import collections
n=int(input())
a=list(map(int, input().split()))

a.sort()
x=0
count=0
b=set(a)
b=list(b)

c = collections.Counter(a)
c=c.most_common()
for i in range(len(c)):
    if c[i][1]!=1:
        x+=c[i][1]-1
    else:
        break
y=len(b)
if x%2==0:
    print(y)

else:
    print(y-1)

Score : 400 points

Problem Statement

Snuke has decided to play a game using cards.
He has a deck consisting of N cards. On the i-th card from the top, an integer A_i is written.

He will perform the operation described below zero or more times, so that the values written on the remaining cards will be pairwise distinct. Find the maximum possible number of remaining cards. Here, N is odd, which guarantees that at least one card can be kept.

Operation: Take out three arbitrary cards from the deck. Among those three cards, eat two: one with the largest value, and another with the smallest value. Then, return the remaining one card to the deck.

Constraints

3 ≦ N ≦ 10^{5}

N is odd.

1 ≦ A_i ≦ 10^{5}

A_i is an integer.

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 A_3 ... A_{N}

Output

Print the answer.

Sample Input 1

5
1 2 1 3 7

Sample Output 1

3

One optimal solution is to perform the operation once, taking out two cards with 1 and one card with 2. One card with 1 and another with 2 will be eaten, and the remaining card with 1 will be returned to deck. Then, the values written on the remaining cards in the deck will be pairwise distinct: 1, 3 and 7.

Sample Input 2

15
1 3 5 2 1 3 2 8 8 6 2 6 11 1 1

Sample Output 2

7


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

Score : 300 points

Problem Statement

Dolphin resides in two-dimensional Cartesian plane, with the positive x-axis pointing right and the positive y-axis pointing up.

Currently, he is located at the point (sx,sy). In each second, he can move up, down, left or right by a distance of 1.

Here, both the x- and y-coordinates before and after each movement must be integers.

He will first visit the point (tx,ty) where sx < tx and sy < ty, then go back to the point (sx,sy), then visit the point (tx,ty) again, and lastly go back to the point (sx,sy).

Here, during the whole travel, he is not allowed to pass through the same point more than once, except the points (sx,sy) and (tx,ty).

Under this condition, find a shortest path for him.

Constraints

-1000 ≤ sx < tx ≤ 1000

-1000 ≤ sy < ty ≤ 1000

sx,sy,tx and ty are integers.

Input

The input is given from Standard Input in the following format:

sx sy tx ty

Output

Print a string S that represents a shortest path for Dolphin.

The i-th character in S should correspond to his i-th movement.

The directions of the movements should be indicated by the following characters:

U: Up

D: Down

L: Left

R: Right

If there exist multiple shortest paths under the condition, print any of them.

Sample Input 1

0 0 1 2

Sample Output 1

UURDDLLUUURRDRDDDLLU

One possible shortest path is:

Going from (sx,sy) to (tx,ty) for the first time: (0,0) → (0,1) → (0,2) → (1,2)

Going from (tx,ty) to (sx,sy) for the first time: (1,2) → (1,1) → (1,0) → (0,0)

Going from (sx,sy) to (tx,ty) for the second time: (0,0) → (-1,0) → (-1,1) → (-1,2) → (-1,3) → (0,3) → (1,3) → (1,2)

Going from (tx,ty) to (sx,sy) for the second time: (1,2) → (2,2) → (2,1) → (2,0) → (2,-1) → (1,-1) → (0,-1) → (0,0)

Sample Input 2

-2 -2 1 1

Sample Output 2

UURRURRDDDLLDLLULUUURRURRDDDLLDL

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.


a,b = map(int,input().split())
val = 0
if a*b < 0:
    val += 1
elif (a>0 and b==0) or (a==0 and b<0):
    val += 1
elif b < a:
    val += 2
val += abs(abs(a)-abs(b))
print(val)

Score : 300 points

Problem Statement

Snuke has a calculator. It has a display and two buttons.

Initially, the display shows an integer x.
Snuke wants to change this value into another integer y, by pressing the following two buttons some number of times in arbitrary order:

Button A: When pressed, the value on the display is incremented by 1.

Button B: When pressed, the sign of the value on the display is reversed.

Find the minimum number of times Snuke needs to press the buttons to achieve his objective.
It can be shown that the objective is always achievable regardless of the values of the integers x and y.

Constraints

x and y are integers.

|x|, |y| ≤ 10^9

x and y are different.

Input

The input is given from Standard Input in the following format:

x y

Output

Print the minimum number of times Snuke needs to press the buttons to achieve his objective.

Sample Input 1

10 20

Sample Output 1

10

Press button A ten times.

Sample Input 2

10 -10

Sample Output 2

1

Press button B once.

Sample Input 3

-10 -20

Sample Output 3

12

Press the buttons as follows:

Press button B once.

Press button A ten times.

Press button B once.


a=input().split()
print(int(a[0])+int(a[2]) if a[1]=="+" else int(a[0])-int(a[2]))

Score : 100 points

Problem Statement

Joisino wants to evaluate the formula "A op B".
Here, A and B are integers, and the binary operator op is either + or -.
Your task is to evaluate the formula instead of her.

Constraints

1≦A,B≦10^9

op is either + or -.

Input

The input is given from Standard Input in the following format:

A op B

Output

Evaluate the formula and print the result.

Sample Input 1

1 + 2

Sample Output 1

3

Since 1 + 2 = 3, the output should be 3.

