l,r,x = map(int, input().split())
z=b[(l,r)]
if f and z==[]:
b[(l,r)]=x
j = l - 1
while j < r:
if a[j] != -1:
if a[j]!=x and not x:
f = 0
break
else:
j=v[j] - 1
a[j]=x
v[j]=r
j += 1
elif f and z!=[] and z!=x:
f=0
if f:
print("YES")
print(*a)
else:
print("NO")
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
main()
```
No
| 98,237 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Tags: dp, implementation
Correct Solution:
```
from bisect import bisect_left as bl
from bisect import bisect_right as br
import heapq
import math
from collections import *
from functools import reduce,cmp_to_key
import sys
input = sys.stdin.readline
# M = mod = 998244353
def factors(n):return sorted(list(set(reduce(list.__add__,([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))))
# def inv_mod(n):return pow(n, mod - 2, mod)
def li():return [int(i) for i in input().rstrip('\n').split(' ')]
def st():return input().rstrip('\n')
def val():return int(input().rstrip('\n'))
def li2():return [i for i in input().rstrip('\n').split(' ')]
def li3():return [int(i) for i in input().rstrip('\n')]
def find(a,mod,n):
rem = n - len(bin(a)[2:])
ans = 0
while rem:
temp = min(rem,50)
ans = (ans + 2**temp)%mod
rem -= temp
return ans
n, k, m = li()
f=[0 for i in range(k)]
s=0
for v in range(n):
tens = 10**v%k
f=[ (sum( [f[(j+k-(x+1)*tens)%k] for x in range(9)] )+f[j])%m for j in range(k)]
for x in range(9):
f[(x+1)*tens%k]+=1
if n-v-1==0:
s+=(f[0]%m)
else:
s+=f[0]*((10**(n-v-2)*9))%m
f[0]=0
print(s%m)
```
| 98,238 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Tags: dp, implementation
Correct Solution:
```
def get_input():
hahaha=input()
(n,k,m)=hahaha.split(sep=None, maxsplit=1000)
return (int(n),int(k),int(m))
(n,k,m)=get_input()
f=[0 for i in range(k)]
s=0
for v in range(n):
tens = 10**v%k
f=[ (sum( [f[(j+k-(x+1)*tens)%k] for x in range(9)] )+f[j])%m for j in range(k)]
for x in range(9):
f[(x+1)*tens%k]+=1
if n-v-1==0:
s+=(f[0]%m)
else:
s+=f[0]*((10**(n-v-2)*9))%m
f[0]=0
print(s%m)
```
| 98,239 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Tags: dp, implementation
Correct Solution:
```
n,k,m=map(int,input().split())
d,r,p,P=0,0,1%k,(10**(n-1))*9
F=[0]*k
F[0]=1
while d<n:
d+=1
P//=10
E=[0]*k
if P==0:P=1
i=1
# print("E=",E)
# print("F=",F)
while i<10:
j=(-i*p)%k
f=0
while f<k:
E[f]+=F[j]
f+=1
j+=1
if j==k:j=0
i+=1
r+=E[0]*P
p=p*10%k
E[0]=0
i=1
while i<k:
F[i]=(F[i]+E[i])%m
i+=1
# print(E,P)
F[0]=1
#print("r=",r)
print(r%m)
#i=10**n
#j=10**(n-1)
#r=0
#F=[0]*k
#while j<i:
# x=str(j)
# l=len(x)
# a=l
# while a:
# a-=1
# s=int(x[a:l])
# if s>0 and s%k==0:
# r+=1
# break
# j+=1
#print()
#print(r)
"""
3 6 9
13 16 19 12 15 18
23 26 29 21 24 27
33 36 39 30
43 46 49 42 45 48
53 56 59 51 54 57
63 66 69 60
73 76 79 72 75 78
83 86 89 81 84 87
93 96 99 90
"""
```
| 98,240 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Submitted Solution:
```
s = input().split()
n,k,m = int(s[0]),int(s[1]),int(s[2])
b = [0 for x in range(n+1)]
ans,p = 0,0
s = 0
for i in range(1,n+1):
b[i] = ((10**i-1)//k - s)%m
s += b[i]
for i in range(1,n+1):
if i == n:
ans += b[i]%m
break
ans += b[i]* 9*10**(n-i)%m
print(ans%m)
```
No
| 98,241 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Submitted Solution:
```
n,k,m=map(int,input().split())
d,r,p,P=0,0,1%k,(10**(n-1))*9
F=[0]*k
F[0]=1
while d<n:
d+=1
P//=10
E=[0]*k
if P==0:P=1
i=1
# print("E=",E)
# print("F=",F)
while i<10:
j=(-i*p)%k
f=0
while f<k:
E[f]+=F[j]
f+=1
j+=1
if j==k:j=0
i+=1
r+=E[0]*P
p=p*10%k
E[0]=0
i=1
while i<k:
F[i]=(F[i]+E[i])%m
i+=1
# print(E,P)
F=E
F[0]=1
#print("r=",r)
print(r%m)
#i=10**n
#j=10**(n-1)
#r=0
#F=[0]*k
#while j<i:
# x=str(j)
# l=len(x)
# a=l
# while a:
# a-=1
# s=int(x[a:l])
# if s>0 and s%k==0:
# r+=1
# break
# j+=1
#print()
#print(r)
"""
3 6 9
13 16 19 12 15 18
23 26 29 21 24 27
33 36 39 30
43 46 49 42 45 48
53 56 59 51 54 57
63 66 69 60
73 76 79 72 75 78
83 86 89 81 84 87
93 96 99 90
"""
```
No
| 98,242 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Amr doesn't like Maths as he finds it really boring, so he usually sleeps in Maths lectures. But one day the teacher suspected that Amr is sleeping and asked him a question to make sure he wasn't.
