n = int(input())
x_list = list(map(int,input().split()))
#print(x_list)
#print(sum(x_list))
#(len(x_list))

#p = sum(x_list) // (len(x_list))
#p = (min(x_list) + max(x_list)) // 2
#print(p)

ans_list = [0] * (max(x_list)+1)

for j in range(1, max(x_list)+1):
    ans = 0
    for i in range(0, n):
        ans += (x_list[i] - j) ** 2
        #print(ans)
    ans_list[j] = ans

ans_list.remove(0)
print(min(ans_list))
#print(ans)n = int(input())
x_list = list(map(int,input().split()))
#print(x_list)
#print(sum(x_list))
#(len(x_list))

#p = sum(x_list) // (len(x_list))
#p = (min(x_list) + max(x_list)) // 2
#print(p)

ans_list = [0] * (max(x_list)+1)

for j in range(1, max(x_list)+1):
    ans = 0
    for i in range(0, n):
        ans += (x_list[i] - j) ** 2
        #print(ans)
    ans_list[j] = ans

ans_list.remove(0)
print(min(ans_list))
#print(ans)

Score : 300 points

Problem Statement

There are N people living on a number line.

The i-th person lives at coordinate X_i.

You are going to hold a meeting that all N people have to attend.

The meeting can be held at any integer coordinate. If you choose to hold the meeting at coordinate P, the i-th person will spend (X_i - P)^2 points of stamina to attend the meeting.

Find the minimum total points of stamina the N people have to spend.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq X_i \leq 100

Input

Input is given from Standard Input in the following format:

N
X_1 X_2 ... X_N

Output

Print the minimum total stamina the N people have to spend.

Sample Input 1

2
1 4

Sample Output 1

5

Assume the meeting is held at coordinate 2. In this case, the first person will spend (1 - 2)^2 points of stamina, and the second person will spend (4 - 2)^2 = 4 points of stamina, for a total of 5 points of stamina. This is the minimum total stamina that the 2 people have to spend.

Note that you can hold the meeting only at an integer coordinate.

Sample Input 2

7
14 14 2 13 56 2 37

Sample Output 2

2354


from operator import mul
from functools import reduce

n, a, b = map(int, input().split())

def cmb(n, r):
    r = min(n-r, r)
    if r == 0:
        return 1
    over = reduce(mul, range(n, n-r, -1))
    under = reduce(mul, range(1, r+1))
    return over // under

mod = 10**9+7

ans = (pow(2, n, mod) -1 - cmb(n, a) % mod - cmb(n, b) % mod) % mod

print(ans)

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 400 points

Problem Statement

Akari has n kinds of flowers, one of each kind.

She is going to choose one or more of these flowers to make a bouquet.

However, she hates two numbers a and b, so the number of flowers in the bouquet cannot be a or b.

How many different bouquets are there that Akari can make?

Find the count modulo (10^9 + 7).

Here, two bouquets are considered different when there is a flower that is used in one of the bouquets but not in the other bouquet.

Constraints

All values in input are integers.

2 \leq n \leq 10^9

1 \leq a < b \leq \textrm{min}(n, 2 \times 10^5)

Input

Input is given from Standard Input in the following format:

n a b

Output

Print the number of bouquets that Akari can make, modulo (10^9 + 7). (If there are no such bouquets, print 0.)

Sample Input 1

4 1 3

Sample Output 1

7

In this case, Akari can choose 2 or 4 flowers to make the bouquet.

There are 6 ways to choose 2 out of the 4 flowers, and 1 way to choose 4, so there are a total of 7 different bouquets that Akari can make.

Sample Input 2

1000000000 141421 173205

Sample Output 2

34076506

Print the count modulo (10^9 + 7).

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


import sys
input=sys.stdin.readline
#from collections import defaultdict
#d = defaultdict(int)
#import fractions
#import math
#import collections
#from collections import deque
#from bisect import bisect_left
#N = int(input())
#A = list(map(int,input().split()))
#S = list(input())
#S.remove("\n")
#N,M = map(int,input().split())
#S,T = map(str,input().split())
#A = [int(input()) for _ in range(N)]
#S = [input() for _ in range(N)]
#A = [list(map(int,input().split())) for _ in range(N)]
#import itertools
MOD = 10**9 + 7
def comb(n, k, p):
    if n<0 or k<0 or n<k:
        return 0
    if n==0 or k==0:
        return 1
    k=k if k<=n-k else n-k
    for i in range(k):
        x=x*(n-i)%mod
        y=y*(i+1)%mod
    return (x*pow(y,mod-2,mod))%mod
n,a,b = map(int,input().split())
A = comb(n,a,MOD)
B = comb(n,b,MOD)
C = pow(2,n,MOD)
ans = C -A-B-1
print(ans%MOD)


n, a, b = map(int, input().split())
mod = 10**9+7

#print(x*pow(y,(mod-2),mod)%mod)

def conv(n, x, mod):
    X = 1
    for i in range(x):
        k = i+1
        X *= (n-k+1)*pow(k,(mod-2),mod)%mod
        X %= mod
    return X

def power_func(a,n,p):
  bi=str(format(n,"b"))#2進表現に
  res=1
  for i in range(len(bi)):
    res=(res*res) %p
    if bi[i]=="1":
      res=(res*a) %p
  return res

#print(conv(n, a, mod))
#print(conv(n, b, mod))

ans = power_func(2, n, mod) - conv(n, a, mod) - conv(n, b, mod) - 1
ans %= mod
print(ans)


# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import math

#############
# Constants #
#############

MOD = 10**9 +7
INF = float('inf')

#############
# Functions #
#############

kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n):
    return kaijo_memo[n]
  if(len(kaijo_memo) == 0):
    kaijo_memo.append(1)
  while(len(kaijo_memo) <= n):
    kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n):
    return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0):
    gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n):
    gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if(n == r):
    return 1
  if(n < r or r < 0):
    return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

#############
# Main Code #
#############

k,q = map(int, input().split())
d = list(map(int, input().split()))

def query(n,x,m):
  x %= m
  ans = n-1
  n -= 1
  s1 = 0
  t1 = 0
  for i in range(k):
    s1 += d[i] % m
    if d[i] % m == 0:
      t1 += 1
  s2 = 0
  t2 = 0
  for i in range(n%k):
    s2 += d[i] % m
    if d[i] % m == 0:
      t2 += 1
  x += s1 * (n//k) + s2
  ans -= x//m
  ans -= t1 * (n//k) + t2

  return ans

for _ in range(q):
  n,x,m = map(int, input().split())
  print(query(n,x,m))

Score : 600 points

Problem Statement

We have a sequence of k numbers: d_0,d_1,...,d_{k - 1}.

Process the following q queries in order:

The i-th query contains three integers n_i, x_i, and m_i.
Let a_0,a_1,...,a_{n_i - 1} be the following sequence of n_i numbers: \begin{eqnarray} a_j = \begin{cases} x_i & ( j = 0 ) \\ a_{j - 1} + d_{(j - 1)~\textrm{mod}~k} & ( 0 < j \leq n_i - 1 ) \end{cases}\end{eqnarray}
Print the number of j~(0 \leq j < n_i - 1) such that (a_j~\textrm{mod}~m_i) < (a_{j + 1}~\textrm{mod}~m_i).

Here (y~\textrm{mod}~z) denotes the remainder of y divided by z, for two integers y and z~(z > 0).

Constraints

All values in input are integers.

1 \leq k, q \leq 5000

0 \leq d_i \leq 10^9

2 \leq n_i \leq 10^9

0 \leq x_i \leq 10^9

2 \leq m_i \leq 10^9

Input

Input is given from Standard Input in the following format:

k q
d_0 d_1 ... d_{k - 1}
n_1 x_1 m_1
n_2 x_2 m_2
:
n_q x_q m_q

Output

Print q lines.

The i-th line should contain the response to the i-th query.

Sample Input 1

3 1
3 1 4
5 3 2

Sample Output 1

1

For the first query, the sequence {a_j} will be 3,6,7,11,14.

(a_0~\textrm{mod}~2) > (a_1~\textrm{mod}~2)

(a_1~\textrm{mod}~2) < (a_2~\textrm{mod}~2)

(a_2~\textrm{mod}~2) = (a_3~\textrm{mod}~2)

(a_3~\textrm{mod}~2) > (a_4~\textrm{mod}~2)

Thus, the response to this query should be 1.

Sample Input 2

7 3
27 18 28 18 28 46 1000000000
1000000000 1 7
1000000000 2 10
1000000000 3 12

Sample Output 2

224489796
214285714
559523809


A,B,C=map(int, input().split())

if A==B or B==C or C==A:
  print('Yes')
else:
  print('No')

Score: 100 points

Problem Statement

A triple of numbers is said to be poor when two of those numbers are equal but the other number is different from those two numbers.

You will be given three integers A, B, and C. If this triple is poor, print Yes; otherwise, print No.

Constraints

A, B, and C are all integers between 1 and 9 (inclusive).

Input

Input is given from Standard Input in the following format:

A B C

Output

If the given triple is poor, print Yes; otherwise, print No.

Sample Input 1

5 7 5

Sample Output 1

Yes

A and C are equal, but B is different from those two numbers, so this triple is poor.

Sample Input 2

4 4 4

Sample Output 2

No

A, B, and C are all equal, so this triple is not poor.

Sample Input 3

4 9 6

Sample Output 3

No

Sample Input 4

3 3 4

Sample Output 4

Yes


N = int(input())
A_list = list(map(int, input().split()))
isDenied = False
for num in A_list:
    if num %2 == 0:
        if not (num%3 == 0 or num%5 == 0):
            print('DENIED')
            isDenied = True

if not isDenied:
    print('APPROVED')

Score: 200 points

Problem Statement

You are an immigration officer in the Kingdom of AtCoder. The document carried by an immigrant has some number of integers written on it, and you need to check whether they meet certain criteria.

According to the regulation, the immigrant should be allowed entry to the kingdom if and only if the following condition is satisfied:

All even numbers written on the document are divisible by 3 or 5.

If the immigrant should be allowed entry according to the regulation, output APPROVED; otherwise, print DENIED.

Notes

The condition in the statement can be rephrased as "If x is an even number written on the document, x is divisible by 3 or 5".
Here "if" and "or" are logical terms.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \dots A_N

Output

If the immigrant should be allowed entry according to the regulation, print APPROVED; otherwise, print DENIED.

Sample Input 1

5
6 7 9 10 31

Sample Output 1

APPROVED

The even numbers written on the document are 6 and 10.

All of them are divisible by 3 or 5, so the immigrant should be allowed entry.

Sample Input 2

3
28 27 24

Sample Output 2

DENIED

28 violates the condition, so the immigrant should not be allowed entry.


n=int(input())
a=list(map(int,input().split()))
tmp=0
for i in range(len(a)):
  tmp=a[i]
  if tmp%2==0:
    if tmp%3!=0:
      if tmp%5!=0:
        print("DENIED")
        break
  if i==n-1:
    print("APPROVED")


N = int(input())
A = list(map(int, input().split()))
A.sort()
B = []
ans='APPROVED'

for i in A:
    if (i % 2 == 0):
        B.append(i)

for i in B:
    if (i % 3 != 0 and i % 5 != 0):
        ans = 'DENIED'
        break

print(ans)


inpvals = []
lopvals = []
outvals = []

max_loop = int(input())
idx_loop = 0
while idx_loop < max_loop :
  inpvals.append(input())
  idx_loop = idx_loop+1

max_cnt = 0
lopvals = set(inpvals)
for inpval in lopvals :
  cnt = inpvals.count(inpval)

  if max_cnt <= cnt :
    if max_cnt != cnt :
      outvals = []
    max_cnt = cnt
    outvals.append(inpval)

outvals.sort()

for outval in outvals :
  print(outval)

Score: 300 points

Problem Statement

We have N voting papers. The i-th vote (1 \leq i \leq N) has the string S_i written on it.

Print all strings that are written on the most number of votes, in lexicographical order.

Constraints

1 \leq N \leq 2 \times 10^5

S_i (1 \leq i \leq N) are strings consisting of lowercase English letters.

The length of S_i (1 \leq i \leq N) is between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

Print all strings in question in lexicographical order.

Sample Input 1

7
beat
vet
beet
bed
vet
bet
beet

Sample Output 1

beet
vet

beet and vet are written on two sheets each, while beat, bed, and bet are written on one vote each. Thus, we should print the strings beet and vet.

