Codeforces Round #591 (Div. 2, based on Technocup 2020 Elimination Round 1) A. CME

链接:

https://codeforces.com/contest/1241/problem/A

题意:

Let's denote correct match equation (we will denote it as CME) an equation $$(a + b = c)$$ there all integers $$(a)$$, $$(b)$$ and $$(c)$$ are greater than zero.

For example, equations $$(2 + 2 = 4)$$ (||+||=||||) and $$(1 + 2 = 3)$$ (|+||=|||) are CME but equations $$(1 + 2 = 4)$$ (|+||=||||), $$(2 + 2 = 3)$$ (||+||=|||), and $$(0 + 1 = 1)$$ (+|=|) are not.

Now, you have $$(n)$$ matches. You want to assemble a CME using all your matches. Unfortunately, it is possible that you can't assemble the CME using all matches. But you can buy some extra matches and then assemble CME!

For example, if $$(n = 2)$$, you can buy two matches and assemble |+|=||, and if $$(n = 5)$$ you can buy one match and assemble ||+|=|||.

Calculate the minimum number of matches which you have to buy for assembling CME.

Note, that you have to answer $$(q)$$ independent queries.

思路:

超过2的偶数可以拆成1 1 2的比例, 奇数加一个也可以满足.

代码:

#include <bits/stdc++.h>
using namespace std;

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        if (n == 2)
            cout << 2 << endl;
        else if (n%2 == 0)
            cout << 0 << endl;
        else
            cout << 1 << endl;
    }

    return 0;
}