First he gave Amr two positive integers n and k. Then he asked Amr, how many integer numbers x > 0 exist such that:
* Decimal representation of x (without leading zeroes) consists of exactly n digits;
* There exists some integer y > 0 such that:
* <image>;
* decimal representation of y is a suffix of decimal representation of x.
As the answer to this question may be pretty huge the teacher asked Amr to output only its remainder modulo a number m.
Can you help Amr escape this embarrassing situation?
Input
Input consists of three integers n, k, m (1 ≤ n ≤ 1000, 1 ≤ k ≤ 100, 1 ≤ m ≤ 109).
Output
Print the required number modulo m.
Examples
Input
1 2 1000
Output
4
Input
2 2 1000
Output
45
Input
5 3 1103
Output
590
Note
A suffix of a string S is a non-empty string that can be obtained by removing some number (possibly, zero) of first characters from S.
Submitted Solution:
```
n,k,m=map(int,input().split())
d,r,p,P=0,0,1%k,(10**(n-1))*9
F=[0]*k
F[0]=1
while d<n:
d+=1
P//=10
E=[0]*k
if P==0:P=1
i=1
while i<10:
j=0
while j<k:
E[j]+=F[(j-i*p)%k]
j+=1
i+=1
r+=E[0]*P
p=p*10%k
E[0]=0
F=E
F[0]=1
print(r%m)
```
No
| 98,243 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
from collections import Counter
_ = input()
s = input().strip()
sequences = Counter()
cur_seq_len = 1
for p, n in zip(s, s[1:] + '*'):
if p == n:
cur_seq_len += 1
else:
if cur_seq_len > 3:
cur_seq_len = 3
sequences[cur_seq_len] += 1
cur_seq_len = 1
basic = sum(sequences.values())
if 3 in sequences or sequences[2] >= 2:
print(basic + 2)
elif sequences[2] == 1:
print(basic + 1)
else:
print(basic)
```
| 98,244 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
if __name__ == "__main__":
n = int( input().strip() )
s = input().strip()
segs = []
start = 0
i = 1
while i <= n:
if i == n or s[i] != s[start]:
segs.append( i - start )
start = i
i = start + 1
else:
i += 1
res = len(segs)
segs.sort()
if segs[-1] >= 3:
res += 2
elif len(segs) >= 2 and segs[-1] >= 2 and segs[-2] >= 2:
res += 2
elif segs[-1] >= 2:
res += 1
print( res )
```
| 98,245 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n = int(input())
s = input()
dl = 1
now = s[0]
k = 0
for i in range(1, n):
if now != s[i]:
now = s[i]
dl += 1
for i in range(n - 1):
if s[i] == s[i + 1] == '0' or s[i] == s[i + 1] == '1':
k += 1
if k >= 2:
print(dl + 2)
elif k == 1:
print(dl + 1)
else:
print(dl)
```
| 98,246 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n = map(int, input())
s = input()
l = []
cnt = 1
for i in range( 1, len(s)):
if s[i] != s[i-1]:
l.append(cnt)
cnt = 1
else:
cnt += 1
l.append(cnt)
result = len(l)
add = 0
if max(l) >= 3:
add = 2
elif max(l) >= 2:
add = 1
if len(l) > 1:
for i in range( 1, len(l)):
if l[i] >= 2 and l[i-1] >= 2:
add = 2
if s[0] != s[len(s)-1] and l[0] >= 2 and l[len(l)-1] >= 2:
add = 2
num = 0
for i in range( 0, len(l)):
if l[i] >= 2:
num += 1
if num >= 2:
add = 2
if add == 0 and (l[0] >= 2 or l[len(l)-1] >= 2):
add = 1
print(result + add)
```
| 98,247 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n=int(input())
s=input()
print(min(s.count('01')+s.count('10')+3,n))
```
| 98,248 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n=int(input())
s=input()
c=1
seg=[]
for i in range(1,n):
if s[i]==s[i-1]:
c+=1
else:
seg.append(c)
c=1
seg.append(c)
f=0
for i in range(len(seg)):
if seg[i]>=3:
f=max(f,2)
if seg[0]==1 and seg[len(seg)-1]==2 or seg[0]==2 and seg[len(seg)-1]==1:
f=max(1,f)
if seg[0]==seg[len(seg)-1]==2:
f=max(f,2)
print(min(n,len(seg)+2))
```
| 98,249 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n=int(input())
s=input().strip('\n')
cur=s[0]
l=1
for i in range(1,n):
if s[i]!=cur:
l+=1
cur=s[i]
print(min(l+2,n))
```
| 98,250 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Tags: dp, greedy, math
Correct Solution:
```
n=int(input())
s=input()
k=0
i=0
r=int(s[0])
while(i<n):
if(int(s[i])==r):
r=1-r
k+=1
i+=1
print(min(k+2,n))
```
| 98,251 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n,line=int(input()),input()
ans=1+sum(line[i]!=line[i-1] for i in range(1,n))
print(min(ans+2,n))
```
Yes
| 98,252 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n , s = int(input()), input()
ans = 1
for i in range (1,n):
ans += (s[i] != s[i-1])
print(min(ans+2,n))
```
Yes
| 98,253 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n = int(input())
s = input()
print(min(n,s.count("01")+s.count("10")+3))
```
Yes
| 98,254 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n = int(input())
t = [int(i) for i in input()]
ch = 0
sm = 0
for i in range(n-1):
if t[i] == t[i+1]:
sm+=1
else:
ch+=1
print(ch + min(sm,2) + 1)
```
Yes
| 98,255 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
"""
Codeforces Round #334 (Div. 2)
Problem 604 C.
