Solve with Python [100 selected past questions that beginners and intermediates should solve] (001 --004 All search: All enumeration)

1. Purpose

Solve 100 past questions that beginners and intermediates should solve in Python. The goal is to turn light blue by the time you finish solving everything.

This article is "001 --004 All Search: All Enumeration".

2. Summary

In Python, you can enumerate combinations with itertools, but I wrote it in a for statement for practice. I didn't stumble in particular.

3. Main story

001 --004 All search: All enumeration

1. ITP1_7_B - How Many Ways?

Answer

answer_list = []
while True:
    n, x = map(int, input().split())
    if n == 0 and x == 0:
        break

    count = 0
    for first in range(1, n + 1):
        for second in range(first + 1, n + 1):
            for third in range(second + 1, n + 1):
                if first + second + third == x:
                    count += 1

    answer_list.append(count)

for answer in answer_list:
    print(answer)

Since the problem statement says that there is no duplication, be careful not to duplicate the range of `first```,` `second, and thirdof the for statement. Specifically, set the start of secondto first + 1``` and the start of `` third to` `second + 1.

2. AtCoder Beginner Contest 106 B - 105

Answer

def is_target(num):
    count = 0
    for i in range(1, num+1):
        if num % i == 0:
            count += 1
    if count == 8:
        return True
    else:
        return False

if __name__ == "__main__":
    N = int(input())
    count = 0
    for num in range(1, N+1, 2):
        count += is_target(num)

    print(count)

Create a function called `is_target``` that returns True``` and `` Falseto see if the divisor is 8. After that, check the odd numbers from 1 to N withis_target``` and add them together (True is 1 so you can add them as they are).

3. AtCoder Beginner Contest 122 B - ATCoder

Answer

target = 'ACGT'
S = input()

answer = 0
for start in range(len(S)):
    if S[start] not in target:
        continue
    count = 0
    for end in range(start, len(S)):
        if S[end] not in target:
            break
        count += 1
    
    answer = max(answer, count)

print(answer)

The substring to be extracted from the string S can be written as `S [start: end] ``` using the subscripts start``` and ```end```, so `` start For each of `` andend, check whether it is ACGT```.

When checking, the for statement of `start``` is continue``` if it is not ```ACGT```, and the for statement of ```end``` is ```ACGT Note that if it is not `` , it is` `` break```.

004. Pakencamp 2019 C-Karaoke

Answer

N, M = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(N)]

answer = 0
for song1 in range(M):
    for song2 in range(song1+1, M):
        score = 0
        for i in range(N):
            score += max(A[i][song1], A[i][song2])

        answer = max(answer, score)

print(answer)

song1Whensong2Turn the for statement about. Then, inside that, for each student (subscript i), the larger score of `` `song1and song2``` is adopted.