Sample Input 2

5 - 7

Sample Output 2

-2


import sys
N = int(input())
A = [int(i) for i in input().split()]
A.sort()
mod = 10**9 + 7
ans = 1
if N == 1:
    if A[0] == 0:
        print(ans)
        sys.exit()
    else:
        print(0)
        sys.exit()
if N%2 == 0:
    for i in range(N//2):
        if A[2*i] != 2*i+1 or A[2*i+1] != 2*i+1:
            print(0)
            sys.exit()
        else:
            ans = 2*ans%mod
    print(ans)
else:
    if A.pop(0) != 0:
        print(0)
        sys.exit()
    else:
        for i in range((N-1)//2):
            if A[2*i] != 2*i+2 or A[2*i+1] != 2*i+2:
                print(0)
                sys.exit()
            else:
                ans = 2*ans%mod
    print(ans)

Score : 300 points

Problem Statement

There are N people, conveniently numbered 1 through N.
They were standing in a row yesterday, but now they are unsure of the order in which they were standing.
However, each person remembered the following fact: the absolute difference of the number of the people who were standing to the left of that person, and the number of the people who were standing to the right of that person.
According to their reports, the difference above for person i is A_i.

Based on these reports, find the number of the possible orders in which they were standing.
Since it can be extremely large, print the answer modulo 10^9+7.
Note that the reports may be incorrect and thus there may be no consistent order.
In such a case, print 0.

Constraints

1≦N≦10^5

0≦A_i≦N-1

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the number of the possible orders in which they were standing, modulo 10^9+7.

Sample Input 1

5
2 4 4 0 2

Sample Output 1

4

There are four possible orders, as follows:

2,1,4,5,3

2,5,4,1,3

3,1,4,5,2

3,5,4,1,2

Sample Input 2

7
6 4 0 2 4 0 2

Sample Output 2

0

Any order would be inconsistent with the reports, thus the answer is 0.

Sample Input 3

8
7 5 1 1 7 3 5 3

Sample Output 3

16


print('vowel') if input() in ['a', 'i', 'u', 'e', 'o'] print('consonant')

Score : 100 points

Problem Statement

Given a lowercase English letter c, determine whether it is a vowel. Here, there are five vowels in the English alphabet: a, e, i, o and u.

Constraints

c is a lowercase English letter.

Input

The input is given from Standard Input in the following format:

c

Output

If c is a vowel, print vowel. Otherwise, print consonant.

Sample Input 1

a

Sample Output 1

vowel

Since a is a vowel, print vowel.

Sample Input 2

z

Sample Output 2

consonant

Sample Input 3

s

Sample Output 3

consonant


s = raw_input()
if s =="a" or s=="i" or s=="u" or s=="e" or s=="o":
    print "vowel"
else:
    print "consonant"


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

for i in range(H):
    c = input()
    print(c)
    print(c)


class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n
    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if x == y:
            return
        if self.parents[x] > self.parents[y]:
            x, y = y, x
        self.parents[x] += self.parents[y]
        self.parents[y] = x
    def size(self, x):
        return -self.parents[self.find(x)]
    def same(self, x, y):
        return self.find(x) == self.find(y)
    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]
    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]
    def group_count(self):
        return len(self.roots())
import sys
input=sys.stdin.readline
n,k,l = map(int, input().split())
uf_road = UnionFind(n)
uf_train = UnionFind(n)
for i in range(k):
    p,q=list(map(int, input().split()))
    uf_road.union(p-1, q-1)
for i in range(l):
    p,q=list(map(int, input().split()))
    uf_train.union(p-1, q-1)

ans=[0]*n
for i in range(n):
    if ans[i]!=0:
        continue
    mem1=set(uf_road.members(i))
    mem2=set(uf_train.members(i))
    inter=mem1 & mem2
    int_num=len(inter)
    for tmp in inter:
        ans[tmp]=int_num
print(*ans)

Score : 400 points

Problem Statement

There are N cities. There are also K roads and L railways, extending between the cities.
The i-th road bidirectionally connects the p_i-th and q_i-th cities, and the i-th railway bidirectionally connects the r_i-th and s_i-th cities.
No two roads connect the same pair of cities. Similarly, no two railways connect the same pair of cities.

We will say city A and B are connected by roads if city B is reachable from city A by traversing some number of roads. Here, any city is considered to be connected to itself by roads.
We will also define connectivity by railways similarly.

For each city, find the number of the cities connected to that city by both roads and railways.

Constraints

2 ≦ N ≦ 2*10^5

1 ≦ K, L≦ 10^5

1 ≦ p_i, q_i, r_i, s_i ≦ N

p_i < q_i

r_i < s_i

When i ≠ j, (p_i, q_i) ≠ (p_j, q_j)

When i ≠ j, (r_i, s_i) ≠ (r_j, s_j)

Input

The input is given from Standard Input in the following format:

N K L
p_1 q_1
:
p_K q_K
r_1 s_1
:
r_L s_L

Output

Print N integers. The i-th of them should represent the number of the cities connected to the i-th city by both roads and railways.

Sample Input 1

4 3 1
1 2
2 3
3 4
2 3

Sample Output 1

1 2 2 1

All the four cities are connected to each other by roads.

By railways, only the second and third cities are connected. Thus, the answers for the cities are 1, 2, 2 and 1, respectively.

Sample Input 2

4 2 2
1 2
2 3
1 4
2 3

Sample Output 2

1 2 2 1

Sample Input 3

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

Sample Output 3

1 1 2 1 2 2 2


Score : 100 points

Problem Statement

Snuke is going to open a contest named "AtCoder s Contest".
Here, s is a string of length 1 or greater, where the first character is an uppercase English letter, and the second and subsequent characters are lowercase English letters.

Snuke has decided to abbreviate the name of the contest as "AxC".
Here, x is the uppercase English letter at the beginning of s.

Given the name of the contest, print the abbreviation of the name.

Constraints

The length of s is between 1 and 100, inclusive.

The first character in s is an uppercase English letter.

The second and subsequent characters in s are lowercase English letters.

Input

The input is given from Standard Input in the following format:

AtCoder s Contest

Output

Print the abbreviation of the name of the contest.