Sample Input 2

8
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo
buffalo

Sample Output 2

buffalo

Sample Input 3

7
bass
bass
kick
kick
bass
kick
kick

Sample Output 3

kick

Sample Input 4

4
ushi
tapu
nichia
kun

Sample Output 4

kun
nichia
tapu
ushi


n = int(input())
a = []
for i in range(n):
  si = str(input())
  a.append(si)
a.sort()
mx = 1
w = []
j = 1
for i in range(1,n):
  if a[i] == a[i-1]:
    j += 1
  else:
    if j > mx:
      w = [a[i-1]]
      mx = j
    elif j == mx:
      w.append(a[i-1])
    j = 1
if j > mx:
  w = [a[n-1]]
elif j == mx:
  w.append(a[n-1])
for wi in w:
  print(wi)


from collections import Counter

l=[]
N=int(input())
for _ in range(N):
    l.append(input())

c = Counter(l)
print(c)


S,T = input().split()
A,B = map(int, input().split())
U = input()

if S == U:
  A -= 1
else:
  B -= 1

print(A,B)

Score : 100 points

Problem Statement

We have A balls with the string S written on each of them and B balls with the string T written on each of them.

From these balls, Takahashi chooses one with the string U written on it and throws it away.

Find the number of balls with the string S and balls with the string T that we have now.

Constraints

S, T, and U are strings consisting of lowercase English letters.

The lengths of S and T are each between 1 and 10 (inclusive).

S \not= T

S=U or T=U.

1 \leq A,B \leq 10

A and B are integers.

Input

Input is given from Standard Input in the following format:

S T
A B
U

Output

Print the answer, with space in between.

Sample Input 1

red blue
3 4
red

Sample Output 1

2 4

Takahashi chose a ball with red written on it and threw it away.
Now we have two balls with the string S and four balls with the string T.

Sample Input 2

red blue
5 5
blue

Sample Output 2

5 4

Takahashi chose a ball with blue written on it and threw it away.
Now we have five balls with the string S and four balls with the string T.


input_str = input().split()
input_int = list(map(int, input().split()))
input_line = input()

input_dict = {}

input_dict[input_str[0]] = input_int[0]
input_dict[input_str[1]] = input_int[1]

input_dict[input_line] = input_dict[input_line] - 1

for i in input_dict:
    print(input_dict[i],end=' ')


# a = list(map(int, input().split()))
# N = int(input())

s, t = input().split()
a, b = list(map(int, input().split()))
string = input()
if string == "red":
    print(a-1, b)
else:
    print(a, b-1)


Score : 200 points

Problem Statement

Given is a string S. Replace every character in S with x and print the result.

Constraints

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Replace every character in S with x and print the result.

Sample Input 1

sardine

Sample Output 1

xxxxxxx

Replacing every character in S with x results in xxxxxxx.

Sample Input 2

xxxx

Sample Output 2

xxxx

Sample Input 3

gone

Sample Output 3

xxxx


S = input()

ls = len(S)

ans = ''
for i in range(ls):
    ans += 'x'

print(ans)


N = int(input())
A = list(map(int, input().split()))

if len(set(A)) == len(A):
	print("YES")
else:
	print("NO")

Score : 300 points

Problem Statement

Given is a sequence of integers A_1, A_2, ..., A_N.
If its elements are pairwise distinct, print YES; otherwise, print NO.

Constraints

2 ≤ N ≤ 200000

1 ≤ A_i ≤ 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

If the elements of the sequence are pairwise distinct, print YES; otherwise, print NO.

Sample Input 1

5
2 6 1 4 5

Sample Output 1

YES

The elements are pairwise distinct.

Sample Input 2

6
4 1 3 1 6 2

Sample Output 2

NO

The second and fourth elements are identical.

Sample Input 3

2
10000000 10000000

Sample Output 3

NO


N = input()
M = list(map(int,input().split()))
X = set(M)
Y = len(X)
Z = len(M)
if Y == Z:
    print("YES")

else:
    print("NO")


def expected_value(arr):
    return ([sum(range(1, ele+1)) / ele for ele in arr])

n, k = [int(x) for x in input().split()]
arr = list(map(int,input().split()))
sum_tmp = [sum(arr[i:i+k]) for i in range(n - (k - 1))]
max_v = max(sum_tmp)
max_i = sum_tmp.index(max_v)
print(sum(expected_value(arr[max_i:max_i + k])))

Score : 400 points

Problem Statement

We have N dice arranged in a line from left to right. The i-th die from the left shows p_i numbers from 1 to p_i with equal probability when thrown.

We will choose K adjacent dice, throw each of them independently, and compute the sum of the numbers shown. Find the maximum possible value of the expected value of this sum.

Constraints

1 ≤ K ≤ N ≤ 200000

1 ≤ p_i ≤ 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 ... p_N

Output

Print the maximum possible value of the expected value of the sum of the numbers shown.

Your output will be considered correct when its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

5 3
1 2 2 4 5

Sample Output 1

7.000000000000

When we throw the third, fourth, and fifth dice from the left, the expected value of the sum of the numbers shown is 7. This is the maximum value we can achieve.

Sample Input 2

4 1
6 6 6 6

Sample Output 2

3.500000000000

Regardless of which die we choose, the expected value of the number shown is 3.5.

Sample Input 3

10 4
17 13 13 12 15 20 10 13 17 11

Sample Output 3

32.000000000000


N,K=map(int, input().split())
p=list(map(int, input().split()))
kitai=[0]*N
kitaiwa=[0]*(N-K+1)
for i in range(N):
    kitai[i]=(p[i]+1)/2
for i in range(N-K+1):
    kitaiwa[i] = sum(kitai[i:i+K])
print(max(kitaiwa))


Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300


from sys import stdin
import sys
import numpy as np
import collections
from functools import cmp_to_key
import heapq

##  input functions for me
def rsa(sep = ''):
    if sep == '' :
        return input().split()
    else: return input().split(sep)
def rip(sep = ''):
    if sep == '' :
        return map(int, input().split())
    else: return map(int, input().split(sep))
def ria(sep = ''):
    return list(rip(sep))
def ri(): return int(input())
def rd(): return float(input())
def rs(): return input()
##

def inv(v, mod):
    return pow(v, mod-2, mod)
def main():
    r1, c1, r2, c2 = rip()

    MM = int(2e6+10)
    fact = [0] * MM
    finv = [0] * MM
    fact[0] = 1
    finv[0] = 1
    mod = int(1e9) + 7
    for i in range(1,MM): fact[i] = (fact[i-1] * i % mod)
    '''
    finv[MM-1] = inv(fact[MM-1], mod)
    for i in reversed(range(1,MM-1)):
        finv[i] = finv[i + 1] * (i + 1) % mod
    def sum_naive(r, c):
        #[0, r) * [0, c)
        ret = 0
        for i in range(r):
            ret += fact[i + 1 + c - 1] * finv[i + 1] * finv[c - 1] % mod
        return ret
    def sum(r, c):
        #[0, r) * [0, c)
        ret = fact[r + c] * finv[r] * finv[c] % mod
        ret += -1 + mod
        ret %= mod
        return ret
    '''
    def sum(r, c):
        #[0, r) * [0, c)
        ret = fact[r + c] * inv(fact[r], mod) * inv(fact[c], mod) % mod
        ret += -1 + mod
        ret %= mod
        return ret
    ans = 0
    ans += sum(r2 + 1, c2 + 1)
    ans -= sum(r2 + 1, c1)
    ans -= sum(r1, c2 + 1)
    ans += sum(r1, c1)
    ans %= mod
    print(ans)

if __name__ == "__main__":
    main()

Score : 600 points

Problem Statement

Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.

Let us define a function f(r, c) as follows:

f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)

Given are integers r_1, r_2, c_1, and c_2.
Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).

Constraints

1 ≤ r_1 ≤ r_2 ≤ 10^6

1 ≤ c_1 ≤ c_2 ≤ 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

r_1 c_1 r_2 c_2

Output

Print the sum of f(i, j) modulo (10^9+7).

Sample Input 1

1 1 2 2

Sample Output 1

14

For example, there are two paths from the point (0, 0) to the point (1, 1): (0,0) → (0,1) → (1,1) and (0,0) → (1,0) → (1,1), so f(1,1)=2.

Similarly, f(1,2)=3, f(2,1)=3, and f(2,2)=6. Thus, the sum is 14.

Sample Input 2

314 159 2653 589

Sample Output 2

602215194


H, A = map(int, input().split())
if A > H:
    print(1)
elif H % A == 0:
    print(H//A)
else:
    print((H//A)+1)

Score : 100 points

Problem Statement

Serval is fighting with a monster.

The health of the monster is H.

In one attack, Serval can decrease the monster's health by A.
There is no other way to decrease the monster's health.

Serval wins when the monster's health becomes 0 or below.

Find the number of attacks Serval needs to make before winning.

Constraints

1 \leq H \leq 10^4

1 \leq A \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H A

Output

Print the number of attacks Serval needs to make before winning.

Sample Input 1

10 4

Sample Output 1

3

After one attack, the monster's health will be 6.

After two attacks, the monster's health will be 2.

After three attacks, the monster's health will be -2.

Thus, Serval needs to make three attacks to win.

Sample Input 2

1 10000

Sample Output 2

1

Sample Input 3

10000 1

Sample Output 3

10000


H,A= map(int, input().split())

if H%A==0:
    print(int(H/A))
else:
    print(int(H/A)+1)


h,n = map(int,input().split())
a = list(map(int,input().split()))
ans = 0
for i in range(n):
  ans += a[i]
if(ans>=h):
  print("Yes")
else:
  print("No")

Score : 200 points

Problem Statement

Raccoon is fighting with a monster.

The health of the monster is H.

Raccoon can use N kinds of special moves. Using the i-th move decreases the monster's health by A_i.
There is no other way to decrease the monster's health.

Raccoon wins when the monster's health becomes 0 or below.

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Constraints

1 \leq H \leq 10^9

1 \leq N \leq 10^5

1 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H N
A_1 A_2 ... A_N

Output

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Sample Input 1

10 3
4 5 6

Sample Output 1

Yes

The monster's health will become 0 or below after, for example, using the second and third moves.

Sample Input 2

20 3
4 5 6

Sample Output 2

No

Sample Input 3

210 5
31 41 59 26 53

Sample Output 3

Yes

Sample Input 4

211 5
31 41 59 26 53

Sample Output 4

No


def main():
    H, N = map(int, input().split())
    A = list(map(int, input().split()))
    s = sum(A)
    if H > s:
        print('No')
    else:
        print('Yes')
main()


a = input()
l = a.split()
h = int(l[0])
n = int(l[1])
b = input()
ll = b.split()
for i in range(0,n):
    ll[i] = int(ll[i])

sum = sum(ll)
if h<sum+1:
    print("Yes")
else:
    print("No")


h, n = (int(x) for x in input().split())
A = list(map(int, input().split()))

if n == len(A):
    if sum(A) > h:
        print('yes')
    else:
        print('No')


#!/usr/bin/env python3
from collections import defaultdict, Counter
from itertools import product, groupby, count, permutations, combinations
from math import pi, sqrt
from collections import deque
from bisect import bisect, bisect_left, bisect_right
from string import ascii_lowercase
from functools import lru_cache
import sys
sys.setrecursionlimit(10000)
INF = float("inf")
YES, Yes, yes, NO, No, no = "YES", "Yes", "yes", "NO", "No", "no"
dy4, dx4 = [0, 1, 0, -1], [1, 0, -1, 0]
dy8, dx8 = [0, -1, 0, 1, 1, -1, -1, 1], [1, 0, -1, 0, 1, 1, -1, -1]

def inside(y, x, H, W):
    return 0 <= y < H and 0 <= x < W

def ceil(a, b):
    return (a + b - 1) // b

def sum_of_arithmetic_progression(s, d, n):
    return n * (2 * s + (n - 1) * d) // 2

def gcd(a, b):
    if b == 0:
        return a
    return gcd(b, a % b)

def lcm(a, b):
    g = gcd(a, b)
    return a / g * b

def solve():
    N, K = map(int, input().split())
    H = list(sorted(list(map(int, input().split())), reverse=True))

    ans = 0
    for i in range(N):
        if i < K:
            continue
        else:
            ans += H[i]
    print(ans)

def main():
    solve()

if __name__ == '__main__':
    main()

Score : 300 points

Problem Statement

Fennec is fighting with N monsters.

