Sol

分情况讨论

1. \(\lfloor\frac{a}{b}\rfloor+1\le \lceil\frac{c}{d}\rceil-1\)
   
   直接取 \(q=1,p=\lfloor\frac{a}{b}\rfloor+1\)
2. \(a=0\)
   
   那么 \(q> \frac{pd}{c}\)
   
   直接取 \(p=1,q=\lfloor\frac{d}{c}\rfloor+1\)
3. \(a<b\) 且 \(c\le d\)
   
   那么递归处理 \(\frac{d}{c} < \frac{q}{p} < \frac{b}{a}\)
4. \(a\ge b\)
   
   那么递归处理 \(\frac{a~mod~b}{b} < \frac{p}{q}-\lfloor\frac{a}{b}\rfloor < \frac{c}{d}-\lfloor\frac{a}{b}\rfloor\)

形式上类似于欧几里得算法，把 \((a,b)\) 转化成 \((a~mod~b,b)\) 可能就是是类欧几里得算法的精髓

# include <bits/stdc++.h>

using namespace std;

typedef long long ll;

inline ll Gcd(ll x, ll y) {

	if (!x || !y) return x | y;

	return !y ? x : Gcd(y, x % y);

}

inline void Solve(ll a, ll b, ll c, ll d, ll &x, ll &y) {

	register ll g = Gcd(a, b), nx, ny;

	a /= g, b /= g, g = Gcd(c, d), c /= g, d /= g;

	nx = a / b + 1, ny = c / d + (c % d > 0) - 1;

	if (nx <= ny) x = nx, y = 1;

	else if (!a) x = 1, y = d / c + 1;

	else if (a < b && c <= d) Solve(d, c, b, a, y, x);

	else Solve(a % b, b, c - d * (a / b), d, x, y), x += y * (a / b);

}

ll a, b, c, d, x, y;

int main() {

	while (scanf("%lld%lld%lld%lld", &a, &b, &c, &d) != EOF) Solve(a, b, c, d, x, y), printf("%lld/%lld\n", x, y);

    return 0;

}