@author yamaton
@date 2015-12-01
"""
import itertools as it
import functools
import operator
import collections
import math
import sys
def solve(s, n):
if n == 1:
return 1
alts = sum(a != b for (a, b) in zip(s, s[1:])) + 1
lens = [sum(1 for i in iterable) for (_, iterable) in it.groupby(s)]
maxlen = max(lens)
if maxlen == 1:
return alts
elif maxlen >= 3:
return alts + 2
else:
if len(lens) == 1:
return alts + 1
if lens[0] == 2 and max(lens[1:]) == 1:
return alts + 1
if lens[-1] == 2 and max(lens[:(n-1)]) == 1:
return alts + 1
if any((a == b == 2) for (a, b) in zip(lens, lens[1:])):
return alts + 2
else:
return alts
# def pp(*args, **kwargs):
# return print(*args, file=sys.stderr, **kwargs)
def main():
n = int(input())
s = input().strip()
assert len(s) == n
result = solve(s, n)
print(result)
if __name__ == '__main__':
main()
```
No
| 98,256 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n = map(int, input())
s = input()
l = []
cnt = 1
for i in range( 1, len(s)):
if s[i] != s[i-1]:
l.append(cnt)
cnt = 1
else:
cnt += 1
l.append(cnt)
result = len(l)
add = 0
if len(l) > 1:
for i in range( 1, len(l)):
if l[i] >= 2 and l[i-1] >= 2:
add = 2
if s[0] != s[len(s)-1] and l[0] >= 2 and l[len(l)-1] >= 2:
add = 2
if max(l) >= 3:
add = 2
if add == 0 and (l[0] >= 2 or l[len(l)-1] >= 2):
add = 1
print(result + add)
```
No
| 98,257 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n = int(input())
ch = str(input())
B = [0 for _ in range(n)]
B[0] = 1
c = ch[0]
for k in range(1,n):
if ch[k] == c:
B[k] = B[k-1]
else:
B[k] = B[k-1] + 1
c = ch[k]
T = [[0 for _ in range(n)] for _ in range(n)]
F = [[0 for _ in range(n)] for _ in range(n)]
for i in range(n-1):
T[i][i+1] = B[i]+1
for i in range(n-1):
for j in range(i+2,n):
c = 0
if ch[j] == ch[j-1] and not(F[i][j-1]):
temp = max([T[i][j-1], B[j-1] + 1])
if temp == 1 + B[j-1]:
c = 1
elif not(F[i][j-1]):
temp = T[i][j-1] + 1
elif ch[j] == ch[j-1]:
temp = T[i][j-1] + 1
else:
temp = max([T[i][j-1], B[j-1] + 1])
T[i][j] = temp
F[i][j] = c
print(max([T[k][n-1] for k in range(n)]))
"""
B = [0 for _ in range(n)]
B[0] = 1
c = ch[0]
for k in range(1,n):
if ch[k] == c:
B[k] = B[k-1]
else:
B[k] = B[k-1] + 1
c = ch[k]
T = [0 for _ in range(n)]
F = [0 for _ in range(n)]
T[1] = 2
if ch[0] == ch[1]:
F[1] = 1
for k in range(2,n):
c = 0
if ch[k] == ch[k-1]:
temp = max([T[k-1], B[k-1] + 1])
if temp == 1 + B[k-1]:
c = 1
elif not(F[k-1]):
temp = T[k-1] + 1
elif ch[k] == ch[k-1]:
temp = T[k-1] + 1
else:
temp = max([T[k-1], B[k-1] + 1])
T[k] = temp
F[k] = c
print(B)
print(T)
print(F)
print(T[n-1])
"""
```
No
| 98,258 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
Input
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
Examples
Input
8
10000011
Output
5
Input
2
01
Output
2
Note
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
Submitted Solution:
```
n = map(int, input())
s = input()
l = []
cnt = 1
for i in range( 1, len(s)):
if s[i] != s[i-1]:
l.append(cnt)
cnt = 1
else:
cnt += 1
l.append(cnt)
result = len(l)
add = 0
if max(l) >= 3:
add = 2
if len(l) > 1:
for i in range( 1, len(l)):
if l[i] >= 2 and l[i-1] >= 2:
add = 2
if l[0] >= 2 and l[len(l)-1] >= 2:
add = 2
if add == 0 and (l[0] >= 2 or l[len(l)-1] >= 2):
add = 1
print(result + add)
```
No
| 98,259 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.
If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?
Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.
Input
The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.
The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.
Output
Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo 109 + 7.
Examples
Input
3 2
2 4 5
Output
3
Input
4 3
7 8 9 10
Output
13
Input
4 0
5 10 20 21
Output
1
Note
In the first sample, we have three options:
* The first and second students form a group, and the third student forms a group. Total imbalance is 2 + 0 = 2.
* The first student forms a group, and the second and third students form a group. Total imbalance is 0 + 1 = 1.
* All three students form their own groups. Total imbalance is 0.
In the third sample, the total imbalance must be 0, so each student must work individually.