The health of the i-th monster is H_i.

Fennec can do the following two actions:

Attack: Fennec chooses one monster. That monster's health will decrease by 1.

Special Move: Fennec chooses one monster. That monster's health will become 0.

There is no way other than Attack and Special Move to decrease the monsters' health.

Fennec wins when all the monsters' healths become 0 or below.

Find the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning when she can use Special Move at most K times.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 2 \times 10^5

1 \leq H_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
H_1 ... H_N

Output

Print the minimum number of times Fennec needs to do Attack (not counting Special Move) before winning.

Sample Input 1

3 1
4 1 5

Sample Output 1

5

By using Special Move on the third monster, and doing Attack four times on the first monster and once on the second monster, Fennec can win with five Attacks.

Sample Input 2

8 9
7 9 3 2 3 8 4 6

Sample Output 2

0

She can use Special Move on all the monsters.

Sample Input 3

3 0
1000000000 1000000000 1000000000

Sample Output 3

3000000000

Watch out for overflow.


h,n=map(int,input().split())
m=list(map(int, input().split()))
m.sort(reverse=True)
del m[:n]
a=0

for i in range(len(m)):
  a=a+m[i]

print(a)


# import bisect
# import heapq
# from copy import deepcopy
# from collections import deque
# from collections import Counter
# from itertools import accumulate
# from itertools import permutations
# import numpy as np
# import math
# mod = 10**9 + 7

# n = int(input())
h, n = map(int, input().split())
# a = list(map(int, input().split()))
a, b = [0] * n, [0] * n

for i in range(n):
    a[i], b[i] = map(int,input().split())

dp = [10**9+7] * (h+1)
dp[0] = 0

for i in range(n):

    for j in range(h):

        dp[min(j+a[i], h)] = min(dp[min(j+a[i],h)], dp[j]+b[i])

print(dp[h])

Score : 500 points

Problem Statement

Ibis is fighting with a monster.

The health of the monster is H.

Ibis can cast N kinds of spells. Casting the i-th spell decreases the monster's health by A_i, at the cost of B_i Magic Points.

The same spell can be cast multiple times. There is no way other than spells to decrease the monster's health.

Ibis wins when the health of the monster becomes 0 or below.

Find the minimum total Magic Points that have to be consumed before winning.

Constraints

1 \leq H \leq 10^4

1 \leq N \leq 10^3

1 \leq A_i \leq 10^4

1 \leq B_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H N
A_1 B_1
:
A_N B_N

Output

Print the minimum total Magic Points that have to be consumed before winning.

Sample Input 1

9 3
8 3
4 2
2 1

Sample Output 1

4

First, let us cast the first spell to decrease the monster's health by 8, at the cost of 3 Magic Points. The monster's health is now 1.

Then, cast the third spell to decrease the monster's health by 2, at the cost of 1 Magic Point. The monster's health is now -1.

In this way, we can win at the total cost of 4 Magic Points.

Sample Input 2

100 6
1 1
2 3
3 9
4 27
5 81
6 243

Sample Output 2

100

It is optimal to cast the first spell 100 times.

Sample Input 3

9999 10
540 7550
691 9680
700 9790
510 7150
415 5818
551 7712
587 8227
619 8671
588 8228
176 2461

Sample Output 3

139815


H,N = map(int,input().split())
MAXH = H+10000
DP = [float("inf")]*MAXH
DP[0] = 0
for i in range(N):
    a,b = map(int,input().split())
    for j in range(MAXH):
        if j+a>MAXH-1:break
        DP[j+a] = min(DP[j+a],DP[j]+b)
print(min(DP[H:]))


n,m = map(int,input().split())
print('Yes' if n == m else 'No')

Score : 100 points

Problem Statement

Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.

The problem has N test cases, all of which must be passed to get an AC verdict.

Takahashi's submission has passed M cases out of the N test cases.

Determine whether Takahashi's submission gets an AC.

Constraints

1 \leq N \leq 100

0 \leq M \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

If Takahashi's submission gets an AC, print Yes; otherwise, print No.

Sample Input 1

3 3

Sample Output 1

Yes

All three test cases have been passed, so his submission gets an AC.

Sample Input 2

3 2

Sample Output 2

No

Only two out of the three test cases have been passed, so his submission does not get an AC.

Sample Input 3

1 1

Sample Output 3

Yes


a, b = map(int, input().split())

A = str(a) * b
B = str(b) * a

AB = [A,B]

print(min(AB))

Score : 200 points

Problem Statement

Given are 1-digit positive integers a and b. Consider these two strings: the concatenation of b copies of the digit a, and the concatenation of a copies of the digit b. Which of these is lexicographically smaller?

Constraints

1 \leq a \leq 9

1 \leq b \leq 9

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the lexicographically smaller of the two strings. (If the two strings are equal, print one of them.)

Sample Input 1

4 3

Sample Output 1

3333

We have two strings 444 and 3333. Between them, 3333 is the lexicographically smaller.

Sample Input 2

7 7

Sample Output 2

7777777


L, R = map(str,input().split())
list = [L, R]
sorted(list)
print(list[0])


N = int(input())
P = [int(x) for x in input().split()]

ans = 1
for i in range(1, N):
    tmp = 1
    j = 0
    while j < i:
        if P[i] > P[j]:
            tmp = 0
            break
        j += 1
    ans += tmp
print(ans)

Score : 300 points

Problem Statement

Given is a permutation P_1, \ldots, P_N of 1, \ldots, N.
Find the number of integers i (1 \leq i \leq N) that satisfy the following condition:

For any integer j (1 \leq j \leq i), P_i \leq P_j.

Constraints

1 \leq N \leq 2 \times 10^5

P_1, \ldots, P_N is a permutation of 1, \ldots, N.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
P_1 ... P_N

Output

Print the number of integers i that satisfy the condition.

Sample Input 1

5
4 2 5 1 3

Sample Output 1

3

i=1, 2, and 4 satisfy the condition, but i=3 does not - for example, P_i > P_j holds for j = 1.

Similarly, i=5 does not satisfy the condition, either. Thus, there are three integers that satisfy the condition.

Sample Input 2

4
4 3 2 1

Sample Output 2

4

All integers i (1 \leq i \leq N) satisfy the condition.

Sample Input 3

6
1 2 3 4 5 6

Sample Output 3

1

Only i=1 satisfies the condition.

Sample Input 4

8
5 7 4 2 6 8 1 3

Sample Output 4

4

Sample Input 5

1
1

Sample Output 5

1


from functools import reduce
def lcm_base(x, y):
    return (x * y) // gcd(x, y)

def lcm_list(numbers):
    return reduce(lcm_base, numbers, 1)

def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

n = int(input())
A = [int(i) for i in input().split()]
k = lcm_list(A)
r = 0
for i in range(n):
    r += k//A[i]
    if r > 10**9+7:
        r %= 10**9+7
print(r)

Score : 400 points

Problem Statement

Given is a positive integer N.

Find the number of pairs (A, B) of positive integers not greater than N that satisfy the following condition:

When A and B are written in base ten without leading zeros, the last digit of A is equal to the first digit of B, and the first digit of A is equal to the last digit of B.

Constraints

1 \leq N \leq 2 \times 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

25

Sample Output 1

17

The following 17 pairs satisfy the condition: (1,1), (1,11), (2,2), (2,22), (3,3), (4,4), (5,5), (6,6), (7,7), (8,8), (9,9), (11,1), (11,11), (12,21), (21,12), (22,2), and (22,22).

Sample Input 2

1

Sample Output 2

1

Sample Input 3

100

Sample Output 3

108

Sample Input 4

2020

Sample Output 4

40812

Sample Input 5

200000

Sample Output 5

400000008


"""
最小公倍数LCMを求めて、LCMを各Aで除した商を合計すればよい
"""
from math import gcd
def lcm(a,b):
    return a*b//gcd(a,b)
def main():
    mod = 10**9 +7
    N = int(input())
    A = list(map(int,input().split()))
    l = A[0]
    for i in range(1,N):
        l = lcm(l,A[i])

    ans = 0
    for a in A:
        ans += l//a
        ans %= mod
    print(ans)
main()

Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).


from functools import reduce
import fractions

N = int(input())

A = list(map(int, input().split()))

def lcm(a):
  x = a[0]
  for i in range(1, len(a)):
    x = (x * a[i]) // fractions.gcd(x, a[i])
  return x

l = lcm(A)
num=0
for a in A:
  num += l/a
print(int(num % (10e9 +7)))


import sys
stdin = sys.stdin
sys.setrecursionlimit(10**6)
ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
nn = lambda: list(stdin.readline().split())
ns = lambda: stdin.readline().rstrip()

# cook your dish here
def dijkstra_back(s,t,n,w,cost):
    #sからtへの最短経路の経路復元
    prev = [s] * n #最短経路の直前の頂点
    d = [float("inf")] * n
    used = [False] * n
    d[s] = 0

    while True:
        v = -1
        for i in range(n):
            if (not used[i]) and (v == -1):
               v = i
            elif (not used[i]) and d[i] < d[v]:
                v = i
        if v == -1:
               break
        used[v] = True

        for i in range(n):
            if d[i] > d[v] + cost[v][i]:
                d[i] = d[v] + cost[v][i]
                prev[i] = v

    path = [t]
    while prev[t] != s:
        path.append(prev[t])
        prev[t] = prev[prev[t]]
    path.append(s)
    path = path[::-1]
    return path

################################

n = ni()
cost = [[float("inf") for i in range(n)] for i in range(n)]
#cost[u][v] : 辺uvのコスト(存在しないときはinf この場合は10**10)
for i in range(n-1):
    a,b = na()
    cost[a-1][b-1] = 1
    cost[b-1][a-1] = 1

dp = set()
m = ni()
for i in range(m):
    u,v = na()
    d = dijkstra_back(u-1,v-1,n,n-1,cost)
    for j in range(len(d)-1):
        a,b = min(d[j],d[j+1]),max(d[j],d[j+1])
        dp.add((a,b))

a = len(dp)
print(2**n-2**(n-1-a))

Score : 600 points

Problem Statement

We have a tree with N vertices numbered 1 to N.
The i-th edge in this tree connects Vertex a_i and Vertex b_i.

Consider painting each of these edges white or black. There are 2^{N-1} such ways to paint the edges. Among them, how many satisfy all of the following M restrictions?

The i-th (1 \leq i \leq M) restriction is represented by two integers u_i and v_i, which mean that the path connecting Vertex u_i and Vertex v_i must contain at least one edge painted black.

Constraints

2 \leq N \leq 50

1 \leq a_i,b_i \leq N

The graph given in input is a tree.

1 \leq M \leq \min(20,\frac{N(N-1)}{2})

1 \leq u_i < v_i \leq N

If i \not= j, either u_i \not=u_j or v_i\not=v_j

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}
M
u_1 v_1
:
u_M v_M

Output

Print the number of ways to paint the edges that satisfy all of the M conditions.

Sample Input 1

3
1 2
2 3
1
1 3

Sample Output 1

3

The tree in this input is shown below:

All of the M restrictions will be satisfied if Edge 1 and 2 are respectively painted (white, black), (black, white), or (black, black), so the answer is 3.

Sample Input 2

2
1 2
1
1 2

Sample Output 2

1

The tree in this input is shown below:

All of the M restrictions will be satisfied only if Edge 1 is painted black, so the answer is 1.

Sample Input 3

5
1 2
3 2
3 4
5 3
3
1 3
2 4
2 5

Sample Output 3

9

The tree in this input is shown below:

Sample Input 4

8
1 2
2 3
4 3
2 5
6 3
6 7
8 6
5
2 7
3 5
1 6
2 8
7 8

Sample Output 4

62

The tree in this input is shown below:


h = int(input())
w = int(input())
n = int(input())
c = 0

if h >= w:
  if n % h == 0:
    c = n // h
  else:
    c = n // h + 1
else:
  if n % w == 0:
    c = n // w
  else:
    c = n // w + 1
print(c)

Score : 100 points

Problem Statement

We have a grid with H rows and W columns, where all the squares are initially white.

You will perform some number of painting operations on the grid.
In one operation, you can do one of the following two actions:

Choose one row, then paint all the squares in that row black.

Choose one column, then paint all the squares in that column black.