Sample Input 1

AtCoder Beginner Contest

Sample Output 1

ABC

The contest in which you are participating now.

Sample Input 2

AtCoder Snuke Contest

Sample Output 2

ASC

This contest does not actually exist.

Sample Input 3

AtCoder X Contest

Sample Output 3

AXC


a, b, x = map(int, input().split())
print(round(b // x) - round((a - 1) // x))

Score : 200 points

Problem Statement

You are given nonnegative integers a and b (a ≤ b), and a positive integer x.
Among the integers between a and b, inclusive, how many are divisible by x?

Constraints

0 ≤ a ≤ b ≤ 10^{18}

1 ≤ x ≤ 10^{18}

Input

The input is given from Standard Input in the following format:

a b x

Output

Print the number of the integers between a and b, inclusive, that are divisible by x.

Sample Input 1

4 8 2

Sample Output 1

3

There are three integers between 4 and 8, inclusive, that are divisible by 2: 4, 6 and 8.

Sample Input 2

0 5 1

Sample Output 2

6

There are six integers between 0 and 5, inclusive, that are divisible by 1: 0, 1, 2, 3, 4 and 5.

Sample Input 3

9 9 2

Sample Output 3

0

There are no integer between 9 and 9, inclusive, that is divisible by 2.

Sample Input 4

1 1000000000000000000 3

Sample Output 4

333333333333333333

Watch out for integer overflows.


s = input()

i = 1
ans = 0
while i < (len(s) - 1):
    if (s[i - 1] != s[i + 1]):
        s = s[:i] + s[i + 1:]
        ans = (ans + 1) % 2
        i = 0
    i += 1

if (ans + 1) % 2 == 0:
    print('First')
else:
    print('Second')

Score : 500 points

Problem Statement

There is a string s of length 3 or greater.
No two neighboring characters in s are equal.

Takahashi and Aoki will play a game against each other.
The two players alternately performs the following operation, Takahashi going first:

Remove one of the characters in s, excluding both ends. However, a character cannot be removed if removal of the character would result in two neighboring equal characters in s.

The player who becomes unable to perform the operation, loses the game. Determine which player will win when the two play optimally.

Constraints

3 ≤ |s| ≤ 10^5

s consists of lowercase English letters.

No two neighboring characters in s are equal.

Input

The input is given from Standard Input in the following format:

s

Output

If Takahashi will win, print First. If Aoki will win, print Second.

Sample Input 1

aba

Sample Output 1

Second

Takahashi, who goes first, cannot perform the operation, since removal of the b, which is the only character not at either ends of s, would result in s becoming aa, with two as neighboring.

Sample Input 2

abc

Sample Output 2

First

When Takahashi removes b from s, it becomes ac.
Then, Aoki cannot perform the operation, since there is no character in s, excluding both ends.

Sample Input 3

abcab

Sample Output 3

First


class UnionFind:
    def __init__(self, n):
        self.par = [i for i in range(n+1)]
        self.rank = [0] * (n+1)

    # 検索
    def find(self, x):
        if self.par[x] == x:
            return x
        else:
            self.par[x] = self.find(self.par[x])
            return self.par[x]

    # 併合
    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)
        if self.rank[x] < self.rank[y]:
            self.par[x] = y
        else:
            self.par[y] = x
            if self.rank[x] == self.rank[y]:
                self.rank[x] += 1

    # 同じ集合に属するか判定
    def same_check(self, x, y):
        return self.find(x) == self.find(y)

    def ans(self):
        print(len(set(self.par))-1)

n,m=map(int,input().split())
uf=UnionFind(n)
l=[[] for i in range(m)]
for i in range(n):
    k=list(map(int,input.split()))
    k=[1:]
    for j in k:
        l[j-1].append(i)

for i in range(m):
    if len(l[i])>0:
        ll=l[i][0]
        for j in range(len(l[i])-1):
            uf.union(ll,l[i][j+1])

for i in range(1,n+1):
    uf.find(i)

if uf.ans()==1:
    print('YES')
else:
    print('NO')

Score : 400 points

Problem Statement

On a planet far, far away, M languages are spoken. They are conveniently numbered 1 through M.

For CODE FESTIVAL 20XX held on this planet, N participants gathered from all over the planet.

The i-th (1≦i≦N) participant can speak K_i languages numbered L_{i,1}, L_{i,2}, ..., L_{i,{}K_i}.

Two participants A and B can communicate with each other if and only if one of the following conditions is satisfied:

There exists a language that both A and B can speak.

There exists a participant X that both A and B can communicate with.

Determine whether all N participants can communicate with all other participants.

Constraints

2≦N≦10^5

1≦M≦10^5

1≦K_i≦M

(The sum of all K_i)≦10^5

1≦L_{i,j}≦M

L_{i,1}, L_{i,2}, ..., L_{i,{}K_i} are pairwise distinct.

Partial Score

200 points will be awarded for passing the test set satisfying the following: N≦1000, M≦1000 and (The sum of all K_i)≦1000.

Additional 200 points will be awarded for passing the test set without additional constraints.

Input

The input is given from Standard Input in the following format:

N M
K_1 L_{1,1} L_{1,2} ... L_{1,{}K_1}
K_2 L_{2,1} L_{2,2} ... L_{2,{}K_2}
:
K_N L_{N,1} L_{N,2} ... L_{N,{}K_N}

Output

If all N participants can communicate with all other participants, print YES. Otherwise, print NO.

Sample Input 1

4 6
3 1 2 3
2 4 2
2 4 6
1 6

Sample Output 1

YES

Any two participants can communicate with each other, as follows:

Participants 1 and 2: both can speak language 2.

Participants 2 and 3: both can speak language 4.

Participants 1 and 3: both can communicate with participant 2.

Participants 3 and 4: both can speak language 6.

Participants 2 and 4: both can communicate with participant 3.