Submitted Solution:
```
# You lost the game.
def binom(n, k):
res = 1
# Since C(n, k) = C(n, n-k)
if ( k > n - k ):
k = n - k
# Calculate value of [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
for i in range(k):
res *= (n - i)
res /= (i + 1)
return res;
n, k = map(int, input().split())
L = list(map(int, input().split()))
L.sort()
B = [1,1,2,5,15,52,203,877]
for k in range(8,n):
r = 0
for i in range(1,k):
r += binom(k,i)*B[i]
B[k] = r
T = [0 for _ in range(n)]
f = 0
d = 0
while d < n:
if f == n or L[f]-L[d] > k:
T[d] = f
d += 1
else:
f += 1
R = []
r = B[T[0]]
for i in range(1,d):
if i > T[i-1]:
R += [r]
r = B[T[i]-i]
else:
r += B[T[i]-i]-B[T[i-1]-i]
print(max(R+[r]))
```
No
| 98,260 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.
If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?
Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.
Input
The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.
The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.
Output
Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo 109 + 7.
Examples
Input
3 2
2 4 5
Output
3
Input
4 3
7 8 9 10
Output
13
Input
4 0
5 10 20 21
Output
1
Note
In the first sample, we have three options:
* The first and second students form a group, and the third student forms a group. Total imbalance is 2 + 0 = 2.
* The first student forms a group, and the second and third students form a group. Total imbalance is 0 + 1 = 1.
* All three students form their own groups. Total imbalance is 0.
In the third sample, the total imbalance must be 0, so each student must work individually.
Submitted Solution:
```
q,w=map(int,input().split())
a=list(map(int,input().split()))
a.sort()
r=0
import math
for i in range(0,q-1):
j=i
while (j<q):
if (a[j]-a[i])<=w:
j+=1
else:
break
m=j-i-1
n=math.factorial(m)
for j in range(1,m+1):
r+=n//(math.factorial(j)*math.factorial(m-j))
print(r+1)
```
No
| 98,261 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
There are n students in a class working on group projects. The students will divide into groups (some students may be in groups alone), work on their independent pieces, and then discuss the results together. It takes the i-th student ai minutes to finish his/her independent piece.
If students work at different paces, it can be frustrating for the faster students and stressful for the slower ones. In particular, the imbalance of a group is defined as the maximum ai in the group minus the minimum ai in the group. Note that a group containing a single student has an imbalance of 0. How many ways are there for the students to divide into groups so that the total imbalance of all groups is at most k?
Two divisions are considered distinct if there exists a pair of students who work in the same group in one division but different groups in the other.
Input
The first line contains two space-separated integers n and k (1 ≤ n ≤ 200, 0 ≤ k ≤ 1000) — the number of students and the maximum total imbalance allowed, respectively.
The second line contains n space-separated integers ai (1 ≤ ai ≤ 500) — the time it takes the i-th student to complete his/her independent piece of work.
Output
Print a single integer, the number of ways the students can form groups. As the answer may be large, print its value modulo 109 + 7.
Examples
Input
3 2
2 4 5
Output
3
Input
4 3
7 8 9 10
Output
13
Input
4 0
5 10 20 21
Output
1
Note
In the first sample, we have three options:
* The first and second students form a group, and the third student forms a group. Total imbalance is 2 + 0 = 2.
* The first student forms a group, and the second and third students form a group. Total imbalance is 0 + 1 = 1.
* All three students form their own groups. Total imbalance is 0.
In the third sample, the total imbalance must be 0, so each student must work individually.
Submitted Solution:
```
import math
modulo = math.pow(10, 9) + 7
def findways(arr, curr, k, dp):
if(curr >= len(arr) - 1):
return 0
if(dp[curr][k] != -1):
return dp[curr][k]
c = 0
for i in range(curr + 1, len(arr)):
if(arr[i] - arr[curr] <= k):
c = c + 1
dp[i][k - (arr[i] - arr[curr])] = findways(arr, i, k - (arr[i] - arr[curr]), dp)
c = c + dp[i][k - (arr[i] - arr[curr])]
else:
break
dp[curr + 1][k] = z = findways(arr, curr + 1, k, dp)
dp[curr][k] = (c + z) % modulo
return int((c + z) % modulo)
n, k = list(map(int, input().split(" ")))
arr = list(map(int, input().split(" ")))
arr.sort()
dp = [[-1 for i in range(k + 1)] for j in range(n + 1)]
s = findways(arr, 0, k, dp)
print(s + 1)
```
No
| 98,262 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
n, t = map(int, input().split())
a = [[None]] + [[0] * i for i in range(1, n + 1)]
c = 2 ** (n + 1)
for _ in range(t):
a[1][0] += c
for i in range(1, n + 1):
for j in range(i):
if a[i][j] > c:
diff = a[i][j] - c
a[i][j] = c
if i < n:
a[i + 1][j] += diff // 2
a[i + 1][j + 1] += diff // 2
full = 0
for i in range(1, n + 1):
for j in range(i):
full += (a[i][j] == c)
print(full)
```
| 98,263 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
n, t = map(int, input().split())
t = min(2000, t)
L = [ [0.0]*n for i in range(n) ]