At least how many operations do you need in order to have N or more black squares in the grid?
It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

Constraints

1 \leq H \leq 100

1 \leq W \leq 100

1 \leq N \leq H \times W

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H
W
N

Output

Print the minimum number of operations needed.

Sample Input 1

3
7
10

Sample Output 1

2

You can have 14 black squares in the grid by performing the "row" operation twice, on different rows.

Sample Input 2

14
12
112

Sample Output 2

8

Sample Input 3

2
100
200

Sample Output 3

2


n = int(input())
robot = []
for i in range(n):
    x, l = map(int,input().split())
    robot.append([x - l, x + l]) #left, right
robot.sort()
preright = robot[0][0]
ans = 0
for i in range(n):
    left = robot[i][0]
    #print(preright, left)
    if preright <= left:
        ans += 1
        preright = robot[i][1]
print(ans)

Score : 200 points

Problem Statement

In a factory, there are N robots placed on a number line.
Robot i is placed at coordinate X_i and can extend its arms of length L_i in both directions, positive and negative.

We want to remove zero or more robots so that the movable ranges of arms of no two remaining robots intersect.
Here, for each i (1 \leq i \leq N), the movable range of arms of Robot i is the part of the number line between the coordinates X_i - L_i and X_i + L_i, excluding the endpoints.

Find the maximum number of robots that we can keep.

Constraints

1 \leq N \leq 100,000

0 \leq X_i \leq 10^9 (1 \leq i \leq N)

1 \leq L_i \leq 10^9 (1 \leq i \leq N)

If i \neq j, X_i \neq X_j.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 L_1
X_2 L_2
\vdots
X_N L_N

Output

Print the maximum number of robots that we can keep.

Sample Input 1

4
2 4
4 3
9 3
100 5

Sample Output 1

3

By removing Robot 2, we can keep the other three robots.

Sample Input 2

2
8 20
1 10

Sample Output 2

1

Sample Input 3

5
10 1
2 1
4 1
6 1
8 1

Sample Output 3

5


n = int(input())
dic = {}

for i in range(n):
    inp = list(map(int, input().split()))
    dic[inp[0]] = inp[1]

dic = sorted(dic.items())
pre = dic[0][0]+dic[0][1]
ans = 1

for i in range(n-1):
    if pre <= dic[i+1][0]-dic[i+1][1]:
        pre = dic[i+1][0]+dic[i+1][1]
        ans += 1

print(ans)


MIN, MAX = 1, 10 ** 9

n, k, s = map(int, input().split())

def solve(n, k, s):
    mid = MAX // 2
    a = s
    b = MIN if s > mid else MAX
    return [a] * k + [b] * (n - k)

print(*solve(n, k, s))

Score : 400 points

Problem Statement

Given are three integers N, K, and S.

Find a sequence A_1, A_2, ..., A_N of N integers between 1 and 10^9 (inclusive) that satisfies the condition below.
We can prove that, under the conditions in Constraints, such a sequence always exists.

There are exactly K pairs (l, r) of integers such that 1 \leq l \leq r \leq N and A_l + A_{l + 1} + \cdots + A_r = S.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N

1 \leq S \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K S

Output

Print a sequence satisfying the condition, in the following format:

A_1 A_2 ... A_N

Sample Input 1

4 2 3

Sample Output 1

1 2 3 4

Two pairs (l, r) = (1, 2) and (3, 3) satisfy the condition in the statement.

Sample Input 2

5 3 100

Sample Output 2

50 50 50 30 70


n=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
inf=100000000000
ans=inf
for i in range(1<<n):#for i in range(2**n)と同じ
    x=[[a[j],j,0] if ((i >> j) & 1) == 1 else [b[j],j,1] for j in range(n)]
    y=[x[i][0] for i in range(n)]
    x.sort()
    for i in range(n):
        if abs(x[i][1]-i)%2!=x[i][2]:
            break
    else:
        c=0
        for i in range(1,n):
            for j in range(i,0,-1):
                if y[j] >= y[j-1]:
                    break
                else:
                    y[j],y[j-1]=y[j-1],y[j]
                    c+=1
        ans=min(ans,c)
if ans==inf:
    print(-1)
else:
    print(ans)

Score : 700 points

Problem Statement

We have N cards numbered 1, 2, ..., N.
Card i (1 \leq i \leq N) has an integer A_i written in red ink on one side and an integer B_i written in blue ink on the other side.
Initially, these cards are arranged from left to right in the order from Card 1 to Card N, with the red numbers facing up.

Determine whether it is possible to have a non-decreasing sequence facing up from left to right (that is, for each i (1 \leq i \leq N - 1), the integer facing up on the (i+1)-th card from the left is not less than the integer facing up on the i-th card from the left) by repeating the operation below. If the answer is yes, find the minimum number of operations required to achieve it.

Choose an integer i (1 \leq i \leq N - 1).
Swap the i-th and (i+1)-th cards from the left, then flip these two cards.

Constraints

1 \leq N \leq 18

1 \leq A_i, B_i \leq 50 (1 \leq i \leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N

Output

If it is impossible to have a non-decreasing sequence, print -1.
If it is possible, print the minimum number of operations required to achieve it.

Sample Input 1

3
3 4 3
3 2 3

Sample Output 1

1

By doing the operation once with i = 1, we have a sequence [2, 3, 3] facing up, which is non-decreasing.

Sample Input 2

2
2 1
1 2

Sample Output 2

-1

After any number of operations, we have the sequence [2, 1] facing up, which is not non-decreasing.

Sample Input 3

4
1 2 3 4
5 6 7 8

Sample Output 3

0

No operation may be required.

Sample Input 4

5
28 15 22 43 31
20 22 43 33 32

Sample Output 4

-1

Sample Input 5

5
4 46 6 38 43
33 15 18 27 37

Sample Output 5

3


Score : 100 points

Problem Statement

Given is a lowercase English letter C that is not z. Print the letter that follows C in alphabetical order.

Constraints

C is a lowercase English letter that is not z.

Input

Input is given from Standard Input in the following format:

C

Output

Print the letter that follows C in alphabetical order.

Sample Input 1

a

Sample Output 1

b

a is followed by b.

Sample Input 2

y

Sample Output 2

z

y is followed by z.


def line(n):
  for _ in range(n):
    p,S = map(str,input().split())
    yield p,S

def main():
  N,M = map(int,input().split())
  X = list(line(M))
  li = []
  AC = 0
  WA = 0
  for p,S in X:
    if S == 'AC':
      if not p in li:
        li.append(p)
        AC += 1
    if S == 'WA' and not p in li:
      WA += 1
  print(int(AC),int(WA))

main()

Score : 300 points

Problem Statement

Takahashi participated in a contest on AtCoder.

The contest had N problems.

Takahashi made M submissions during the contest.

The i-th submission was made for the p_i-th problem and received the verdict S_i (AC or WA).

The number of Takahashi's correct answers is the number of problems on which he received an AC once or more.

The number of Takahashi's penalties is the sum of the following count for the problems on which he received an AC once or more: the number of WAs received before receiving an AC for the first time on that problem.

Find the numbers of Takahashi's correct answers and penalties.

Constraints

N, M, and p_i are integers.

1 \leq N \leq 10^5

0 \leq M \leq 10^5

1 \leq p_i \leq N

S_i is AC or WA.

Input

Input is given from Standard Input in the following format:

N M
p_1 S_1
:
p_M S_M

Output

Print the number of Takahashi's correct answers and the number of Takahashi's penalties.

Sample Input 1

2 5
1 WA
1 AC
2 WA
2 AC
2 WA

Sample Output 1

2 2

In his second submission, he received an AC on the first problem for the first time. Before this, he received one WA on this problem.

In his fourth submission, he received an AC on the second problem for the first time. Before this, he received one WA on this problem.

Thus, he has two correct answers and two penalties.

Sample Input 2

100000 3
7777 AC
7777 AC
7777 AC

Sample Output 2

1 0

Note that it is pointless to get an AC more than once on the same problem.

Sample Input 3

6 0

Sample Output 3

0 0


listA=[] #appendのために宣言が必要
while True:
    try:
        listA.append(input().split())
    except:
        break;

N = int(listA[0][0])
M = int(listA[0][1])

ans = [0] * N
pena = [0] * N

for i in range(1, M):
    num=int(listA[i][0]) -1
    res=listA[i][1]
    if res=="WA" and ans[num] == 0:
        pena[num] =  pena[num]+1
    if res=="AC" and ans[num] == 0:
        ans[num] = ans[num] + 1

print('%d %d' % (sum(ans), sum(pena) ))


from collections import deque

def bfs(yx, H, W, S):
    dirs = (1, W+2, -1, -W-2)
    visited = [0] * ((H+2) * (W+2))
    q = deque()
    q.append((yx, 0))
    max_depth = 0
    while q:
        yx, depth = q.popleft()
        visited[yx] = 1
        max_depth = max(max_depth, depth)
        for dyx in dirs:
            nyx = yx + dyx
            if S[nyx] == '.' and visited[nyx] == 0:
                q.append((nyx, depth+1))
    return max_depth

def solve():
    H, W = map(int, input().split())
    S = deque()
    S.append('#' * (W+2))
    for h in range(H):
        buf = input()
        S.append('#' + buf + '#')
    S.append('#' * (W+2))
    S = ''.join(S)

    dist_max = 0
    for yx in range((H+2) * (W+2)):
        if S[yx] == '.':
            d = bfs(yx, H, W, S)
            dist_max = max(d, dist_max)
    print(dist_max)

if __name__ == '__main__':
    solve()

Score : 400 points

Problem Statement

Takahashi has a maze, which is a grid of H \times W squares with H horizontal rows and W vertical columns.

The square at the i-th row from the top and the j-th column is a "wall" square if S_{ij} is #, and a "road" square if S_{ij} is ..

From a road square, you can move to a horizontally or vertically adjacent road square.

You cannot move out of the maze, move to a wall square, or move diagonally.

Takahashi will choose a starting square and a goal square, which can be any road squares, and give the maze to Aoki.

Aoki will then travel from the starting square to the goal square, in the minimum number of moves required.

In this situation, find the maximum possible number of moves Aoki has to make.

Constraints

1 \leq H,W \leq 20

S_{ij} is . or #.

S contains at least two occurrences of ..

Any road square can be reached from any road square in zero or more moves.

Input

Input is given from Standard Input in the following format:

H W
S_{11}...S_{1W}
:
S_{H1}...S_{HW}

Output

Print the maximum possible number of moves Aoki has to make.

Sample Input 1

3 3
...
...
...

Sample Output 1

4

If Takahashi chooses the top-left square as the starting square and the bottom-right square as the goal square, Aoki has to make four moves.

Sample Input 2

3 5
...#.
.#.#.
.#...

Sample Output 2

10

If Takahashi chooses the bottom-left square as the starting square and the top-right square as the goal square, Aoki has to make ten moves.


def mod_inv(a, m):
    return pow(a, m - 2, m)

def combination(a, b):
    if b > a - b:
        return combination(a, a - b)
    if b < 0:
        return 0
    ans_mul = 1
    ans_div = 1
    for i in range(b):
        ans_mul *= a - i
        ans_div *= i + 1
        ans_mul %= MOD
        ans_div %= MOD
    ans = ans_mul * mod_inv(ans_div, MOD)
    return ans

N, K = map(int, input().split())
A = sorted(map(int, input().split()))
MOD = 10 ** 9 + 7
# max
mx = 0
mn = 0
for i in range(N):
    mx += A[i] * combination(i, K - 1)
    mn += A[i] * combination(N - i - 1, K - 1)
    mx %= MOD
    mn %= MOD
print((mx - mn) % MOD)

Score : 500 points

Problem Statement

For a finite set of integers X, let f(X)=\max X - \min X.

Given are N integers A_1,...,A_N.

We will choose K of them and let S be the set of the integers chosen. If we distinguish elements with different indices even when their values are the same, there are {}_N C_K ways to make this choice. Find the sum of f(S) over all those ways.

Since the answer can be enormous, print it \bmod (10^9+7).

Constraints

1 \leq N \leq 10^5

1 \leq K \leq N

|A_i| \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 ... A_N

Output

Print the answer \bmod (10^9+7).

Sample Input 1

4 2
1 1 3 4

Sample Output 1

11

There are six ways to choose S: \{1,1\},\{1,3\},\{1,4\},\{1,3\},\{1,4\}, \{3,4\} (we distinguish the two 1s). The value of f(S) for these choices are 0,2,3,2,3,1, respectively, for the total of 11.