Participants 1 and 4: both can communicate with participant 2.

Note that there can be languages spoken by no participant.

Sample Input 2

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

Sample Output 2

NO

For example, participants 1 and 3 cannot communicate with each other.


h,w = map(int,input().split())
a = [list(input()) for nesya in range(h)]
if a.count('#') == h+w-1:
  print('Possible')
else:
  print('Impossible')

Score : 200 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

We have a grid of H rows and W columns. Initially, there is a stone in the top left cell. Shik is trying to move the stone to the bottom right cell. In each step, he can move the stone one cell to its left, up, right, or down (if such cell exists). It is possible that the stone visits a cell multiple times (including the bottom right and the top left cell).

You are given a matrix of characters a_{ij} (1 \leq i \leq H, 1 \leq j \leq W). After Shik completes all moving actions, a_{ij} is # if the stone had ever located at the i-th row and the j-th column during the process of moving. Otherwise, a_{ij} is .. Please determine whether it is possible that Shik only uses right and down moves in all steps.

Constraints

2 \leq H, W \leq 8

a_{i,j} is either # or ..

There exists a valid sequence of moves for Shik to generate the map a.

Input

The input is given from Standard Input in the following format:

H W
a_{11}a_{12}...a_{1W}
:
a_{H1}a_{H2}...a_{HW}

Output

If it is possible that Shik only uses right and down moves, print Possible. Otherwise, print Impossible.

Sample Input 1

4 5
##...
.##..
..##.
...##

Sample Output 1

Possible

The matrix can be generated by a 7-move sequence: right, down, right, down, right, down, and right.

Sample Input 2

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

Sample Output 2

Impossible

Sample Input 3

4 5
##...
.###.
.###.
...##

Sample Output 3

Impossible


def main():
    H, W = map(int, input().split())
    A = [input() for _ in range(H)]
    i = 0
    j = 0
    visited = set([(0, 0)])
    while i < H and j < H:
        if i + 1 < H and A[i+1][j] == "#":
            i += 1
        elif j + 1 < W and A[i][j+1] == "#":
            j += 1
        else:
            print("Impossible")
            return
        visited.add((i, j))
    for i in range(H):
        for j in range(W):
            if A[i][j] == "#" and (i, j) not in visited:
                print("Impossible")
                return
    print("Possible")

if __name__ == "__main__":
    main()


n = int(input())
list = input().split()
for i in range(0,n):
   if int(list[i]) == n:
      p = i
      break

list1 = []
list2 = []
for k in range(0,n):
   list1.append(n* int ( list[k] ) - n + 1)
for k in range(0,n):
   list2.append( max ( int(list1[p] ) + 1 - p + k - int( list1[k] ), n - int ( list[k] ) +1))

a = ""
b = ""

list2.sort(reverse = True)
list1.sort()
for l in range (0,n):
   a += str(list1[l])
   a += " "
   b += str(list2[l])
   b += " "

print(a)
print(b)

Score : 400 points

Problem Statement

#nck {
width: 30px;
height: auto;
}

You are given a permutation p of the set {1, 2, ..., N}. Please construct two sequences of positive integers a_1, a_2, ..., a_N and b_1, b_2, ..., b_N satisfying the following conditions:

1 \leq a_i, b_i \leq 10^9 for all i

a_1 < a_2 < ... < a_N

b_1 > b_2 > ... > b_N

a_{p_1}+b_{p_1} < a_{p_2}+b_{p_2} < ... < a_{p_N}+b_{p_N}

Constraints

2 \leq N \leq 20,000

p is a permutation of the set {1, 2, ..., N}

Input

The input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

The output consists of two lines. The first line contains a_1, a_2, ..., a_N seperated by a space. The second line contains b_1, b_2, ..., b_N seperated by a space.

It can be shown that there always exists a solution for any input satisfying the constraints.

Sample Input 1

2
1 2

Sample Output 1

1 4
5 4

a_1 + b_1 = 6 and a_2 + b_2 = 8. So this output satisfies all conditions.

Sample Input 2

3
3 2 1

Sample Output 2

1 2 3
5 3 1

Sample Input 3

3
2 3 1

Sample Output 3

5 10 100
100 10 1


#!/usr/bin/env python3
import sys, math, itertools, heapq, collections, bisect, fractions
input = lambda: sys.stdin.buffer.readline().rstrip().decode('utf-8')
sys.setrecursionlimit(10**8)
inf = float('inf')
ans = count = 0

# a,b,c=map(int,input().split())
# for i in itertools.permutations([a,b,c],3):
#     if i[0]==sum(i[1:]) or sum(i[:2])==i[2]:
#         print("Yes")
#         exit()

# print("No")

Score : 100 points

Problem Statement

Two students of AtCoder Kindergarten are fighting over candy packs.

There are three candy packs, each of which contains a, b, and c candies, respectively.

Teacher Evi is trying to distribute the packs between the two students so that each student gets the same number of candies. Determine whether it is possible.

Note that Evi cannot take candies out of the packs, and the whole contents of each pack must be given to one of the students.

Constraints

1 ≦ a, b, c ≦ 100

Input

The input is given from Standard Input in the following format:

a b c

Output

If it is possible to distribute the packs so that each student gets the same number of candies, print Yes. Otherwise, print No.

Sample Input 1

10 30 20

Sample Output 1

Yes

Give the pack with 30 candies to one student, and give the two packs with 10 and 20 candies to the other. Then, each gets 30 candies.

Sample Input 2

30 30 100

Sample Output 2

No

In this case, the student who gets the pack with 100 candies always has more candies than the other.

Note that every pack must be given to one of them.