for _ in range(t) :
L[0][0] += 1.0
for i in range(n-1) :
for j in range(i+1) :
if L[i][j] > 1.0 :
x = (L[i][j] - 1.0) / 2
L[i+1][j] += x
L[i+1][j+1] += x
L[i][j] = 1.0
ans = 0
for i in range(n) :
for j in range(i+1) :
if L[i][j] > 0.9999999999 :
ans += 1
print(ans)
```
| 98,264 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
n,t=list(map(int,input().split()))
li=[[t-1]]
cnt=1 if t-1>=0 else 0
for i in range(2,n+1):
tr=li[-1][0]/2-1
if tr>=0:
cnt+=2
li.append([tr])
for j in range(1,i-1):
a=li[-2][j-1]
b=li[-2][j]
temp=-1
if a >0 :
temp+=a/2
if b>0:
temp+=b/2
if temp>=0:
cnt+=1
li[-1].append(temp)
li[-1].append(tr)
print(cnt)
```
| 98,265 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
a = 1<<10
bocal = [[0]*(i+1) for i in range(10)]
ans = [[0]*(i+1) for i in range(10)]
x = input().split()
n = int(x[0])
t = int(x[1])
bocal[0][0]=a
ans[0][0]=1
for z in range(2,2000):
bocal[0][0]+=a
for i in range(10):
for j in range(i+1):
if ans[i][j]==0 and bocal[i][j]>=a:
ans[i][j]=z
if bocal[i][j]>a:
x = bocal[i][j]-a
if i<9:
bocal[i+1][j]+=x//2
bocal[i+1][j+1]+=x//2
bocal[i][j]=a
a=0
for i in range(n):
for j in range(i+1):
if ans[i][j]<=t:
a+=1
print(a)
```
| 98,266 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
n,t=map(int, input().split())
m=[[0 for _ in range(11)]for _ in range(11)]
m[1][1]=t
for i in range(1, n):
for j in range(1, n + 1):
if m[i][j] > 1:
m[i + 1][j] += (m[i][j] - 1) / 2
m[i + 1][j + 1] += (m[i][j] - 1) / 2
print(sum([sum(map(lambda x: x >= 1, m[i])) for i in range(11)]))
```
| 98,267 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
inin=input().split(' ')
n=int(inin[0])
t=int(inin[1])
mat=[]
for i in range(n+2):
mat.append([0.0]*(n+2))
# def add(i,j,amt):
# if i>=n or j<0 or j>=n:
# return
# mat[i][j]+=amt
# if mat[i][j]>1:
# over=mat[i][j]-1
# mat[i][j]-=over
# add(i+1,j,over/2)
# add(i+1,j+1,over/2)
for time in range(t):
# add(0,0,1)
mat[1][1]+=1.0
for i in range(1,n+1):
for j in range(1,i+1):
if mat[i][j]>1.0:
over=mat[i][j]-1.0
mat[i+1][j]+=over/2
mat[i+1][j+1]+=over/2
mat[i][j]=1.0
result=0
for i in range(1,n+1):
for j in range(1,i+1):
if mat[i][j]>=1.0:
result+=1
print(result)
# for line in mat:
# # print(line)
# for b in line:
# print(str(b),end='\t')
# print()
```
| 98,268 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
#676B
arr = list(map(int, input().split(" ")))
n = arr[0]
t = arr[1]
glasses = 0
a = [[0] * (i + 1) for i in range(n+1)]
a[0][0] = t
for j in range(n):
for k in range(j + 1):
if a[j][k] >= 1:
glasses += 1
extra = a[j][k] - 1
a[j][k] = 1
a[j+1][k] += extra / 2.0
a[j+1][k+1] += extra / 2.0
print(min(glasses, n * (n+1)/2))
```
| 98,269 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Tags: implementation, math, math
Correct Solution:
```
import sys,math
res=[0]
n,m=map(int,input().split())
z=[]
need=0
for i in range(1,n+1):
need+=i
for i in range(1,n+1):
z.append([0]*i)
for i in range(1,m+1):
z[0][0]+=2
for i in range(n-1):
for j in range(i+1):
if z[i][j]>=2:
h=z[i][j]-2
z[i+1][j]+=h/2
z[i+1][j+1]+=h/2
z[i][j]=2
for j in range(n):
if z[-1][j]>=2:
z[-1][j]=2
s=0
for i in range(n):
s+=z[i].count(2)
print(s)
```
| 98,270 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
class CodeforcesTask676BSolution:
def __init__(self):
self.result = ''
self.n_t = []
def read_input(self):
self.n_t = [int(x) for x in input().split(" ")]
def process_task(self):
glasses = [[0.0] * (x + 1) for x in range(self.n_t[0])]
for s in range(self.n_t[1]):
glasses[0][0] += 1.0
for x in range(self.n_t[0] - 1):
for y in range(x + 1):
if glasses[x][y] > 1.0:
to_pour = glasses[x][y] - 1.0
glasses[x + 1][y] += to_pour / 2
glasses[x + 1][y + 1] += to_pour / 2
glasses[x][y] = 1.0
full = 0
for g in glasses:
for glass in g:
if glass >= 1.0:
full += 1
self.result = str(full)
def get_result(self):
return self.result
if __name__ == "__main__":
Solution = CodeforcesTask676BSolution()
Solution.read_input()
Solution.process_task()
print(Solution.get_result())
```
Yes
| 98,271 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
n,t=map(int,input().split())
l=[[0]*(i+1) for i in range(n+1)]
l[0][0]=t
ans=0
for i in range(n):
for j in range(i+1):
if l[i][j]>=1:
ans+=1
l[i+1][j]+=(l[i][j]-1)/2
l[i+1][j+1]+=(l[i][j]-1)/2
print(ans)
```
Yes
| 98,272 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
def push(graph, pos, level):
if graph[pos] > 1:
over = graph[pos] - 1
graph[pos] = 1
if level + pos < numberofglasses:
graph[level + pos] += over / 2
if level + pos + 1 < numberofglasses:
graph[level + pos + 1] += over / 2
if level + pos < numberofglasses:
push(graph, level + pos, level + 1)
if level + pos + 1 < numberofglasses:
push(graph, level + pos + 1, level + 1)
n, t = map(int, input().split())
table = dict()
current = 0
for i in range(1, 11):
current += i
table[i] = current
graph = [0] * table[n]
numberofglasses = table[n]
graph[0] += t
push(graph, 0, 1)
counter = 0
for elem in graph:
if elem == 1:
counter += 1
print(counter)
```
Yes
| 98,273 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
n, k = map(int,input().split())
a = [[0]*i for i in range(1,12)]
a[0][0] = k
ans = 0