Sample Input 2

6 3
10 10 10 -10 -10 -10

Sample Output 2

360

There are 20 ways to choose S. In 18 of them, f(S)=20, and in 2 of them, f(S)=0.

Sample Input 3

3 1
1 1 1

Sample Output 3

0

Sample Input 4

10 6
1000000000 1000000000 1000000000 1000000000 1000000000 0 0 0 0 0

Sample Output 4

999998537

Print the sum \bmod (10^9+7).


#coding:utf-8
import sys
sys.setrecursionlimit(10**6)
write = sys.stdout.write
dbg = lambda *something : print(*something) if DEBUG else 0
DEBUG = False
def main(given = sys.stdin.readline):
    input = lambda : given().rstrip()
    LMIIS = lambda : list(map(int,input().split()))
    II = lambda : int(input())
    XLMIIS = lambda x : [LMIIS() for _ in range(x)]
    MOD = 10**9+7
    from math import factorial
    N = II()
    X = LMIIS()
    # f = factorial(N-1)
    # def calc(k,slimes):
    #     # print(slimes)
    #     if k == N-1:
    #         return 0
    #     r = 0
    #     # for i in range(N-k-1):
    #     #     slimes2 = slimes[:]
    #     #     r = (slimes2[i+1]-slimes2[i]) / float(N-k-1)
    #     #     slimes2.pop(i)
    #     #     r += calc(k+1,slimes2) % MOD
    #     i = N-k-2
    #     slimes2 = slimes[:]
    #     print(i,slimes2)
    #     r = (slimes2[i+1]-slimes2[i]) * f / (N-k-1) % MOD
    #     slimes2.pop(i)
    #     r += calc(k+1,slimes2)
    #     return r % MOD
    f = factorial(N-1) % MOD
    def calc(k,slimes):
        r = 0

        for k in range(N-1):
            r += (slimes[-1]-slimes[-2-k]) * f // (N-k-1)
            r %= MOD
        return r
    print(calc(0,X))

if __name__ == '__main__':
    main()

Score : 600 points

Problem Statement

There are N slimes standing on a number line.
The i-th slime from the left is at position x_i.

It is guaruanteed that 1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}.

Niwango will perform N-1 operations. The i-th operation consists of the following procedures:

Choose an integer k between 1 and N-i (inclusive) with equal probability.

Move the k-th slime from the left, to the position of the neighboring slime to the right.

Fuse the two slimes at the same position into one slime.

Find the total distance traveled by the slimes multiplied by (N-1)! (we can show that this value is an integer), modulo (10^{9}+7). If a slime is born by a fuse and that slime moves, we count it as just one slime.

Constraints

2 \leq N \leq 10^{5}

1 \leq x_1 < x_2 < \ldots < x_N \leq 10^{9}

x_i is an integer.

Subtasks

400 points will be awarded for passing the test cases satisfying N \leq 2000.

Input

Input is given from Standard Input in the following format:

N
x_1 x_2 \ldots x_N

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

5

With probability \frac{1}{2}, the leftmost slime is chosen in the first operation, in which case the total distance traveled is 2.

With probability \frac{1}{2}, the middle slime is chosen in the first operation, in which case the total distance traveled is 3.

The answer is the expected total distance traveled, 2.5, multiplied by 2!, which is 5.

Sample Input 2

12
161735902 211047202 430302156 450968417 628894325 707723857 731963982 822804784 880895728 923078537 971407775 982631932

Sample Output 2

750927044

Find the expected value multiplied by (N-1)!, modulo (10^9+7).


k,x = input()

text = "No"

if k*500>=x:
  text = "Yes"

print(text)

Score : 100 points

Problem Statement

Takahashi has K 500-yen coins. (Yen is the currency of Japan.)
If these coins add up to X yen or more, print Yes; otherwise, print No.

Constraints

1 \leq K \leq 100

1 \leq X \leq 10^5

Input

Input is given from Standard Input in the following format:

K X

Output

If the coins add up to X yen or more, print Yes; otherwise, print No.

Sample Input 1

2 900

Sample Output 1

Yes

Two 500-yen coins add up to 1000 yen, which is not less than X = 900 yen.

Sample Input 2

1 501

Sample Output 2

No

One 500-yen coin is worth 500 yen, which is less than X = 501 yen.

Sample Input 3

4 2000

Sample Output 3

Yes

Four 500-yen coins add up to 2000 yen, which is not less than X = 2000 yen.


k, x = map(int, input().split())
if k * 500 >= x:
    print("Yes")
else:
    print("No")


Score : 200 points

Problem Statement

We have a string S of length N consisting of uppercase English letters.

How many times does ABC occur in S as contiguous subsequences (see Sample Inputs and Outputs)?

Constraints

3 \leq N \leq 50

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print number of occurrences of ABC in S as contiguous subsequences.

Sample Input 1

10
ZABCDBABCQ

Sample Output 1

2

Two contiguous subsequences of S are equal to ABC: the 2-nd through 4-th characters, and the 7-th through 9-th characters.

Sample Input 2

19
THREEONEFOURONEFIVE

Sample Output 2

0

No contiguous subsequences of S are equal to ABC.

Sample Input 3

33
ABCCABCBABCCABACBCBBABCBCBCBCABCB

Sample Output 3

5


from math import gcd
from functools import reduce
import sys
input = sys.stdin.readline

def lcm(a, b):
    return a*b // gcd(a, b)

def count_factor_2(num):
    count = 0
    while num % 2 == 0:
        num //= 2
        count += 1
    return count

def main():
    n, m = map(int, input().split())
    A = list(map(lambda x: int(x)//2, input().split()))

    check = len(set(map(count_factor_2, A)))
    if check != 1:
        print(0)
        return

    lcm_a = reduce(lcm, A)
    step = lcm_a * 2
    ans = (m + lcm_a) // step
    print(ans)

if __name__ == "__main__":
    main()

Score : 400 points

Problem Statement

Given are a sequence A= {a_1,a_2,......a_N} of N positive even numbers, and an integer M.

Let a semi-common multiple of A be a positive integer X that satisfies the following condition for every k (1 \leq k \leq N):

There exists a non-negative integer p such that X= a_k \times (p+0.5).

Find the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Constraints

1 \leq N \leq 10^5

1 \leq M \leq 10^9

2 \leq a_i \leq 10^9

a_i is an even number.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
a_1 a_2 ... a_N

Output

Print the number of semi-common multiples of A among the integers between 1 and M (inclusive).

Sample Input 1

2 50
6 10

Sample Output 1

2

15 = 6 \times 2.5

15 = 10 \times 1.5

45 = 6 \times 7.5

45 = 10 \times 4.5

Thus, 15 and 45 are semi-common multiples of A. There are no other semi-common multiples of A between 1 and 50, so the answer is 2.

Sample Input 2

3 100
14 22 40

Sample Output 2

0

The answer can be 0.

Sample Input 3

5 1000000000
6 6 2 6 2

Sample Output 3

166666667


from collections import deque

n,m=map(int,input().split())
l=list(map(int,input().split()))
l=[i/2 for i in l]
if n==1:
  if (m//l[0])%2==0:
    print(int((m//l[0])/2))
  else:
    print(int((m//l[0]+1)/2))
  exit()
def gcd(a,b):
  while b!=0:
    a,b=b,a%b
  return a

l=set(l)
a1=l.pop()
a2=l.pop()
g=gcd(a1,a2)
lcm=g*(a1/g)*(a2/g)
d=1
while l:
  a=l.pop()
  d=gcd(g,a)
  lcm=d*(a/d)*(g/d)
  g=lcm

if lcm>m:
  print(0)
  exit()
for a in l:
  if (lcm/a)%2==0:
    print(0)
    break
else:
  if (m//lcm)%2==0:
    print(int((m//lcm)/2))
  else:
    print(int((m//lcm+1)/2))


x,y = input().split()
result = y+x

print(result)

Score : 100 points

Problem Statement

Given are two strings S and T consisting of lowercase English letters. Concatenate T and S in this order, without space in between, and print the resulting string.

Constraints

S and T are strings consisting of lowercase English letters.

The lengths of S and T are between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S T

Output

Print the resulting string.

Sample Input 1

oder atc

Sample Output 1

atcoder

When S = oder and T = atc, concatenating T and S in this order results in atcoder.

Sample Input 2

humu humu

Sample Output 2

humuhumu


#  Copyright: this code was written at Dec 30th, 2019. for AtCoder, by Silviase
s,t = input().split()
print(t+s)


a,b,k=map(int,input().split())

if a+b<=k:
  print(0,0)
elif a<=k<a+b:
  print(0,b-(k-a))
else:
  print(a-k+b,b)

Score : 200 points

Problem Statement

Takahashi has A cookies, and Aoki has B cookies.
Takahashi will do the following action K times:

If Takahashi has one or more cookies, eat one of his cookies.

Otherwise, if Aoki has one or more cookies, eat one of Aoki's cookies.

If they both have no cookies, do nothing.

In the end, how many cookies will Takahashi and Aoki have, respectively?

Constraints

0 \leq A \leq 10^{12}

0 \leq B \leq 10^{12}

0 \leq K \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the numbers of Takahashi's and Aoki's cookies after K actions.

Sample Input 1

2 3 3

Sample Output 1

0 2

Takahashi will do the following:

He has two cookies, so he eats one of them.

Now he has one cookie left, and he eats it.

Now he has no cookies left, but Aoki has three, so Takahashi eats one of them.

Thus, in the end, Takahashi will have 0 cookies, and Aoki will have 2.

Sample Input 2

500000000000 500000000000 1000000000000

Sample Output 2

0 0

Watch out for overflows.


X = int(input())

r = X

while(True):
	for i in range(2, r):
		if r%i == 0:
			r += 1
			break
	else:
		break
print(r)

Score: 300 points

Problem Statement

Find the minimum prime number greater than or equal to X.

Notes

A prime number is an integer greater than 1 that cannot be evenly divided by any positive integer except 1 and itself.

For example, 2, 3, and 5 are prime numbers, while 4 and 6 are not.

Constraints

2 \le X \le 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the minimum prime number greater than or equal to X.

Sample Input 1

20

Sample Output 1

23

The minimum prime number greater than or equal to 20 is 23.

Sample Input 2

2

Sample Output 2

2

X itself can be a prime number.

Sample Input 3

99992

Sample Output 3

100003


#%%
def count(a, k, n):
    l = []
    i, j, count = 0, n-1, 0
    while True:
        tmp = a[i] + a[j]
        if tmp >= k:
            if j > 0:
                j -= 1
        elif tmp < k:
            l.append(n-j-1)
            if j < n - 1:
                j += 1
                count += n - j
            else:
                count += n - j - 1
            i += 1
        if i >= n:
            break
        tmp = a[i] + a[j]
        if j == 0 and tmp >= k:
            l.append(n-j)
            count += n - j
            i += 1
        if i >= n:
            break
    return [count, l]

n, m = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
l, r = 10**6, 1

while l - r > 1:
    mid = (l+r)//2
    if count(a, mid, n)[0] < m:
        l = mid
    elif count(a, mid, n)[0] > m:
        r = mid
    else :
        l = mid
        r = mid
    print(l, r)

ans = 0
if l == r:
    tmp = count(a, l, n)
    for i in range(n):
        ans += a[i] * tmp[1][i] * 2
else:
    p = count(a, l, n)
    q = count(a, r, n)
    check = -1
    for i in range(n):
        if p[1][i] != q[1][i]:
            check = i
            break
    ans += (a[i] + a[n - p[1][check] - 1]) * (m - p[0])
    tmp = count(a, l, n)
    for i in range(n):
        ans += a[i] * tmp[1][i] * 2
print(ans)

Score : 500 points

Problem Statement

Takahashi has come to a party as a special guest.
There are N ordinary guests at the party. The i-th ordinary guest has a power of A_i.

Takahashi has decided to perform M handshakes to increase the happiness of the party (let the current happiness be 0).
A handshake will be performed as follows:

Takahashi chooses one (ordinary) guest x for his left hand and another guest y for his right hand (x and y can be the same).

Then, he shakes the left hand of Guest x and the right hand of Guest y simultaneously to increase the happiness by A_x+A_y.

However, Takahashi should not perform the same handshake more than once. Formally, the following condition must hold:

Assume that, in the k-th handshake, Takahashi shakes the left hand of Guest x_k and the right hand of Guest y_k. Then, there is no pair p, q (1 \leq p < q \leq M) such that (x_p,y_p)=(x_q,y_q).

