Sumitomo Mitsui Trust Bank Programming Contest 2019

https://atcoder.jp/contests/sumitrust2019

I found a website that help promotes logical thinking , there will be a period of competition , you can register get start for the competition and get points , Let's take a look at recent matches .

---

A - November 30

Solution

2019-M2-D2 follows 2019-M1-D1 so we just check M1 & M2 are they equal

m1, d1 = map(int, input().split())
m2, d2 = map(int, input().split())
 
if m1 == m2:
 print(0)
else:
 print(1)

---

B - Tax Rate

Solution

this question the point is rounded down to the nearest integer , math.ceil(N / 1.08) is before tax  ,  round down math.ceil(N / 1.08) times 1.08 to an integer if result equal N Takahashi's memory is true  print(result) else false print(":(")

import math
N = int(input())
 
B = math.ceil(N / 1.08)
if math.floor(B * 1.08) == N:
  print(B)
else:
  print(':(')

---

C - 100 to 105

Solution

if one items cost range is 100~105 , two items is 200 ~210 , three items is 300~315 , and so on ,
each item plus 5 tens digit , so we can get the rules

x = int(input())
count = x//100
res = x%100
if count*5 >= res:
    print(1)
else:
    print(0)

---

D - Lucky PIN

Solution

Use three loops to get three unique values. Loops meet the condition index value +1 to find the value of the next position

N = int(input())
S = input()
cnt = 0

 
for i in range(10):
    for j in range(10):
        for k in range(10):
            idx = S.find(str(k))
            if idx != -1:
                idx = S.find(str(j),idx+1)
                if idx != -1:
                    idx = S.find(str(i),idx+1)
                    if idx != -1:
                        cnt += 1
                        print(k,j,i)
 
print(cnt)

to be continued....