for i in range(n):
for j in range(len(a[i])):
if a[i][j] >= 1:
ans += 1
rem = a[i][j]-1
a[i][j] = 1
a[i+1][j] += rem/2
a[i+1][j+1] += rem/2
print(ans)
```
Yes
| 98,274 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
n, t = list(map(int, input().split()))
a = []
for i in range(n):
a.append([0] * (n + 1))
a[0][1] = t
ans = 0
for i in range(1, n):
for j in range(1, n + 1):
a[i][j] = max(0, (a[i - 1][j] - 1) / 2) + max(0, (a[i - 1][j - 1] - 1) / 2)
if a[i][j] >= 1:
ans += 1
print(ans + 1)
```
No
| 98,275 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
n, t = map(int,input().split())
g = [[0.0] * i for i in range(1,n+1)]
for _ in range(t):
g[0][0] += 1.0
for i in range(n - 1):
for j in range(i+1):
spill = max(0, g[i][j] - 1.0)
g[i][j] -= spill
if i < n - 1:
g[i + 1][j] += spill / 2
g[i + 1][j + 1] += spill / 2
cnt = 0
for i in range(n):
for j in range(i + 1):
if g[i][j] == 1.0:
cnt += 1
print(cnt)
```
No
| 98,276 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
n=input().split()
p=0
m=0
l=int(n[0])
t=int(n[1])
for j in range (l):
m=m+(j+1)
for i in range(t):
if i==0 :
result=1
elif result == m:
break
elif i==1:
next
elif i%2==0:
result += 2
elif i % 2 == 1:
p += 1
result += p
print (result)
```
No
| 98,277 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Mary has just graduated from one well-known University and is now attending celebration party. Students like to dream of a beautiful life, so they used champagne glasses to construct a small pyramid. The height of the pyramid is n. The top level consists of only 1 glass, that stands on 2 glasses on the second level (counting from the top), then 3 glasses on the third level and so on.The bottom level consists of n glasses.
Vlad has seen in the movies many times how the champagne beautifully flows from top levels to bottom ones, filling all the glasses simultaneously. So he took a bottle and started to pour it in the glass located at the top of the pyramid.
Each second, Vlad pours to the top glass the amount of champagne equal to the size of exactly one glass. If the glass is already full, but there is some champagne flowing in it, then it pours over the edge of the glass and is equally distributed over two glasses standing under. If the overflowed glass is at the bottom level, then the champagne pours on the table. For the purpose of this problem we consider that champagne is distributed among pyramid glasses immediately. Vlad is interested in the number of completely full glasses if he stops pouring champagne in t seconds.
Pictures below illustrate the pyramid consisting of three levels.
<image> <image>
Input
The only line of the input contains two integers n and t (1 ≤ n ≤ 10, 0 ≤ t ≤ 10 000) — the height of the pyramid and the number of seconds Vlad will be pouring champagne from the bottle.
Output
Print the single integer — the number of completely full glasses after t seconds.
Examples
Input
3 5
Output
4
Input
4 8
Output
6
Note
In the first sample, the glasses full after 5 seconds are: the top glass, both glasses on the second level and the middle glass at the bottom level. Left and right glasses of the bottom level will be half-empty.
Submitted Solution:
```
# your code goes here
n,k1=map(int,input().split())
if(n==1):
print(1)
else:
print(abs(n-k1)+2)
```
No
| 98,278 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
n,m = map(int,input().split())
l = []
for i in range(m):
la,r,t,c = map(int,input().split())
l.append([la,r,t,c])
cost = 0
for i in range(n):
min_time = 10**12
for j in range(m):
if l[j][0]<=i+1<=l[j][1]:
if l[j][2]<min_time:
min_time = l[j][2]
ans = j
if min_time!=10**12:
cost+=l[ans][3]
print(cost)
```
| 98,279 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
#N sections each with m participants
#[li,ri] sections for ith participant
n,m=map(int,input().split())
play=[]
for i in range(m):
l,r,t,c=map(int,input().split())
play.append([l,r,t,c])
sm=0
for i in range(1,n+1):
mini=10**9
get=0
for j in range(m):
l,r,t,c=play[j][0],play[j][1],play[j][2],play[j][3]
if l<=i<=r :
if t<mini:
mini=t
get=c
sm+=get
print(sm)
```
| 98,280 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
n,m=map(int,input().split(" "));
a=[]
for i in range(m):
k=list(map(int,input().split(" ")))
a.append(k)
total=0;
for i in range(1,n+1):
temp=0
ans=0;
for j in a:
if i>=j[0] and i<=j[1]:
if ans==0:
ans=j[2];
temp=j[3]
elif j[2]<ans:
ans=j[2]
temp=j[3]
total=total+temp
print(total)
```
| 98,281 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
import sys
inp = sys.stdin.readlines()
all_lines = []
for line in inp:
all_lines.append(list(map(int, line.split(' '))))
num_athletes = all_lines[0][1]
num_sections = all_lines[0][0]
profit = 0
for section in range(num_sections):
least_time = 0
i = 0
earning = 0
for athlete in all_lines[1:]:
if athlete[0]-1 <= section and athlete[1]-1 >= section:
if i == 0:
least_time =athlete[2]
earning = athlete[3]
elif athlete[2]<least_time:
least_time = athlete[2]
earning = athlete[3]
i += 1
profit += earning
print(profit)
```
| 98,282 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