Sample Input 3

56 25 31

Sample Output 3

Yes


import sys
sys.setrecursionlimit(10 ** 5 + 10)
def input(): return sys.stdin.readline().strip()

def resolve():
    def main():
        s=list(input())
        ls=len(s)
        for i in range(ls-1):
            for j in range(i+1,ls):
                if s[i]=='C' and s[j]=='F':
                    return 'Yes'
        return 'No'
    print(main())

resolve()

Score : 100 points

Problem Statement

This contest is CODEFESTIVAL, which can be shortened to the string CF by deleting some characters.

Mr. Takahashi, full of curiosity, wondered if he could obtain CF from other strings in the same way.

You are given a string s consisting of uppercase English letters.
Determine whether the string CF can be obtained from the string s by deleting some characters.

Constraints

2 ≤ |s| ≤ 100

All characters in s are uppercase English letters (A-Z).

Input

The input is given from Standard Input in the following format:

s

Output

Print Yes if the string CF can be obtained from the string s by deleting some characters.
Otherwise print No.

Sample Input 1

CODEFESTIVAL

Sample Output 1

Yes

CF is obtained by deleting characters other than the first character C and the fifth character F.

Sample Input 2

FESTIVALCODE

Sample Output 2

No

FC can be obtained but CF cannot be obtained because you cannot change the order of the characters.

Sample Input 3

CF

Sample Output 3

Yes

It is also possible not to delete any characters.

Sample Input 4

FCF

Sample Output 4

Yes

CF is obtained by deleting the first character.


A,B,C = map(int,input().split())
K = 1
if A != B:
    K = K + 1
else:
    pass
if A != C:
    if B != C:
        K = K + 1
    else:
        pass
else:
    pass
print(int(K))

Score : 100 points

Problem Statement

AtCoDeer the deer recently bought three paint cans.
The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c.
Here, the color of each paint can is represented by an integer between 1 and 100, inclusive.

Since he is forgetful, he might have bought more than one paint can in the same color.
Count the number of different kinds of colors of these paint cans and tell him.

Constraints

1≦a,b,c≦100

Input

The input is given from Standard Input in the following format:

a b c

Output

Print the number of different kinds of colors of the paint cans.

Sample Input 1

3 1 4

Sample Output 1

3

Three different colors: 1, 3, and 4.

Sample Input 2

3 3 33

Sample Output 2

2

Two different colors: 3 and 33.


A,B,C = map(int,input().split())
ans =1
if A != B:
  ans += 1
if A != C and B != C:
  ans += 1
print(ANS)


def main():
    a,b,c=map(int, input().split())
    number=1
    if a==b:
        if b==c:
            number=1
        else:
            number=2
    else:
        if b==c:
            number=2
        elif a==c:
            number=2
        else:
            number=3
    print(number)

if __name__ == "__main__":
    main()


import sys

readline = sys.stdin.readline
MOD = 10 ** 9 + 7
INF = float('INF')
sys.setrecursionlimit(10 ** 5)

def main():
    s = list(input())
    k = int(readline())
    l = len(s)
    rem = k

    for i in range(l):
        char = s[i]
        diff = (ord("z") - ord(s[i]) + 1) % 26
        if rem >= diff:
            rem -= diff
            s[i] = "a"

    rem = rem % 26
    s[l - 1] = chr(ord(s[l - 1]) + rem)
    print("".join(s))

if __name__ == '__main__':
    main()

Score : 400 points

Problem Statement

Mr. Takahashi has a string s consisting of lowercase English letters.
He repeats the following operation on s exactly K times.

Choose an arbitrary letter on s and change that letter to the next alphabet. Note that the next letter of z is a.

For example, if you perform an operation for the second letter on aaz, aaz becomes abz.
If you then perform an operation for the third letter on abz, abz becomes aba.

Mr. Takahashi wants to have the lexicographically smallest string after performing exactly K operations on s.
Find the such string.

Constraints

1≤|s|≤10^5

All letters in s are lowercase English letters.

1≤K≤10^9

Input

The input is given from Standard Input in the following format:

s
K

Output

Print the lexicographically smallest string after performing exactly K operations on s.

Sample Input 1

xyz
4

Sample Output 1

aya

For example, you can perform the following operations: xyz, yyz, zyz, ayz, aya.

Sample Input 2

a
25

Sample Output 2

z

You have to perform exactly K operations.

Sample Input 3

codefestival
100

Sample Output 3

aaaafeaaivap


from itertools import combinations, product

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

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

def get(fmt):
    ps = input().split()
    r = []
    for p, t in zip(ps, fmt):
        if t == "i":
            r.append(int(p))
        if t == "f":
            r.append(float(p))
        if t == "s":
            r.append(p)
    if len(r) == 1:
        r = r[0]
    return r

def put(*args, **kwargs):
    print(*args, **kwargs)

def rep(n, f, *args, **kwargs):
    return [f(*args, **kwargs) for _ in range(n)]

def rep_im(n, v):
    return rep(n, lambda: v)

YES_NO = ["NO", "YES"]

a, b, h = rep(3, get, "i")

put((a + b) * h // 2)

Score : 100 points

Problem Statement

You are given a trapezoid. The lengths of its upper base, lower base, and height are a, b, and h, respectively.

An example of a trapezoid

Find the area of this trapezoid.

Constraints

1≦a≦100

1≦b≦100

1≦h≦100

All input values are integers.

h is even.

Input

The input is given from Standard Input in the following format:

a
b
h

Output

Print the area of the given trapezoid. It is guaranteed that the area is an integer.

Sample Input 1

3
4
2

Sample Output 1

7

When the lengths of the upper base, lower base, and height are 3, 4, and 2, respectively, the area of the trapezoid is (3+4)×2/2 = 7.

