BZOJ2187：fraction

Sol

分情况讨论

1. (lfloorfrac{a}{b} floor+1le lceilfrac{c}{d} ceil-1)
   直接取 (q=1,p=lfloorfrac{a}{b} floor+1)
2. (a=0)
   那么 (q> frac{pd}{c})
   直接取 (p=1,q=lfloorfrac{d}{c} floor+1)
3. (a<b) 且 (cle d)
   那么递归处理 (frac{d}{c} < frac{q}{p} < frac{b}{a})
4. (age b)
   那么递归处理 (frac{a~mod~b}{b} < frac{p}{q}-lfloorfrac{a}{b} floor < frac{c}{d}-lfloorfrac{a}{b} floor)

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
", x, y);
    return 0;
}