n,m=map(int,input().split())
vis=[0]*(n+1)
ans=0
arr=[]
for i in range(m):
arr.append(list(map(int,input().split())))
arr.sort(key=lambda i:i[2])
for i in range(m):
for j in range(arr[i][0],arr[i][1]+1):
if not vis[j]:
vis[j]=1
ans+=arr[i][3]
print(ans)
```
| 98,283 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
import sys
from array import array # noqa: F401
def input():
return sys.stdin.buffer.readline().decode('utf-8')
n, m = map(int, input().split())
a = [tuple(map(int, input().split())) for _ in range(m)]
ans = 0
for i in range(1, n + 1):
time, profit = 10**9, 0
for j in range(m):
if a[j][0] <= i <= a[j][1] and time > a[j][2]:
time = a[j][2]
profit = a[j][3]
ans += profit
print(ans)
```
| 98,284 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
sections, n_p = [int(x) for x in input().split(' ')]
participants = []
paritipants_on_sections = [(0, 1000000000000000000)] * (sections + 1) # profit + time
for i in range(n_p):
# participants.append([int(x) for x in input().split(' ')])
start, end, time, profit = [int(x) for x in input().split(' ')]
# print('---')
for j in range(start, end+1):
tmp_time = j * time
# print(tmp_time)
if tmp_time < paritipants_on_sections[j][1]:
paritipants_on_sections[j] = (profit, tmp_time)
print(sum(x[0] for x in paritipants_on_sections))
```
| 98,285 |
Provide tags and a correct Python 3 solution for this coding contest problem.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Tags: greedy, implementation
Correct Solution:
```
m, n = map(int, input().split())
p = [(1001, 0, 0)] * m
for i in range(n):
l, r, t, c = map(int, input().split())
p[l - 1: r] = [min((x, y, z), (t, i, - c)) for x, y, z in p[l - 1: r]]
print(- sum(z for x, y, z in p))
```
| 98,286 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
n = [int(i) for i in input().split()]
batata = [1001]*(n[0])
score = [0]*(n[0])
for i in range(n[1]):
all = [int(i) for i in input().split()]
for b in range(all[0]-1, all[1]):
if batata[b] > all[2]:
batata[b] = all[2]
score[b] = all[3]
sum = 0
for i in range(n[0]):
sum += score[i]
print(sum)
```
Yes
| 98,287 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
from sys import stdin,stdout
import math
from collections import Counter,deque
L=lambda:list(map(int, stdin.readline().strip().split()))
M=lambda:map(int, stdin.readline().strip().split())
I=lambda:int(stdin.readline().strip())
IN=lambda:stdin.readline().strip()
C=lambda:stdin.readline().strip().split()
mod=1000000007
#Keymax = max(Tv, key=Tv.get)
def s(a):print(" ".join(list(map(str,a))))
#______________________-------------------------------_____________________#
#I_am_pavan
def solve():
n,m=M()
a=[[] for i in range(n+1)]
for i in range(m):
l,r,t,c=M()
for j in range(l,r+1):
a[j].append((t,c))
count=0
for i in range(1,n+1):
x=len(a[i])
if x==0:
continue
t=1001
for j in range(x):
if t>a[i][j][0]:
ans=a[i][j][1]
t=a[i][j][0]
count+=ans
print(count)
t=1
for i in range(t):
solve()
```
Yes
| 98,288 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
n,m=map(int,input().split())
racer=[-1]*(n+1);profit=[0]*(n+1);
for i in range(m):
l,r,c,p=map(int,input().split())
for i in range(l,r+1):
if(racer[i]==-1 or racer[i]>c):profit[i]=p;racer[i]=c;
print(sum(profit))
```
Yes
| 98,289 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
def solve(sections):
profit = 0
for i in sections:
if not i:
continue
min = (i[0][0], i[0][1])
for j in i:
if j[0] < min[0]:
min = j
profit += min[1]
return profit
n, m = [int(i) for i in input().split()]
sections = [[] for i in range(n)]
for i in range(m):
l, r, t, c = [int(j) for j in input().split()]
for j in range(l-1, r):
sections[j] += [(t, c)]
print(solve(sections))
```
Yes
| 98,290 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
n = [int(i) for i in input().split()]
batata = [1000]*(n[0])
score = [0]*(n[0])
for i in range(n[1]):
all = [int(i) for i in input().split()]
for b in range(all[0]-1, all[1]):
if batata[b] > all[2]:
batata[b] = all[2]
score[b] = all[3]
sum = 0
for i in range(n[0]):
sum += score[i]
print(sum)
```
No
| 98,291 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
n,m=map(int,input().split())
list1=[111]*(n+1)
list2=[0]*(n+1)
for i in range(m):
l,r,t,c=map(int,input().split())
x=r-l+1
for j in range(l,r+1):
if(list1[j]>t):
list1[j]=t
list2[j]=c
elif(list1[j]==t and list2[j]<c):
list2[j]=c
print(sum(list2))
```
No
| 98,292 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
n, m = map(int, input().split())
tab = []
for _ in range(m):
l = list(map(int, input().split()))
tab.append(l)
tab = list(map(list, tab))
p = 0
for i in range(1, n+1):
s = 0
t = 1000
a = ''
for j in range(m):
if i in range(tab[j][0], tab[j][1]+1):
if tab[j][2] < t:
s = tab[j][3]
t = tab[j][2]
a = j
p += s
print(p)
```
No
| 98,293 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
In Chelyabinsk lives a much respected businessman Nikita with a strange nickname "Boss". Once Nikita decided to go with his friend Alex to the Summer Biathlon World Cup. Nikita, as a very important person, received a token which allows to place bets on each section no more than on one competitor.
