background

I decided to study for the programming test preparation for the entrance exam. Therefore, I will study because AtCoder was a site with many explanations in Japanese.

Purpose

Study for programming test preparation

Introduction

I thought that a site with many explanations would lead to my own study when taking a programming test. Therefore, when I was looking for a place where many programming tests were conducted in the Japanese community, I was studying because there was AtCoder.

Requirement definition

Solve AtCoder Beginners Selection
Solve related problems
Solve 60 LeetCode questions

environment

Windows10Pro
Python3.7
Anaconda

Result of doing

Question 1: ABC 086 A --Product

a,b=map(int,input().split())
if a*b%2==0:
   print('Even')
else :
    print('Odd')

Question 2: ABC 081 A --Placing Marbles

s=input()
count=0
for i in s:
    # print(i)
    if int(i)==1:
        count+=1
    # print(count)

print(count)

Alternative 1

It's simple and I prefer this one. If I used count on 04/19, it failed with RunTimeError, so it is recommended to use the first answer.

python print(input().count('1'))

###Alternative 2
How to guide your math skills
I don't really recommend it.
If you are good at math, you may like to write it.

print(int(input())%9)



##Question 3: [ABC 081 B - Shift Only](https://atcoder.jp/contests/abc081/tasks/abc081_b)
```python
N = int(input())
A = map(int, input().split())

count = 0

while all(a % 2 == 0 for a in A):
    A = [a / 2 for a in A]
    count += 1

print(count)

##Question 4: ABC 087 B - Coins

A=int(input())
B=int(input())
C=int(input())
X=int(input())

count=0

for a in range(A+1):
    for b in range(B+1):
        for c in range(C+1):
            if X==500*a+100*b+50*c:
                    count+=1

print(count)

##Question 5: ABC 083 B - Some Sums I learned that using a map can be used for convenient sorting. I used it for the first time.

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

def FindSumOfDigits(num):
    count = 0
    while num > 0:
        count += num % 10
 # This is different / missing
        num=num // 10

    return count

for n in range(1, N+1):
    count = FindSumOfDigits(n)
    if A<= count <=B:
        ans += n
print(ans)

##Question 6: ABC 088 B - Card Game for Two

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

Alice=Bob=0

for i in range(N):
    if i%2==0:
        Alice+=A[i]
    else:
        Bob+=A[i]

print(Alice-Bob)

##Question 7: ABC 085 B - Kagami Mochi

N=int(input())
d=[input() for _ in range(N)]
print(len(set(d)))

##Question 8: ABC 085 C - Otoshidama It was a problem that used mathematical skills such as solving simultaneous equations.

N, Y = map(int, input().split())

res10000 = res5000 = res1000 = -1

for a in range(N + 1):
    for b in range(N + 1 - a):
        c = N - a - b
        if Y == 10000 * a + 5000 * b + 1000 * c:
            res10000 = a
            res5000 = b
            res1000 = c

print(a, b, c)

##Question 9: ABC 049 C - Daydream

s = input().replace("eraser","").replace("erase","").replace("dreamer","").replace("dream","")

# Rewriting if statement
 print('NO' if s else 'YES')

if s:
  print("NO")
else:
  print("YES")

The idea of searching from behind and erasing

def main():
    S = input()

    while len(S) >= 5:
        if len(S) >= 7 and S[-7:] == "dreamer":
            S = S[:-7]
            continue

        if len(S) >= 6 and S[-6:] == "eraser":
            S = S[:-6]
            continue

        elif S[-5:] == "dream" or S[-5:] == "erase":
            S = S[:-5]
            continue

        else:
            break

    if len(S) == 0:
        print("YES")
    else:
        print("NO")
main()

##Question 10: ABC 086 C - Traveling Mathematical solution

n=int(input())
for i in range(n):
    t,x,y=map(int,input().split())
    if (x+y) > t or (x+y+t)%2:
        print('No')
        exit()

print('Yes')

###Alternative 1 The royal way to solve


N = int(input())
t = [0] * (N+1)
x = [0] * (N+1)
y = [0] * (N+1)
for i in range(N):
    t[i+1], x[i+1], y[i+1] = map(int, input().split())

f = True
for i in range(N):
    dt = t[i+1] - t[i]
    dist = abs(x[i+1]-x[i]) + abs(y[i+1]-y[i])
    if dt < dist:
        f = False
    if dist%2 != dt%2:
        f = False

print('Yes' if f else 'No')

###Betsukai 2

import sys

def main():
    N = int(input())
    t, x, y = 0, 0, 0

    for _ in range(N):
        next_t, next_x, next_y = [int(__) for __ in input().split()]

        delta_t = next_t - t
        distance = abs(next_x - x) + abs(next_y - y)

        if distance > delta_t:
            print("No")
            sys.exit()

        if (delta_t - distance) % 2 != 0:
            print("No")
            sys.exit()

        t, x, y = next_t, next_x, next_y

    print("Yes")

main()

#Impressions As for how to solve mathematics, I thought it would be a good idea for people who are not afraid to try to find the law by doing math firmly at school. Mathematically, I feel that I don't use it much because I feel that I have too much thought in competitive programming. #What I can do 1.I learned the basic idea of AtCoder 1.Ability to translate from esoteric problem sentences #Task 1.Try to solve a similar problem from the reference link 1.Challenge Leet Code #References

[GO] [Python] Solve 10 past elite problems of Atcoder