乱写了一发然后过了

做法就是令 $c = 1$ 然后枚举 $a$，所以 $b = n - a - b$，直接判断是否满足 $\gcd(a, b) = 1$，满足就输出然后结束。

void solve() {
  int n;
  iocin >> n;
  rep (i, 1, n) {
    int a, b, c;
    for (a = 2; a <= n; ++a) {
      c = 1, b = n - a - 1;
      if (a >= 1 && b >= 1 && c >= 1) {
        if (__gcd(a, b) == c) {
          cout << a << ' ' << b << ' ' << c << endl;
          goto exit;       
        }
      }
    }
  }
  exit:;
}