What is the maximum possible happiness after M handshakes?

Constraints

1 \leq N \leq 10^5

1 \leq M \leq N^2

1 \leq A_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_N

Output

Print the maximum possible happiness after M handshakes.

Sample Input 1

5 3
10 14 19 34 33

Sample Output 1

202

Let us say that Takahashi performs the following handshakes:

In the first handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 4.

In the second handshake, Takahashi shakes the left hand of Guest 4 and the right hand of Guest 5.

In the third handshake, Takahashi shakes the left hand of Guest 5 and the right hand of Guest 4.

Then, we will have the happiness of (34+34)+(34+33)+(33+34)=202.

We cannot achieve the happiness of 203 or greater, so the answer is 202.

Sample Input 2

9 14
1 3 5 110 24 21 34 5 3

Sample Output 2

1837

Sample Input 3

9 73
67597 52981 5828 66249 75177 64141 40773 79105 16076

Sample Output 3

8128170


import sys
n,m,v,p=map(int,sys.stdin.readline().split())
a=list(map(int, sys.stdin.readline().split()))
#a=[int(i) for i in sys.stdin.readline().split()]
a.sort(reverse=True)
a=a[p-1:]
l,r=0,n-p

def adopt(x):
    global a,n,m,v
    b=a[x]+m
    if b<a[0]:
        return False
    al=(v-(n-x))*m
    s=x*b-sum(a[:x])
    return al<=s

while l+1<r:
    k=(l+r)//2
    if adopt(k):
        l=k
    else:
        r=k

if adopt(r):
    print(p+r)
else:
    print(p+l)

Score : 700 points

Problem Statement

N problems are proposed for an upcoming contest. Problem i has an initial integer score of A_i points.

M judges are about to vote for problems they like. Each judge will choose exactly V problems, independently from the other judges,
and increase the score of each chosen problem by 1.

After all M judges cast their vote, the problems will be sorted in non-increasing order of score, and the first P problems will be chosen for the problemset.
Problems with the same score can be ordered arbitrarily, this order is decided by the chief judge.

How many problems out of the given N have a chance to be chosen for the problemset?

Constraints

2 \le N \le 10^5

1 \le M \le 10^9

1 \le V \le N - 1

1 \le P \le N - 1

0 \le A_i \le 10^9

Input

Input is given from Standard Input in the following format:

N M V P
A_1 A_2 ... A_N

Output

Print the number of problems that have a chance to be chosen for the problemset.

Sample Input 1

6 1 2 2
2 1 1 3 0 2

Sample Output 1

5

If the only judge votes for problems 2 and 5, the scores will be 2 2 1 3 1 2.
The problemset will consist of problem 4 and one of problems 1, 2, or 6.

If the only judge votes for problems 3 and 4, the scores will be 2 1 2 4 0 2.
The problemset will consist of problem 4 and one of problems 1, 3, or 6.

Thus, problems 1, 2, 3, 4, and 6 have a chance to be chosen for the problemset. On the contrary, there is no way for problem 5 to be chosen.

Sample Input 2

6 1 5 2
2 1 1 3 0 2

Sample Output 2

3

Only problems 1, 4, and 6 have a chance to be chosen.

Sample Input 3

10 4 8 5
7 2 3 6 1 6 5 4 6 5

Sample Output 3

8


import numpy as np

if __name__ == "__main__":
    x,y = [int(x) for x in input().split()]
    print(np.lcm(x,y))

Score : 300 points

Problem Statement

Takahashi is organizing a party.

At the party, each guest will receive one or more snack pieces.

Takahashi predicts that the number of guests at this party will be A or B.

Find the minimum number of pieces that can be evenly distributed to the guests in both of the cases predicted.

We assume that a piece cannot be divided and distributed to multiple guests.

Constraints

1 \leq A, B \leq 10^5

A \neq B

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of pieces that can be evenly distributed to the guests in both of the cases with A guests and B guests.

Sample Input 1

2 3

Sample Output 1

6

When we have six snack pieces, each guest can take three pieces if we have two guests, and each guest can take two if we have three guests.

Sample Input 2

123 456

Sample Output 2

18696

Sample Input 3

100000 99999

Sample Output 3

9999900000


# -*- coding: utf-8 -*-

N = int(input())
a = list(map(int, input().split()))

ans = 0
pos = 1
for i in range(N):
    if a[i] != pos:
        ans += 1
    else:
        pos += 1

print(ans if ans < N else -1)

Score : 400 points

Problem Statement

We have N bricks arranged in a row from left to right.

The i-th brick from the left (1 \leq i \leq N) has an integer a_i written on it.

Among them, you can break at most N-1 bricks of your choice.

Let us say there are K bricks remaining. Snuke will be satisfied if, for each integer i (1 \leq i \leq K), the i-th of those brick from the left has the integer i written on it.

Find the minimum number of bricks you need to break to satisfy Snuke's desire. If his desire is unsatisfiable, print -1 instead.

Constraints

All values in input are integers.

1 \leq N \leq 200000

1 \leq a_i \leq N

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of bricks that need to be broken to satisfy Snuke's desire, or print -1 if his desire is unsatisfiable.

Sample Input 1

3
2 1 2

Sample Output 1

1

If we break the leftmost brick, the remaining bricks have integers 1 and 2 written on them from left to right, in which case Snuke will be satisfied.

Sample Input 2

3
2 2 2

Sample Output 2

-1

In this case, there is no way to break some of the bricks to satisfy Snuke's desire.

Sample Input 3

10
3 1 4 1 5 9 2 6 5 3

Sample Output 3

7

Sample Input 4

1
1

Sample Output 4

0

There may be no need to break the bricks at all.


n,u,v = map(int,input().split())

edges = [[] for _ in range(n)]

for _ in range(n-1):
    a,b = map(int,input().split())
    edges[a-1].append(b-1)
    edges[b-1].append(a-1)

visit1 = [False]*n
visit2 = [False]*n

visit1[u-1] = True
visit2[v-1] = True

stack1 = [u-1]
stack2 = [v-1]

dis1 = [-1] * n
dis2 = [-1] * n
dis1[u-1] = 0
dis2[v-1] = 0

while stack1:
    for _ in range(len(stack1)):
        q1 = stack1.pop(-1)
        for i in edges[q1]:
            if not visit1[i] and not visit2[i]:
                visit1[i] = True
                dis1[i] = dis1[q1] + 1
                stack1.append(i)

    for _ in range(len(stack2)):
        q2 = stack2.pop(-1)
        for i in edges[q2]:
            if not visit2[i]:
                visit2[i] = True
                dis2[i] = dis2[q2] + 1
                stack2.append(i)

while stack2:
    q = stack2.pop(-1)
    for i in edges[q]:
        if not visit2[i]:
            visit2[i] = True
            dis2[i] = dis2[q] + 1
            stack2.append(i)

ans = max(dis2)

print(ans)

Score : 600 points

Problem Statement

We have a tree with N vertices. The i-th edge connects Vertex A_i and B_i bidirectionally.

Takahashi is standing at Vertex u, and Aoki is standing at Vertex v.

Now, they will play a game of tag as follows:

1. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Takahashi moves to a vertex of his choice that is adjacent to his current vertex.

2. If Takahashi and Aoki are standing at the same vertex, the game ends. Otherwise, Aoki moves to a vertex of his choice that is adjacent to his current vertex.

3. Go back to step 1.

Takahashi performs his moves so that the game ends as late as possible, while Aoki performs his moves so that the game ends as early as possible.

Find the number of moves Aoki will perform before the end of the game if both Takahashi and Aoki know each other's position and strategy.

It can be proved that the game is bound to end.

Constraints

2 \leq N \leq 10^5

1 \leq u,v \leq N

u \neq v

1 \leq A_i,B_i \leq N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N u v
A_1 B_1
:
A_{N-1} B_{N-1}

Output

Print the number of moves Aoki will perform before the end of the game.

Sample Input 1

5 4 1
1 2
2 3
3 4
3 5

Sample Output 1

2

If both players play optimally, the game will progress as follows:

Takahashi moves to Vertex 3.

Aoki moves to Vertex 2.

Takahashi moves to Vertex 5.

Aoki moves to Vertex 3.

Takahashi moves to Vertex 3.

Here, Aoki performs two moves.

Note that, in each move, it is prohibited to stay at the current vertex.

Sample Input 2

5 4 5
1 2
1 3
1 4
1 5

Sample Output 2

1

Sample Input 3

2 1 2
1 2

Sample Output 3

0

Sample Input 4

9 6 1
1 2
2 3
3 4
4 5
5 6
4 7
7 8
8 9

Sample Output 4

5


a = list(map(int,input().split()))

if a[0]+a[1]+a[2] >= 22:
	print("bust")
else:
	print("win")

Score : 100 points

Problem Statement

Given are three integers A_1, A_2, and A_3.

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Constraints

1 \leq A_i \leq 13 \ \ (i=1,2,3)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A_1 A_2 A_3

Output

If A_1+A_2+A_3 is greater than or equal to 22, print bust; otherwise, print win.

Sample Input 1

5 7 9

Sample Output 1

win

5+7+9=21, so print win.

Sample Input 2

13 7 2

Sample Output 2

bust

13+7+2=22, so print bust.


S = input()
k = len(S)//2
res = 0
for i in range(k):
  if S[i]!=S[len(S)-1-i]:
    res += 1
print(res)

Score : 200 points

Problem Statement

Takahashi loves palindromes. Non-palindromic strings are unacceptable to him. Each time he hugs a string, he can change one of its characters to any character of his choice.

Given is a string S. Find the minimum number of hugs needed to make S palindromic.

Constraints

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Print the minimum number of hugs needed to make S palindromic.

Sample Input 1

redcoder

Sample Output 1

1

For example, we can change the fourth character to o and get a palindrome redooder.

Sample Input 2

vvvvvv

Sample Output 2

0

We might need no hugs at all.

Sample Input 3

abcdabc

Sample Output 3

2


s=input()
s1=s[:len(s)//2]
s2=s[len(s)//2:][::-1]
ans=0
for i in range(len(s)//2):
    if s1[i]!=s2[i]:
        ans+=1
print(ans)


N = int(input())

notes = []
for _ in range(N):
    note = []
    for _ in range(int(input())):
        x, y = map(int, input().split())
        note.append((x-1, y))
    notes.append(note)

ans = 0
for state in range(1 << N):
    flag = True
    for i in range(N):
        if state >> i & 1:
            for x, y in notes[i]:
                flag &= (y == 1 and state >> x & 1) or (y == 0 and (state >> x & 1) == 0)

    if flag:
        ans = max(ans, sum(state >> i & 1 for i in range(N)))
print(ans)

Score : 300 points

Problem Statement

There are N people numbered 1 to N. Each of them is either an honest person whose testimonies are always correct or an unkind person whose testimonies may be correct or not.

Person i gives A_i testimonies. The j-th testimony by Person i is represented by two integers x_{ij} and y_{ij}. If y_{ij} = 1, the testimony says Person x_{ij} is honest; if y_{ij} = 0, it says Person x_{ij} is unkind.

How many honest persons can be among those N people at most?

Constraints

All values in input are integers.

1 \leq N \leq 15

0 \leq A_i \leq N - 1

1 \leq x_{ij} \leq N

x_{ij} \neq i

x_{ij_1} \neq x_{ij_2} (j_1 \neq j_2)

y_{ij} = 0, 1

Input

Input is given from Standard Input in the following format:

N
A_1
x_{11} y_{11}
x_{12} y_{12}
:
x_{1A_1} y_{1A_1}
A_2
x_{21} y_{21}
x_{22} y_{22}
:
x_{2A_2} y_{2A_2}
:
A_N
x_{N1} y_{N1}
x_{N2} y_{N2}
:
x_{NA_N} y_{NA_N}

Output

Print the maximum possible number of honest persons among the N people.

Sample Input 1

3
1
2 1
1
1 1
1
2 0

Sample Output 1

2

If Person 1 and Person 2 are honest and Person 3 is unkind, we have two honest persons without inconsistencies, which is the maximum possible number of honest persons.

Sample Input 2

3
2
2 1
3 0
2
3 1
1 0
2
1 1
2 0

Sample Output 2

0

Assuming that one or more of them are honest immediately leads to a contradiction.

