Codeforces Round 410 Div2 Problem C

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Mike and gcd problem

Solution sketch

這個講解比官方解析棒很多!

大體而言，全部都是偶數，一定ok。其餘，觀察兩奇數需要一次操作即可變成偶數偶數，一奇數一偶數需要兩次操作。

注意 gcd(all number) != 1 這個特判不能省下 (Wrong answer on test 40)。

AC code

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#include <bits/stdc++.h>
using namespace std;
void change(int &a, int &b)
{
    int tmp = b;
    b = a + b;
    a = a - tmp;
}
int gcd(int a, int b)
{
    return a == 0 ? b : gcd(b % a, a);
}
int main()
{
    int n;
    scanf("%d", &n);
    vector<int> inp;
    int g = 0; // default value can't be 1, come on bro
    for (int i = 0; i < n; i++) {
        int tmp;
        scanf("%d", &tmp);
        inp.push_back(tmp);
        g = gcd(g, tmp);
    }
    int cnt = 0;
    for (int i = 1; i < n; i++) {
        if ((inp[i - 1] % 2 == 1) && (inp[i] % 2 == 1)) {
            change(inp[i - 1], inp[i]);
            cnt++;
        }
    }
    for (int i = 1; i < n; i++) {
        if ((inp[i - 1] % 2 == 1) || (inp[i] % 2 == 1)) {
            change(inp[i - 1], inp[i]);
            change(inp[i - 1], inp[i]);
            cnt += 2;
        }
    }
    printf("YES\n");
    printf("%d\n", g > 1 ? 0 : cnt);
    return 0;
}