Sample Input 2

4
4
4

Sample Output 2

16

In this case, a parallelogram is given, which is also a trapezoid.


import math
A, B, C = map(int, input().split())
l = [A,B,C]

max_index = l.index(max(l))
if max_index == 0:
    large = A * B * math.floor(C/2)
elif max_index == 1:
    large = B * C * math.floor(A/2)
else:
    large = C * A * math.floor(B/2)
# print(A*B*C)
# print(large)
ans = A*B*C - 2*large
print(ans)

Score : 200 points

Problem Statement

We have a rectangular parallelepiped of size A×B×C, built with blocks of size 1×1×1. Snuke will paint each of the A×B×C blocks either red or blue, so that:

There is at least one red block and at least one blue block.

The union of all red blocks forms a rectangular parallelepiped.

The union of all blue blocks forms a rectangular parallelepiped.

Snuke wants to minimize the difference between the number of red blocks and the number of blue blocks. Find the minimum possible difference.

Constraints

2≤A,B,C≤10^9

Input

The input is given from Standard Input in the following format:

A B C

Output

Print the minimum possible difference between the number of red blocks and the number of blue blocks.

Sample Input 1

3 3 3

Sample Output 1

9

For example, Snuke can paint the blocks as shown in the diagram below.
There are 9 red blocks and 18 blue blocks, thus the difference is 9.

Sample Input 2

2 2 4

Sample Output 2

0

For example, Snuke can paint the blocks as shown in the diagram below.
There are 8 red blocks and 8 blue blocks, thus the difference is 0.

Sample Input 3

5 3 5

Sample Output 3

15

For example, Snuke can paint the blocks as shown in the diagram below.
There are 45 red blocks and 30 blue blocks, thus the difference is 9.


N,K,X,Y= map(int, input().split())
if N<=K:
  print(N*X)
else:
  print(N*X+(N-K)*Y)

Score : 100 points

Problem Statement

There is a hotel with the following accommodation fee:

X yen (the currency of Japan) per night, for the first K nights

Y yen per night, for the (K+1)-th and subsequent nights

Tak is staying at this hotel for N consecutive nights.
Find his total accommodation fee.

Constraints

1 \leq N, K \leq 10000

1 \leq Y < X \leq 10000

N,\,K,\,X,\,Y are integers.

Input

The input is given from Standard Input in the following format:

N
K
X
Y

Output

Print Tak's total accommodation fee.

Sample Input 1

5
3
10000
9000

Sample Output 1

48000

The accommodation fee is as follows:

10000 yen for the 1-st night

10000 yen for the 2-nd night

10000 yen for the 3-rd night

9000 yen for the 4-th night

9000 yen for the 5-th night

Thus, the total is 48000 yen.

Sample Input 2

2
3
10000
9000

Sample Output 2

20000


n=int(input())
k=int(input())
x=int(input())
y=int(input())

ans=0
if n<=k:
    ans=x*n
else:
    ans=x*k+y*(n-k)
print(ans)


from collections import defaultdict

def main():
    S = input()
    d = defaultdict(int)
    for c in S:
        d[c] += 1
    for n in d.values():
        if n % 2 == 1:
            print('No')
            exit()
    print('Yes')

if __name__ == '__main__':
    main()

Score : 200 points

Problem Statement

Let w be a string consisting of lowercase letters.
We will call w beautiful if the following condition is satisfied:

Each lowercase letter of the English alphabet occurs even number of times in w.

You are given the string w. Determine if w is beautiful.

Constraints

1 \leq |w| \leq 100

w consists of lowercase letters (a-z).

Input

The input is given from Standard Input in the following format:

w

Output

Print Yes if w is beautiful. Print No otherwise.

Sample Input 1

abaccaba

Sample Output 1

Yes

a occurs four times, b occurs twice, c occurs twice and the other letters occur zero times.

Sample Input 2

hthth

Sample Output 2

No


w = input()
W = list(set(w))
for i in W:
    if w.count(i)%2 == 0:
        continue
    else:
        print("No")
        exit()
print("Yes")


N, A = map(int, input().split())
X = [int(a) for a in input().split()]
PO = [x-A for x in X if x > A]
NE = [A-x for x in X if x < A]
ZE = N - len(PO) - len(NE)

Y = [1] + [0] * 3000
Z = [1] + [0] * 3000

for i in PO:
    for j in range(len(Y) - A)[::-1]:
        Y[i+j] += Y[j]
for i in NE:
    for j in range(len(Z) - A)[::-1]:
        Z[i+j] += Z[j]

ans = -1
for i in range(len(Y)):
    ans += Y[i] * Z[i] * (2**ZE)
print(ans)

Score : 300 points

Problem Statement

Tak has N cards. On the i-th (1 \leq i \leq N) card is written an integer x_i.
He is selecting one or more cards from these N cards, so that the average of the integers written on the selected cards is exactly A.
In how many ways can he make his selection?

Constraints

1 \leq N \leq 50

1 \leq A \leq 50

1 \leq x_i \leq 50

N,\,A,\,x_i are integers.

Partial Score

200 points will be awarded for passing the test set satisfying 1 \leq N \leq 16.

Input

The input is given from Standard Input in the following format:

N A
x_1 x_2 ... x_N

Output

Print the number of ways to select cards such that the average of the written integers is exactly A.

Sample Input 1

4 8
7 9 8 9

Sample Output 1

5

The following are the 5 ways to select cards such that the average is 8:

Select the 3-rd card.

Select the 1-st and 2-nd cards.

Select the 1-st and 4-th cards.

Select the 1-st, 2-nd and 3-rd cards.

Select the 1-st, 3-rd and 4-th cards.

Sample Input 2

3 8
6 6 9

Sample Output 2

0

Sample Input 3

8 5
3 6 2 8 7 6 5 9

Sample Output 3

19

Sample Input 4

33 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Sample Output 4

8589934591

The answer may not fit into a 32-bit integer.


from bisect import bisect_left
N = int(input())
A = [int(input()) for _ in range(N)]
A_sorted = sorted(A)
A = [bisect_left(A_sorted, a) for a in A]
res = 0
for i, a in enumerate(A):
    if (i-a)%2:
        res+=1
print(res//2)

Score : 600 points

Problem Statement

Snuke got an integer sequence of length N from his mother, as a birthday present. The i-th (1 ≦ i ≦ N) element of the sequence is a_i. The elements are pairwise distinct.
He is sorting this sequence in increasing order.
With supernatural power, he can perform the following two operations on the sequence in any order:

Operation 1: choose 2 consecutive elements, then reverse the order of those elements.

