BZOJ2187：fraction

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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;
}