2020牛客暑期多校训练营（第三场）F.Fraction Construction Problem(数学)

F-Fraction Construction Problem

题意：给定a和b，构造$\frac{c}{d}-\frac{e}{f}=\frac{a}{b}$ 且 $d,f<b$
题解：
1.当ab不互质，即ab存在非1公因数时，可以构造出$\frac{a+x}{b}-\frac{x}{b}=\frac{a}{b}$ ,x任取，因为要求$d,f<b$，对等式左边分子分母同除ab的gcd即可（当然其他的非1公因数也可），设g为ab的gcd，构造的等式为$\frac{\frac{a}{g}+1}{\frac{b}{g}}-\frac{1}{\frac{b}{g}}=\frac{a}{b}$
2.当ab互质，但b的互异质因子只有一个时，则无解。
    证明：b的互异质因子只有一个，即$b=p^k$,上式通分后得到$\frac{cf-ed}{df}=\frac{a}{b}$，则$d^{\prime }=p^{k_1},f^{\prime }=p^{k_2}$，那么$d,f$一定包含因子$p^s$，$cf,ed$也一定包含因子$p^s$,则a一定包含因子$p^{ss}$，与ab互质矛盾

3.当ab互质，且b的互异质因子大于1个时，可令$df=b$，找到一组互质的$d,f$带入上式，$cf-ed=a=>fx-dy=gcd(f,d)\ast a$，拓展欧几里得找一组可行解即可，注意找到的$x,y$可能为负，换一下即可

当$b=1$或b为质数时，一定无解，可以用线性筛先将质数筛出来，提高效率。b的范围是2e6，我看成了1e6，没筛完，导致一直wa（自抠双目）

Code：

#include 
using namespace std;
#define ll long long
#define db double
#define ull unsigned ll
#define pii pair
#define pdd pair
#define pll pair
#define mem(a, b) memset(a, b, sizeof(a));
#define lowbit(x) (x & -x)
#define lrt nl, mid, rt << 1
#define rrt mid + 1, nr, rt << 1 | 1
template <typename T>
inline void read(T& t) {
    t = 0;
    int f = 1;
    char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (isdigit(ch)) {
        t = t * 10 + ch - '0';
        ch = getchar();
    }
    t *= f;
}
const int dx[] = {0, 1, 0, -1};
const int dy[] = {1, 0, -1, 0};
const ll Inf = 0x7f7f7f7f7f7f7f7f;
const int inf = 0x7f7f7f7f;
const db eps = 1e-5;
const db Pi = acos(-1);
const int maxn = 2e6 + 10;

int cnt;
bool is_prime[maxn];
int primes[maxn];

void Primes(int n) {
    memset(is_prime, true, sizeof(is_prime));
    is_prime[1] = 0;
    for (int i = 2; i <= n; i++) {
        if (is_prime[i])
            primes[++cnt] = i;
        for (int j = 1; j <= cnt && i * primes[j] <= n; j++) {
            is_prime[i * primes[j]] = 0;
            if (i % primes[j] == 0)
                break;
        }
    }
}

void exgcd(ll a, ll b, ll& x, ll& y) {
    if (!b) {
        y = 0, x = 1;
        return;
    }
    exgcd(b, a % b, y, x);
    y -= a / b * x;
}

int main(void) {
    Primes(2e6);
    int T;
    read(T);
    while (T--) {
        ll a, b, c, d, e, f, k;
        read(a), read(b);
        if (__gcd(a, b) != 1) {
            ll k = __gcd(a, b);
            printf("%lld %lld 1 %lld\n", a / k + 1, b / k, b / k);
            continue;
        }
        if (b == 1 || is_prime[b]) {
            printf("-1 -1 -1 -1\n");
            continue;
        }
        ll tmp = b, op = 1;
        for (int i = 1; primes[i] * primes[i] <= tmp; i++) {
            if (tmp % primes[i] == 0) {
                while (tmp % primes[i] == 0)
                    tmp /= primes[i], op *= primes[i];
                break;
            }
        }
        if (tmp == 1) {
            printf("-1 -1 -1 -1\n");
            continue;
        }
        d = tmp, f = op;
        ll x, y;
        exgcd(f, d, x, y);
        c = x * a, e = y * a;
        if (c < 0 && e > 0)
            printf("%lld %lld %lld %lld\n", e, f, -c, d);
        else
            printf("%lld %lld %lld %lld\n", c, d, -e, f);
    }
    return 0;
}