F-Fraction Construction Problem 2020牛客暑期多校训练营（第三场）（构造，数论，exgcd）
思路：
分一下情况来讨论：
1、(gcd(a,b)>1)：
令(k=gcd(a,b))，则(frac{a+k}{b}-frac{k}{b}=frac{a/k+1}{b/k}-frac{1}{b/k}=frac{a/k}{b/k}=frac{a}{b})
(b/k<b)，所以构造上式即可满足条件。
2、(gcd(a,b)=1)，且(mathit b)至少有2个质因子。
设(b=p_1^{x_1}*p_2^{x_2}dots p_m^{x_m})（唯一分解定律），令(d=p_1^{x_1},f=b/d)，
则(d*f=b)，
将题目中的公式左侧进行通分：
(frac{c*f-e*d}{d*f}=frac{a}{b})，因为(d*f=b)，所以等式转化为(c*f-e*d=a)
又因为(gcd(f,d)=1)（原因可以看他们的构造方法，它们没有相同的质因子。）所以上面的方程一定有解（裴蜀定理 ）。
然后用(exgcd)解上面的不定方程即可。
3、(gcd(a,b)=1,b=p^x)的情况（即b=1或者等于某个质数的幂次方）
这种情况一定是无解的，证明如下：
看这个方程：
(frac{c*f-e*d}{d*f}=frac{a}{b})
因为(gcd(a,b)=1)，所以如果等式成立，要有((d*f)\%b=0)，
设(d*f=p^y,f<d)，则：
(frac{c*f-e*d}{d*f}=frac{f*(c-e*d/f)}{d*f}=frac{(c-e*d/f)}{d})
原方程转为：(frac{(c-e*d/f)}{d}=frac{a}{b})
又因为(d<b)，所以方程一定无解。
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), ' ', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define chu(x) if(DEBUG_Switch) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
#define du3(a,b,c) scanf("%d %d %d",&(a),&(b),&(c))
#define du2(a,b) scanf("%d %d",&(a),&(b))
#define du1(a) scanf("%d",&(a));
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("
");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '
' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '
' : ' ');}}
const int maxn = 2000010;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
#define DEBUG_Switch 0
long long exgcd(long long a, long long b, long long &x, long long &y)
{
long long d = a;
if (b)
{
d = exgcd(b, a % b, y, x);
y -= x * (a / b);
}
else
{
x = 1;
y = 0;
}
return d;
}
bool noprime[maxn + 50];
vector <int> p;
int getPrime()
{
// 华丽的初始化
memset(noprime, false, sizeof(noprime));
p.clear();
int m = (int)sqrt(maxn + 0.5);
// 优化的埃筛
for (int i = 2; i <= m; i++)
{
if (!noprime[i])
{
for (int j = i * i; j <= maxn; j += i)
{
noprime[j] = true;
}
}
}
// 把素数加到vector里
for (int i = 2; i <= maxn; i++)
{
if (!noprime[i])
{
p.push_back(i);
}
}
//返回vector的大小
return p.size();
}
int main()
{
#if DEBUG_Switch
freopen("C:\code\input.txt", "r", stdin);
#endif
//freopen("C:\code\output.txt","w",stdout);
int len = getPrime();
int t;
t = readint();
while (t--)
{
ll a = readint();
ll b = readint();
ll g = gcd(a, b);
if (g != 1)
{
printf("%lld %lld %lld %lld
", a / g + 1, b / g, 1ll, b / g );
} else
{
ll n = b;
ll fac1 = 1ll;
for (int i = 0; 1ll * p[i]*p[i] <= n && i < len; i++)
{
if ( n % p[i] == 0)
{
while (n % p[i] == 0)
{
fac1 *= p[i];
n /= p[i];
}
break;
}
}
if (n > 1 && fac1 > 1)
{
ll d = fac1, f = n;
ll c, e;
exgcd(f, d, c, e);
c *= a;
e *= a;
if (c > 0 && e < 0)
{
printf("%lld %lld %lld %lld
", c, d, -e, f );
} else
{
printf("%lld %lld %lld %lld
", e, f, -c, d );
}
} else
{
printf("-1 -1 -1 -1
");
}
}
}
return 0;
}