公式

B - Poisonous Oyster 解説 by en_translator


If you are new to learning programming and do not know where to start, please try Problem A “Welcome to AtCoder” from practice contest. There you can find a sample code for each language.
Also, if you are not familiar with problems in programming contests, we recommend you to try some problems in “AtCoder Beginners Selection” (https://atcoder.jp/contests/abs).


The answer is as follows:

  • If both Takahashi and Aoki got sick:
    • The answer is \(1\), which is contained in both \(\{1,2\}\) and \(\{1,3\}\).
  • If Takahashi got sick but Aoki did not:
    • The answer is \(2\), which is contained in \(\{1,2\}\) but not in \(\{1,3\}\).
  • If Aoki got sick but Takahashi did not:
    • The answer is \(2\), which is not contained in \(\{1,2\}\) but in \(\{1,3\}\).
  • If both Takahashi and Aoki did not get sick:
    • The answer is \(4\), which is contained in neither \(\{1,2\}\) nor \(\{1,3\}\).

Thus, we can use if statements to implement the casework. For more details on implementation, please refer to the sample code below (C++ and Python).

Sample code (C++):

#include <bits/stdc++.h>

using namespace std;

int main() {
    string s1, s2;
    cin >> s1 >> s2;
    if (s1 == "sick") {
        if (s2 == "sick") cout << 1 << endl;
        else cout << 2 << endl;
    } else {
        if (s2 == "sick") cout << 3 << endl;
        else cout << 4 << endl;
    }
}

Sample code (Python) :

s1, s2 = input().split()

if s1 == "sick":
  if s2 == "sick":
    print(1)
  else:
    print(2)
else:
  if s2 == "sick":
    print(3)
  else:
    print(4)

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