CodeForces - 1366D Two Divisors

题目链接

题目大意

  对于任意一个\(a_i, 1\leq i \leq n\)问是否存在两个\(a_i\)的因数\(d_1, d_2\)满足\(gcd(d_1+d_2, a_i)=1\)

解题思路

  如果\(gcd(x, y) = 1\)，那么\(gcd(x+y, x \times y) = 1\)。如果一个数不是质数，就把他的一个质因子当a，\(a_i / a\)当b，a和b正好满足条件。

代码

const int maxn = 1e7+10;
int u[maxn], p[maxn], a[maxn], b[maxn];
int main() {
    IOS;
    for (int i = 2; i<maxn; ++i) {
        if (!u[i]) u[i] = p[++p[0]] = i;
        for (int j = 1; i*p[j]<maxn; ++j) {
            u[i*p[j]] = p[j];
            if (i%p[j]==0) break;
        }
    }
    int n; cin >> n;
    for (int i = 1; i<=n; ++i) {
        int num; cin >> num;
        vector<int> tmp;
        int f = 1, t = num;
        for (int j = 1; 1LL*p[j]*p[j]<=num; ++j)
            if (num%p[j]==0) {
                while(num%p[j]==0) {
                    num /= p[j];
                    if (tmp.empty()) f *= p[j];
                }
                tmp.push_back(p[j]);
            }
        if (num>1) tmp.push_back(num);
        if (f!=1 && t/f!=1) a[i] = f, b[i] = t/f;
        else a[i] = b[i] = -1;
    }
    for (int i = 1; i<=n; ++i) cout << a[i] << (i==n ? '\n':' ');
    for (int i = 1; i<=n; ++i) cout << b[i] << (i==n ? '\n':' ');
    return 0;
}