Operation 2: choose 3 consecutive elements, then reverse the order of those elements.

Snuke likes Operation 2, but not Operation 1. Find the minimum number of Operation 1 that he has to perform in order to sort the sequence in increasing order.

Constraints

1 ≦ N ≦ 10^5

0 ≦ A_i ≦ 10^9

If i ≠ j, then A_i ≠ A_j.

All input values are integers.

Input

The input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of times Operation 1 that Snuke has to perform.

Sample Input 1

4
2
4
3
1

Sample Output 1

1

The given sequence can be sorted as follows:

First, reverse the order of the last three elements. The sequence is now: 2,1,3,4.

Then, reverse the order of the first two elements. The sequence is now: 1,2,3,4.

In this sequence of operations, Operation 1 is performed once. It is not possible to sort the sequence with less number of Operation 1, thus the answer is 1.

Sample Input 2

5
10
8
5
3
2

Sample Output 2

0


N = int(input())
total = 0

for i in range(N):
    total += i+1

print(total)

Score : 100 points

Problem Statement

There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

Constraints

1≦N≦100

Input

The input is given from Standard Input in the following format:

N

Output

Print the necessary number of candies in total.

Sample Input 1

3

Sample Output 1

6

The answer is 1+2+3=6.

Sample Input 2

10

Sample Output 2

55

The sum of the integers from 1 to 10 is 55.

Sample Input 3

1

Sample Output 3

1

Only one child. The answer is 1 in this case.


n = int(input())
data = list(map(int,input().split()))
mini = 100000
maxi = -100000
for i in range(n):
    if mini > data[i]:
        mini = data[i]
    if maxi < data[i]:
        maxi = data[i]
mincost = 1000000000
ans = 0
for i in range(mini,maxi+1):
    cost = 0
    for j in range(n):
        cost += (data[j] - i)**2
    if mincost > cost:
        ans = i
        mincost = cost
print(mincost)

Score : 200 points

Problem Statement

Evi has N integers a_1,a_2,..,a_N. His objective is to have N equal integers by transforming some of them.

He may transform each integer at most once. Transforming an integer x into another integer y costs him (x-y)^2 dollars. Even if a_i=a_j (i≠j), he has to pay the cost separately for transforming each of them (See Sample 2).

Find the minimum total cost to achieve his objective.

Constraints

1≦N≦100

-100≦a_i≦100

Input

The input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum total cost to achieve Evi's objective.

Sample Input 1

2
4 8

Sample Output 1

8

Transforming the both into 6s will cost (4-6)^2+(8-6)^2=8 dollars, which is the minimum.

Sample Input 2

3
1 1 3

Sample Output 2

3

Transforming the all into 2s will cost (1-2)^2+(1-2)^2+(3-2)^2=3 dollars. Note that Evi has to pay (1-2)^2 dollar separately for transforming each of the two 1s.

Sample Input 3

3
4 2 5

Sample Output 3

5

Leaving the 4 as it is and transforming the 2 and the 5 into 4s will achieve the total cost of (2-4)^2+(5-4)^2=5 dollars, which is the minimum.

Sample Input 4

4
-100 -100 -100 -100

Sample Output 4

0

Without transforming anything, Evi's objective is already achieved. Thus, the necessary cost is 0.


a, b = map(int, input().split())
ans = []

for i in range(a,b+1):
    ans.append(i)

if ans.count(0) == 1:
    print("Zero")
    exit()

c = 0

for i in ans:
    if i < 0:
        c += 1

if c % 2 != 0:
    print("Negative")
else:
    print("Positive")

Problem Statement

You are given two integers a and b (a≤b). Determine if the product of the integers a, a+1, …, b is positive, negative or zero.

Constraints

a and b are integers.

-10^9≤a≤b≤10^9

Partial Score

In test cases worth 100 points, -10≤a≤b≤10.

Input

The input is given from Standard Input in the following format:

a b

Output

If the product is positive, print Positive. If it is negative, print Negative. If it is zero, print Zero.

Sample Input 1

1 3

Sample Output 1

Positive

1×2×3=6 is positive.

Sample Input 2

-3 -1

Sample Output 2

Negative

(-3)×(-2)×(-1)=-6 is negative.

Sample Input 3

-1 1

Sample Output 3

Zero

(-1)×0×1=0.


a,b=[int(i) for i in input().split()]
if 0<a and 0<b:
  print("Positive")
elif a*b<=0:
  print("Zero")
elif (b-a)%2==1:
  print("Positive")
else:
  print("Negative")


N,L = map(int, input().split())
A=list(map(int, input().split()))
choose = -1
for i in range(1, N):
    if A[i - 1] + A[i] >= L:
        choose = i - 1
        break
if choose == -1:
    print('Impossible')
else:
    print('Possible')
    for i in range(0, choose):
        print(i + 1)
    for i in range(N - 1, choose + 1, -1):
        print(i)
    print(choose + 1)

Problem Statement

We have N pieces of ropes, numbered 1 through N. The length of piece i is a_i.

At first, for each i (1≤i≤N-1), piece i and piece i+1 are tied at the ends, forming one long rope with N-1 knots. Snuke will try to untie all of the knots by performing the following operation repeatedly:

Choose a (connected) rope with a total length of at least L, then untie one of its knots.

Is it possible to untie all of the N-1 knots by properly applying this operation? If the answer is positive, find one possible order to untie the knots.