Sample Input 3

2
1
2 0
1
1 0

Sample Output 3

1


n=int(input())
A=list(map(int,input().split()))
mod = 10**9+7;ans=0
for i in range(60):
  cnt=0
  for a in A: cnt+=a>>i&1
  cnt=cnt*(n-cnt)%mod
  for i in range(i): cnt=cnt*2%mod
  ans+=cnt
print(ans%mod)

Score : 400 points

Problem Statement

We have N integers. The i-th integer is A_i.

Find \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

What is \mbox{ XOR }?

The XOR of integers A and B, A \mbox{ XOR } B, is defined as follows:

When A \mbox{ XOR } B is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if either A or B, but not both, has 1 in the 2^k's place, and 0 otherwise.

For example, 3 \mbox{ XOR } 5 = 6. (In base two: 011 \mbox{ XOR } 101 = 110.)

Constraints

2 \leq N \leq 3 \times 10^5

0 \leq A_i < 2^{60}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the value \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} (A_i \mbox{ XOR } A_j), modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

6

We have (1\mbox{ XOR } 2)+(1\mbox{ XOR } 3)+(2\mbox{ XOR } 3)=3+2+1=6.

Sample Input 2

10
3 1 4 1 5 9 2 6 5 3

Sample Output 2

237

Sample Input 3

10
3 14 159 2653 58979 323846 2643383 27950288 419716939 9375105820

Sample Output 3

103715602

Print the sum modulo (10^9+7).


import numpy as np
n = int(input())
mod = 10 ** 9 + 7
a = [list(bin(int(i))[2:].zfill(60)) for i in input().split()]
a = np.array(a)
x = a.T
x = x.tolist()
#print(x)
ans = 0
for i in range(60):
    y = x[i].count('0')
    z = x[i].count('1')
    tmp = ( (n * (n - 1) // 2) - (y * (y - 1) // 2) - (z * (z - 1) // 2) ) % mod
    tmp2 = 1
    for j in range(60 - i - 1):
        tmp2 = tmp2 * 2 % mod
    #print(tmp,tmp2)

    ans = (ans + tmp2 * tmp) % mod
print(ans)


N=int(input())
A=list(map(int,input().split()))
S=0
for i in range(N-1):
  x=format(A[i], 'b')
  x=x.zfill(60)
  for j in range(i+1,N):
    y=format(A[j], 'b')
    y=y.zfill(60)
    for k in range(60):
      if x[k] !=  y[k]:
        S=(S+ 2**(59-k))%(10**9+7)
print(S)


n=int(input())
MOD=10**9+7
a=list(map(int,input().split()))
r=0
for i in range(n):
    for j in range(i,n):
        if(i!=j):
            r+=a[i]^a[j]
print(r%MOD)


H,W = map(int,input().split())

A = [list(map(int,input().split())) for i in range(H)]
B = [list(map(int,input().split())) for i in range(H)]

C = [[0]*W for i in range(H)]

for h in range(H):
    for w in range(W):
        C[h][w] = abs(A[h][w] - B[h][w])

#print(C)

ZZZ = 80*70

DP = [[[0]*ZZZ for i in range(W)] for j in range(H)]

for h in range(H):
    for w in range(W):
        if h==0 and w==0:
            DP[h][w][C[h][w]] = 1
        elif h==0:
            for i in range(ZZZ):
                DP[h][w][i] = DP[h][w-1][abs(i-C[h][w])]| (DP[h][w-1][abs(i+C[h][w])] if abs(i+C[h][w])<ZZZ else 0)
        elif w==0:
            for i in range(ZZZ):
                DP[h][w][i] = DP[h-1][w][abs(i-C[h][w])]|(DP[h-1][w][abs(i+C[h][w])] if abs(i+C[h][w])<ZZZ else 0)
        else:
            for i in range(ZZZ):
                DP[h][w][i] = DP[h-1][w][abs(i-C[h][w])]|(DP[h-1][w][abs(i+C[h][w])] if abs(i+C[h][w])<ZZZ else 0)|DP[h][w-1][abs(i-C[h][w])]|(DP[h][w-1][abs(i+C[h][w])] if abs(i+C[h][w])<ZZZ else 0)

for i in range(ZZZ):
    if DP[H-1][W-1][i]==1:
        ans = i
        break
print(ans)

Score : 500 points

Problem Statement

We have a grid with H horizontal rows and W vertical columns. Let (i,j) denote the square at the i-th row from the top and the j-th column from the left.

The square (i, j) has two numbers A_{ij} and B_{ij} written on it.

First, for each square, Takahashi paints one of the written numbers red and the other blue.

Then, he travels from the square (1, 1) to the square (H, W). In one move, he can move from a square (i, j) to the square (i+1, j) or the square (i, j+1). He must not leave the grid.

Let the unbalancedness be the absolute difference of the sum of red numbers and the sum of blue numbers written on the squares along Takahashi's path, including the squares (1, 1) and (H, W).

Takahashi wants to make the unbalancedness as small as possible by appropriately painting the grid and traveling on it.

Find the minimum unbalancedness possible.

Constraints

2 \leq H \leq 80

2 \leq W \leq 80

0 \leq A_{ij} \leq 80

0 \leq B_{ij} \leq 80

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H W
A_{11} A_{12} \ldots A_{1W}
:
A_{H1} A_{H2} \ldots A_{HW}
B_{11} B_{12} \ldots B_{1W}
:
B_{H1} B_{H2} \ldots B_{HW}

Output

Print the minimum unbalancedness possible.

Sample Input 1

2 2
1 2
3 4
3 4
2 1

Sample Output 1

0

By painting the grid and traveling on it as shown in the figure below, the sum of red numbers and the sum of blue numbers are 3+3+1=7 and 1+2+4=7, respectively, for the unbalancedness of 0.

Sample Input 2

2 3
1 10 80
80 10 1
1 2 3
4 5 6

Sample Output 2

2


H,W=map(int,input().split())
a=[None]*H
b=[None]*H
for i in range(H):
  a[i]=list(map(int,input().split()))
for i in range(H):
  b[i]=list(map(int,input().split()))

dp=[[[False for i in range(240)] for j in range(W)] for p in range(H)]

dp[0][0][abs(a[0][0]-b[0][0])+80]=True
for i in range(H):
  for j in range(W):
    for k in range(240):
      if dp[i][j][k]:
        if i+1<H:
          p=abs(a[i+1][j]-b[i+1][j])
          if k-p>=0:
            dp[i+1][j][k-p]=True
          if k+p<=239:
            dp[i+1][j][k+p]=True
        if j+1<W:
          p=abs(a[i][j+1]-b[i][j+1])
          if k-p>=0:
            dp[i][j+1][k-p]=True
          if k+p<=239:
            dp[i][j+1][k+p]=True
for k in range(240):
  if dp[-1][-1][k]:
    if k-80>=0:
      print(k-80)
      break


import numpy as np
H,W=map(int,input().split())
a = [[] for _ in range(H)]
for i in range(H):
    a[i] = list(map(int, input().split()))
b = [[] for _ in range(H)]
for i in range(H):
    b[i] = list(map(int, input().split()))
#print(a)
#print(b)
for i in range(H):
    for j in range(W):
        a[i][j] =abs(a[i][j]-b[i][j])
#print(a)
dp = [[set() for _ in range(W)] for _ in range(H)]
dp[0][0].add(a[0][0])
dp[0][0].add(-a[0][0])
for i in range(H):
    for j in range(W):
        tmp = set()
        if i!=0:
            for k in dp[i-1][j]:
                tmp.add(k+a[i][j])
                tmp.add(k-a[i][j])
        if j!=0:
            for k in dp[i][j-1]:
                tmp.add(k+a[i][j])
                tmp.add(k-a[i][j])
        if i>0 or j>0:
            dp[i][j] = tmp
        #print(tmp)
#print(dp[H-1][W-1])
ls = list(dp[H-1][W-1])
res = np.abs(np.asarray(ls)).argmin()
print(ls[res])


n=int(input())
ans=0
for i in range(1,50000+1):
    z=i*1.08
    if n<=z<(n+1):
        ans=z
        break
if ans==0:
    print(":(")
else:
    print(ans)

Score: 200 points

Problem Statement

Takahashi bought a piece of apple pie at ABC Confiserie. According to his memory, he paid N yen (the currency of Japan) for it.

The consumption tax rate for foods in this shop is 8 percent. That is, to buy an apple pie priced at X yen before tax, you have to pay X \times 1.08 yen (rounded down to the nearest integer).

Takahashi forgot the price of his apple pie before tax, X, and wants to know it again. Write a program that takes N as input and finds X. We assume X is an integer.

If there are multiple possible values for X, find any one of them. Also, Takahashi's memory of N, the amount he paid, may be incorrect. If no value could be X, report that fact.

Constraints

1 \leq N \leq 50000

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If there are values that could be X, the price of the apple pie before tax, print any one of them.

If there are multiple such values, printing any one of them will be accepted.

If no value could be X, print :(.

Sample Input 1

432

Sample Output 1

400

If the apple pie is priced at 400 yen before tax, you have to pay 400 \times 1.08 = 432 yen to buy one.

Otherwise, the amount you have to pay will not be 432 yen.

Sample Input 2

1079

Sample Output 2

:(

There is no possible price before tax for which you have to pay 1079 yen with tax.

Sample Input 3

1001

Sample Output 3

927

If the apple pie is priced 927 yen before tax, by rounding down 927 \times 1.08 = 1001.16, you have to pay 1001 yen.


n = int(input())
if n%1.08==0:
  print(int(n//1.08))
else:
  x_init=n//1.08 + 1
  x_n = int(1.08*x_init)
  if x_n ==n:
    print(int(x_init))
  else:
    print(':(')


X = int(input())
import sys
q = X // 100
r = X % 100
if r >= 0 and r <= 5 * q:
    print(1)
    sys.exit()
print(0)

Score: 300 points

Problem Statement

AtCoder Mart sells 1000000 of each of the six items below:

Riceballs, priced at 100 yen (the currency of Japan) each

Sandwiches, priced at 101 yen each

Cookies, priced at 102 yen each

Cakes, priced at 103 yen each

Candies, priced at 104 yen each

Computers, priced at 105 yen each

Takahashi wants to buy some of them that cost exactly X yen in total.
Determine whether this is possible.

(Ignore consumption tax.)

Constraints

1 \leq X \leq 100000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

If it is possible to buy some set of items that cost exactly X yen in total, print 1; otherwise, print 0.

Sample Input 1

615

Sample Output 1

1

For example, we can buy one of each kind of item, which will cost 100+101+102+103+104+105=615 yen in total.

Sample Input 2

217

Sample Output 2

0

No set of items costs 217 yen in total.


N = input()

S = input()
len_S = len(S)

key_list = range(0, 1000)

key_list = tuple(map(lambda key: '0' * (3 - len(str(key))) + str(key), key_list))

ans_list = []

for key in key_list:
    target_num = 0
    for s in S:
        if key[target_num] == s:
            target_num += 1
        if target_num == 3:
            ans_list.append(key)
            break

print(len(ans_list))

Score: 400 points

Problem Statement

AtCoder Inc. has decided to lock the door of its office with a 3-digit PIN code.

The company has an N-digit lucky number, S. Takahashi, the president, will erase N-3 digits from S and concatenate the remaining 3 digits without changing the order to set the PIN code.

How many different PIN codes can he set this way?

Both the lucky number and the PIN code may begin with a 0.

Constraints

4 \leq N \leq 30000

S is a string of length N consisting of digits.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of different PIN codes Takahashi can set.

Sample Input 1

4
0224

Sample Output 1

3

Takahashi has the following options:

Erase the first digit of S and set 224.

Erase the second digit of S and set 024.

Erase the third digit of S and set 024.

Erase the fourth digit of S and set 022.

Thus, he can set three different PIN codes: 022, 024, and 224.

Sample Input 2

6
123123

Sample Output 2

17

Sample Input 3

19
3141592653589793238

Sample Output 3

329


import itertools
n = int(input())
seq = list(input())
L = list(itertools.combinations(seq,3))

print(len(set(L)))


n = int(input())

hats = [0, 0, 0]
count = 1
mod = 1000000007

for a in map(int, input().split()):
  count *= hats.count(a)
  count %= mod
  count[count.index(a)] += 1

print(count)

Score: 500 points

Problem Statement

N people are standing in a queue, numbered 1, 2, 3, ..., N from front to back. Each person wears a hat, which is red, blue, or green.

