牛客多校(2020第三场)F Fraction Construction Problem

题目链接：https://ac.nowcoder.com/acm/contest/5668/F

题意：

给定一个分数a/b，请构造俩个分数c/d, e/f满足 c/d - e/f = a/b,且 d < b && f < b

题解:

1. a和b不互质， 设g = gcd(a, b),则 d = f = b / g, c = a/g + 1, e = a/g；

2. a和b互质， 且b的相异质因子不超过1个， b无法符合题意分解，无解。

3. a和b互质， 且b的相异质因子超过1个， 先找到d * f = b， 且 d, f互质， 然后的到 c * f - e * d = a， 使用扩展欧几里得算法。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define ll long long
const int N = 2e6 + 5;

int v[N], prime[N], num_prime[N], vis[N];

void init() {
    memset(v, 0, sizeof(v));
    int m = 0;
    for (int i = 2; i <= sqrt(N); i++) {
        if (!vis[i]) {
            v[i] = i; //没有访问过的，最小质因子就是自己
            for (int j = i*i; j < N; j+=i) { //寻找最小质因子是i的数
                vis[j] = 1;
                if (!v[j])  v[j] = i;
            }
        }
    }

    memset(num_prime, 0, sizeof(num_prime));
    num_prime[1] = 1;
    for (int i = 2; i <= N; i++) {
        if (v[i]) {
            num_prime[i]++;
            int temp = i;
            while (temp % v[i] == 0) {temp /= v[i];}
            if (temp != 1) {
                num_prime[i]++;
            }
        }
    }
}

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

ll gcd(int a, int b) {
    return b? gcd(b, a%b): a;
}

int main() {
    init();
    int t;
    scanf("%d", &t);
    while (t--) {
        int a, b;
        scanf("%d %d", &a, &b);
        int g = gcd(a,b);
        if (g != 1) {
            a = a/g;
            b = b/g;
            printf ("%d %d %d %d
", a+1, b, 1, b);
            continue;
        }
        if (num_prime[b] <= 1)  {printf("-1 -1 -1 -1
"); continue;}
        ll d = 1, f = b;
        while (f % v[b] == 0) {
            f /= v[b];
            d *= v[b];
        }

        exgcd(f, d);
        y = -y;
        while (x < 0 || y < 0)  {x += d, y += f;}
        ll c =  x * a, e =  y * a;
        printf ("%lld %lld %lld %lld
",c,d,e,f);
    }
}