To begin with friends learned the rules: in the race there are n sections of equal length and m participants. The participants numbered from 1 to m. About each participant the following is known:
* li — the number of the starting section,
* ri — the number of the finishing section (li ≤ ri),
* ti — the time a biathlete needs to complete an section of the path,
* ci — the profit in roubles. If the i-th sportsman wins on one of the sections, the profit will be given to the man who had placed a bet on that sportsman.
The i-th biathlete passes the sections from li to ri inclusive. The competitor runs the whole way in (ri - li + 1)·ti time units. It takes him exactly ti time units to pass each section. In case of the athlete's victory on k sections the man who has betted on him receives k·ci roubles.
In each section the winner is determined independently as follows: if there is at least one biathlete running this in this section, then among all of them the winner is the one who has ran this section in minimum time (spent minimum time passing this section). In case of equality of times the athlete with the smaller index number wins. If there are no participants in this section, then the winner in this section in not determined. We have to say that in the summer biathlon all the participants are moving at a constant speed.
We should also add that Nikita can bet on each section and on any contestant running in this section.
Help the friends find the maximum possible profit.
Input
The first line contains two integers n and m (1 ≤ n, m ≤ 100). Then follow m lines, each containing 4 integers li, ri, ti, ci (1 ≤ li ≤ ri ≤ n, 1 ≤ ti, ci ≤ 1000).
Output
Print a single integer, the maximal profit in roubles that the friends can get. In each of n sections it is not allowed to place bets on more than one sportsman.
Examples
Input
4 4
1 4 20 5
1 3 21 10
3 3 4 30
3 4 4 20
Output
60
Input
8 4
1 5 24 10
2 4 6 15
4 6 30 50
6 7 4 20
Output
105
Note
In the first test the optimal bet is: in the 1-2 sections on biathlete 1, in section 3 on biathlete 3, in section 4 on biathlete 4. Total: profit of 5 rubles for 1 section, the profit of 5 rubles for 2 section, profit of 30 rubles for a 3 section, profit of 20 rubles for 4 section. Total profit 60 rubles.
In the second test the optimal bet is: on 1 and 5 sections on biathlete 1, in the 2-4 sections on biathlete 2, in the 6-7 sections on athlete 4. There is no winner in the 8 section. Total: profit of 10 rubles for 1 section, the profit of 15 rubles for 2,3,4 section, profit of 10 rubles for a 5 section, profit of 20 rubles for 6, 7 section. Total profit 105 rubles.
Submitted Solution:
```
m, n = map(int, input().split())
p = [(-1000, 0)] * m
for i in range(n):
l, r, t, c = map(int, input().split())
p[l - 1: r] = [max((i, j), (-t, c)) for i, j in p[l - 1: r]]
print(sum(j for i, j in p))
```
No
| 98,294 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).
<image>
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
Input
2 3
1 2
Output
1
Input
6 1
5 1 2 3 4 1
Output
2
Note
In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
Tags: brute force, math, number theory
Correct Solution:
```
_, n=map(int,input().split())
m={}
ans=0
for x in map(int, input().split()):
ans+=m.get(x^n,0)
m[x]=m.get(x,0)+1
print(ans)
```
| 98,295 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).
<image>
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
Input
2 3
1 2
Output
1
Input
6 1
5 1 2 3 4 1
Output
2
Note
In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
Tags: brute force, math, number theory
Correct Solution:
```
n, x = map(int, input().split())
l, res = [0] * 131073, 0
for a in map(int, input().split()):
res, l[a] = res + l[a ^ x], l[a] + 1
print(res)
```
| 98,296 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).
<image>
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
Input
2 3
1 2
Output
1
Input
6 1
5 1 2 3 4 1
Output
2
Note
In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
Tags: brute force, math, number theory
Correct Solution:
```
a,b = map(int,input().split())
lista = list(map(int,input().split()))
n=0
d = dict()
for x in lista:
# get(a,b) retorna b se n tiver a chave a
n += d.get(x^b,0)
d[x] = d.get(x,0)+1
print(n)
```
| 98,297 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).
<image>
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
Input
2 3
1 2
Output
1
Input
6 1
5 1 2 3 4 1
Output
2
Note
In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
Tags: brute force, math, number theory
Correct Solution:
```
n,x = map(int,input().split())
lis = list(map(int,input().split()))
ans=0
has=[0]*(1000000)
for i in lis:
has[i]+=1
ans=0
if x==0:
for i in range(1000000):
ans+=((has[i])*(has[i]-1))//2
else:
for i in range(n):
a=lis[i]^x
# print(lis[i],a,ans)
# print(has[a],has[lis[i]])
ans+=(has[lis[i]]*has[a])
has[lis[i]]=0
has[a]=0
print(ans)
```
| 98,298 |
Provide tags and a correct Python 3 solution for this coding contest problem.
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).
<image>
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Examples
Input
2 3
1 2
Output
1
Input
6 1
5 1 2 3 4 1
Output
2
Note
In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
Tags: brute force, math, number theory
Correct Solution:
```
n, x = map(int, input().split())
arr = list(map(int, input().split()))
b = [0] * int(2e5)
for i in arr:
b[i] += 1
res = 0
for i in arr:
r = i ^ x
res += b[r]
if x == 0:
res -= len(arr)
print(res // 2)
```
| 98,299 |
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