Constraints

2≤N≤10^5

1≤L≤10^9

1≤a_i≤10^9

All input values are integers.

Input

The input is given from Standard Input in the following format:

N L
a_1 a_2 ... a_n

Output

If it is not possible to untie all of the N-1 knots, print Impossible.

If it is possible to untie all of the knots, print Possible, then print another N-1 lines that describe a possible order to untie the knots. The j-th of those N-1 lines should contain the index of the knot that is untied in the j-th operation. Here, the index of the knot connecting piece i and piece i+1 is i.

If there is more than one solution, output any.

Sample Input 1

3 50
30 20 10

Sample Output 1

Possible
2
1

If the knot 1 is untied first, the knot 2 will become impossible to untie.

Sample Input 2

2 21
10 10

Sample Output 2

Impossible

Sample Input 3

5 50
10 20 30 40 50

Sample Output 3

Possible
1
2
3
4

Another example of a possible solution is 3, 4, 1, 2.


a,b,c =int(x) for in input().split()
if a+b+c==17 and a*b*c==175:
  print('Yes')
else:
  print('No')

Score : 100 points

Problem Statement

Iroha loves Haiku. Haiku is a short form of Japanese poetry. A Haiku consists of three phrases with 5, 7 and 5 syllables, in this order.

To create a Haiku, Iroha has come up with three different phrases. These phrases have A, B and C syllables, respectively. Determine whether she can construct a Haiku by using each of the phrases once, in some order.

Constraints

1≦A,B,C≦10

Input

The input is given from Standard Input in the following format:

A B C

Output

If it is possible to construct a Haiku by using each of the phrases once, print YES (case-sensitive). Otherwise, print NO.

Sample Input 1

5 5 7

Sample Output 1

YES

Using three phrases of length 5, 5 and 7, it is possible to construct a Haiku.

Sample Input 2

7 7 5

Sample Output 2

NO


n,k=[s for s in input().split()]
n=int(n)
k=int(k)
hate=list(map(int, input().split())) #hate取得
like=[]
for i in range(1,10): #like取得
	if i not in hate:
		like.append(i)
pointer=0

for ichi in like: #1桁の場合
	if ichi>=n:
		kane=ichi
		pointer=1
		break

if pointer==0: #2桁の場合
	for ju in like:
		for ichi in like:
			if ju*10+ichi>=n:
				kane=ju*10+ichi
				pointer=1
				break

if pointer==0: #3桁の場合
	for hyaku in like:
		for ju in like:
			for ichi in like:
				if hyaku*100+ju*10+ichi>=n:
					kane=hyaku*100+ju*10+ichi
					pointer=1
					break

if pointer==0: #4桁の場合
	for sen in like:
		for hyaku in like:
			for ju in like:
				for ichi in like:
					if sen*1000+hyaku*100+ju*10+ichi>=n:
						kane=sen*1000+hyaku*100+ju*10+ichi
						pointer=1
						break

if pointer==1:
	print(kane)


a,b,c=[int(x) for x in input().split()]
print("YES") if sorted([a,b,c]) == [5,5,7] else print("NO")


def resolve():
    N, K = list(map(int, input().split()))
    D = input().split()
    i = N
    while True:
        for s in str(i):
            if s in D:
                i += 1
                break
        else:
            print(i)
            return
    raise Exception("unreachable!")

if '__main__' == __name__:
    resolve()

Score : 300 points

Problem Statement

Iroha is very particular about numbers. There are K digits that she dislikes: D_1, D_2, ..., D_K.

She is shopping, and now paying at the cashier.
Her total is N yen (the currency of Japan), thus she has to hand at least N yen to the cashier (and possibly receive the change).

However, as mentioned before, she is very particular about numbers. When she hands money to the cashier, the decimal notation of the amount must not contain any digits that she dislikes. Under this condition, she will hand the minimum amount of money.

Find the amount of money that she will hand to the cashier.

Constraints

1 ≦ N < 10000

1 ≦ K < 10

0 ≦ D_1 < D_2 < … < D_K≦9

\{D_1,D_2,...,D_K\} ≠ \{1,2,3,4,5,6,7,8,9\}

Input

The input is given from Standard Input in the following format:

N K
D_1 D_2 … D_K

Output

Print the amount of money that Iroha will hand to the cashier.

Sample Input 1

1000 8
1 3 4 5 6 7 8 9

Sample Output 1

2000

She dislikes all digits except 0 and 2.

The smallest integer equal to or greater than N=1000 whose decimal notation contains only 0 and 2, is 2000.

Sample Input 2

9999 1
0

Sample Output 2

9999


#!/usr/bin/env python3
from collections import defaultdict
import sys
sys.setrecursionlimit(10**8)
input = sys.stdin.readline

n, k = [int(item) for item in input().split()]
d = [item for item in input().split()]
for i in range(n, n+10**5):
    si = str(i)
    ok = True
    for item in d:
        if item in si:
            ok = False
            break
    if ok:
        print(i)
        exit()


mount_list = []

for i in range(10):
  mount_list.append(int(input()))

mount_list.sort(reverse = True)

for i in range(3):
  print(mount_list[i])

List of Top 3 Hills

There is a data which provides heights (in meter) of mountains. The data is only for ten mountains.

Write a program which prints heights of the top three mountains in descending order.

Input

Height of mountain 1
Height of mountain 2
Height of mountain 3
.
.
Height of mountain 10

Constraints

0 ≤ height of mountain (integer) ≤ 10,000

Output

Height of the 1st mountain
Height of the 2nd mountain
Height of the 3rd mountain

Sample Input 1

1819
2003
876
2840
1723
1673
3776
2848
1592
922

Output for the Sample Input 1

3776
2848
2840

Sample Input 2

100
200
300
400
500
600
700
800
900
900

Output for the Sample Input 2

900
900
800