The person numbered i says:

"In front of me, exactly A_i people are wearing hats with the same color as mine."

Assuming that all these statements are correct, find the number of possible combinations of colors of the N people's hats.

Since the count can be enormous, compute it modulo 1000000007.

Constraints

1 \leq N \leq 100000

0 \leq A_i \leq N-1

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print the number of possible combinations of colors of the N people's hats, modulo 1000000007.

Sample Input 1

6
0 1 2 3 4 5

Sample Output 1

3

We have three possible combinations, as follows:

Red, Red, Red, Red, Red, Red

Blue, Blue, Blue, Blue, Blue, Blue

Green, Green, Green, Green, Green, Green

Sample Input 2

3
0 0 0

Sample Output 2

6

Sample Input 3

54
0 0 1 0 1 2 1 2 3 2 3 3 4 4 5 4 6 5 7 8 5 6 6 7 7 8 8 9 9 10 10 11 9 12 10 13 14 11 11 12 12 13 13 14 14 15 15 15 16 16 16 17 17 17

Sample Output 3

115295190


S = input()
a = 0

if S = 'SUN':
  a = 7
elif S = 'MON':
  a = 6
elif S = 'TUE':
  a = 5
elif S = 'WED':
  a = 4
elif S = 'THU':
  a = 3
elif S = 'FRI':
  a = 2
elif S = 'SAT':
  a = 1

print(a)

Score : 100 points

Problem Statement

Given is a string S representing the day of the week today.

S is SUN, MON, TUE, WED, THU, FRI, or SAT, for Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday, respectively.

After how many days is the next Sunday (tomorrow or later)?

Constraints

S is SUN, MON, TUE, WED, THU, FRI, or SAT.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of days before the next Sunday.

Sample Input 1

SAT

Sample Output 1

1

It is Saturday today, and tomorrow will be Sunday.

Sample Input 2

SUN

Sample Output 2

7

It is Sunday today, and seven days later, it will be Sunday again.


d=input()
if d=='MON':
    print('6')
elif d=='TUE':
    print('5')
elif d=='WED':
    print('4')
elif d=='THU':
    print('3')
elif d=='FRI':
    print('2')
elif d=='SAT':
    print('1')
elif d=='SUN':
    print('7')


N = int(input())
S = list(input())

result = ''
for i in range(0,len(S)):
  num = ord(S[i])
  if num + N > 90:
    p = num + N - 90 + 64
  else:
    p = num + N
  new = chr(p)
  result += new

print(result)

Score : 200 points

Problem Statement

We have a string S consisting of uppercase English letters. Additionally, an integer N will be given.

Shift each character of S by N in alphabetical order (see below), and print the resulting string.

We assume that A follows Z. For example, shifting A by 2 results in C (A \to B \to C), and shifting Y by 3 results in B (Y \to Z \to A \to B).

Constraints

0 \leq N \leq 26

1 \leq |S| \leq 10^4

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the string resulting from shifting each character of S by N in alphabetical order.

Sample Input 1

2
ABCXYZ

Sample Output 1

CDEZAB

Note that A follows Z.

Sample Input 2

0
ABCXYZ

Sample Output 2

ABCXYZ

Sample Input 3

13
ABCDEFGHIJKLMNOPQRSTUVWXYZ

Sample Output 3

NOPQRSTUVWXYZABCDEFGHIJKLM


#初期入力
import sys
input = sys.stdin.readline
import random
N = int(input())
connection   ={i:set() for i in range(1,N+1)}
A            ={i:tuple() for i in range(1,N+1)}
for i in range(N-1):
    a,b = (int(x) for x in input().split())
    connection[a].add(b) # つながりを逆向きも含めて入力
    connection[b].add(a) #
    A[i]=a,b             #色を出力するため

#色の数(各頂点の最大値)
color_max =0
for i in connection.values():
    color_max =max(color_max, len(i) )

#各頂点に色を準備⇒辺の色決定
#color_vertex ={i:set() for i in range(1,N+1)} #頂点に接続している辺に使った色
color_vertex ={i:[] for i in range(1,N+1)} #頂点に接続している辺に使った色
color_side =[0]*(N-1)
for i in range(N-1):
    v_a,v_b = A[i] #i番目の辺とつながっている頂点
    color =0
    while True:
        q,color =divmod(color,color_max)    #0からcolor_max-1、後で+1
        if color not in color_vertex[v_a] and color not in color_vertex[v_b]:  #使った色リストに入っているか？
            color_side[i] =color+1          #辺の色を出力用に決める（2行上の＋１）
            #color_vertex[v_a].add(color)    #前の頂点に使った色として追加
            #color_vertex[v_b].add(color)    #先の頂点に使った色として追加
            color_vertex[v_a].append(color)    #前の頂点に使った色として追加
            color_vertex[v_b].append(color)    #先の頂点に使った色として追加
            break
        else:
            color +=1

#出力
print(color_max)
#for i in color_side:
 #   print(i)
print(*color_side, sep="\n")

Score : 400 points

Problem Statement

Given is a tree G with N vertices.
The vertices are numbered 1 through N, and the i-th edge connects Vertex a_i and Vertex b_i.

Consider painting the edges in G with some number of colors.
We want to paint them so that, for each vertex, the colors of the edges incident to that vertex are all different.

Among the colorings satisfying the condition above, construct one that uses the minimum number of colors.

Constraints

2 \le N \le 10^5

1 \le a_i \lt b_i \le N

All values in input are integers.

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
\vdots
a_{N-1} b_{N-1}

Output

Print N lines.

The first line should contain K, the number of colors used.

The (i+1)-th line (1 \le i \le N-1) should contain c_i, the integer representing the color of the i-th edge, where 1 \le c_i \le K must hold.

If there are multiple colorings with the minimum number of colors that satisfy the condition, printing any of them will be accepted.

Sample Input 1

3
1 2
2 3

Sample Output 1

2
1
2

Sample Input 2

8
1 2
2 3
2 4
2 5
4 7
5 6
6 8

Sample Output 2

4
1
2
3
4
1
1
2

Sample Input 3

6
1 2
1 3
1 4
1 5
1 6

Sample Output 3

5
1
2
3
4
5


from heapq import heappush,heappop,heapify
from collections import deque,defaultdict,Counter
import itertools
from itertools import permutations,combinations
import sys
import bisect
import string
import math
import time
#import random
def I():
    return int(input())
def MI():
    return map(int,input().split())
def LI():
    return [int(i) for i in input().split()]
def LI_():
    return [int(i)-1 for i in input().split()]
def StoI():
    return [ord(i)-97 for i in input()]
def ItoS(nn):
    return chr(nn+97)
def show(*inp,end='\n'):
    if show_flg:
        print(*inp,end=end)
YN=['Yes','No']
mo=10**9+7
inf=float('inf')
l_alp=string.ascii_lowercase
u_alp=string.ascii_uppercase
ts=time.time()
#sys.setrecursionlimit(10**6)
input=lambda: sys.stdin.readline().rstrip()

show_flg=False
show_flg=True

n=I()
g=[[] for i in range(n)]
c=[]
dc={}
for i in range(n-1):
    a,b=LI_()
    c.append((a,b))
    g[a].append(b)
    g[b].append(a)
    dc[(a,b)]=0

mxd=0
mxi=0
for i in range(n):
    m=len(g[i])
    if m==1:
        mxd=m
        mxi=i

#vには最短距離をセット & -1 で未訪問フラグ

v=[-1]*(n)

def bfs(x):
    p=deque()
    p.append(x)
    v[x]=0
    while p:
        c=p.popleft()
        col=0
        for i in g[c]:
            if v[i]==-1:
                if v[c]==col:col+=1
                p.append(i)
                v[i]=col
                dc[(c,i)]=col
                col+=1
    return

bfs(mxi)
print(mxd)
for i,j in c:
    print(dc[(i,j)]+1)

#show(v)


from collections import defaultdict

n, k = map(int, input().split())
a = list(map(lambda x: int(x)-1, input().split()))
s = [0] * (n+1)
for i in range(n):
    s[i+1] = s[i] + a[i]
s = [x % k for x in s]
d = defaultdict(int)
ans = 0

for i in range(n+1):
    j = i - k
    if j >= 0: d[s[j]] -= 1
    ans += d[s[i]]
    d[s[i]] += 1

print(ans)

Score : 500 points

Problem Statement

Given are a sequence of N positive integers A_1, A_2, \ldots, A_N, and a positive integer K.

Find the number of non-empty contiguous subsequences in A such that the remainder when dividing the sum of its elements by K is equal to the number of its elements. We consider two subsequences different if they are taken from different positions, even if they are equal sequences.

Constraints

All values in input are integers.

1 \leq N \leq 2\times 10^5

1 \leq K \leq 10^9

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print the number of subsequences that satisfy the condition.

Sample Input 1

5 4
1 4 2 3 5

Sample Output 1

4

Four sequences satisfy the condition: (1), (4,2), (1,4,2), and (5).

Sample Input 2

8 4
4 2 4 2 4 2 4 2

Sample Output 2

7

(4,2) is counted four times, and (2,4) is counted three times.

Sample Input 3

10 7
14 15 92 65 35 89 79 32 38 46

Sample Output 3

8


try:
	x,y=map(int,input().split())
except Exception:
 	exit(100)
def fun(x):
	price=0
	if x==3:
		price=100000
	elif x==2:
		price=200000
	elif x==1:
		price=300000
	return price
if x ==1 and y ==1:
	print(fun(x)+fun(y)+4Ω00000)
else:
	print(fun(x)+fun(y))

Score: 100 points

Problem Statement

We held two competitions: Coding Contest and Robot Maneuver.

In each competition, the contestants taking the 3-rd, 2-nd, and 1-st places receive 100000, 200000, and 300000 yen (the currency of Japan), respectively. Furthermore, a contestant taking the first place in both competitions receives an additional 400000 yen.

DISCO-Kun took the X-th place in Coding Contest and the Y-th place in Robot Maneuver.
Find the total amount of money he earned.

Constraints

1 \leq X \leq 205

1 \leq Y \leq 205

X and Y are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the amount of money DISCO-Kun earned, as an integer.

Sample Input 1

1 1

Sample Output 1

1000000

In this case, he earned 300000 yen in Coding Contest and another 300000 yen in Robot Maneuver. Furthermore, as he won both competitions, he got an additional 400000 yen.
In total, he made 300000 + 300000 + 400000 = 1000000 yen.

Sample Input 2

3 101

Sample Output 2

100000

In this case, he earned 100000 yen in Coding Contest.

Sample Input 3

4 4

Sample Output 3

0

In this case, unfortunately, he was the highest-ranked contestant without prize money in both competitions.


import sys
from itertools import accumulate
readline = sys.stdin.readline
N = int(readline())
A = list(map(int, readline().split()))
SA = sum(A)
CA = list(accumulate(A))
ans = float('inf')
for ca in CA:
    b = SA - ca
    ans = min(ans, abs(b - ca))
print(ans)

Score: 200 points

Problem Statement

Takahashi, who works at DISCO, is standing before an iron bar.
The bar has N-1 notches, which divide the bar into N sections. The i-th section from the left has a length of A_i millimeters.

Takahashi wanted to choose a notch and cut the bar at that point into two parts with the same length.
However, this may not be possible as is, so he will do the following operations some number of times before he does the cut:

Choose one section and expand it, increasing its length by 1 millimeter. Doing this operation once costs 1 yen (the currency of Japan).

Choose one section of length at least 2 millimeters and shrink it, decreasing its length by 1 millimeter. Doing this operation once costs 1 yen.

Find the minimum amount of money needed before cutting the bar into two parts with the same length.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 2020202020

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_N

Output

Print an integer representing the minimum amount of money needed before cutting the bar into two parts with the same length.

Sample Input 1

3
2 4 3

Sample Output 1

3

The initial lengths of the sections are [2, 4, 3] (in millimeters). Takahashi can cut the bar equally after doing the following operations for 3 yen:

Shrink the second section from the left. The lengths of the sections are now [2, 3, 3].

Shrink the first section from the left. The lengths of the sections are now [1, 3, 3].

Shrink the second section from the left. The lengths of the sections are now [1, 2, 3], and we can cut the bar at the second notch from the left into two parts of length 3 each.

Sample Input 2

12
100 104 102 105 103 103 101 105 104 102 104 101

Sample Output 2

0