题解 [UVA12775]Gift Dilemma

Luogu link

注意到 $C/\gcd (A,B,C)\ge 200$，即 $C\ge 200$。
于是可以枚举 $z$，即 $Ax+By=P-Cz$。因为 $C\ge 200$，所以 $z$ 最多枚举 $5\times 10^5$ 次，复杂度足够。

接下来问题转化为 $Ax+By=K$ 的自然数解。使用 $\text{exgcd}$ 接触该方程一组特解 $x_0,y_0$，通解即为：
$$\{\begin{array}{l}x=x_0+kB/\gcd (A,B)\\ y=y_0-kA/\gcd (A,B)\end{array}$$
此时 $k$ 的个数即为答案。因为 $x,y\ge 0$，所以解不等式，得：
$$\lceil \frac{-x_{0}d}{b}\rceil \le k\le \lfloor \frac{y_{0}d}{a}\rfloor$$

//UVA12775
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;

ll gcd(ll a, ll b){ return b ? gcd(b,a%b) : a; }
void exgcd(ll a, ll b, ll &x, ll &y){
	if(b == 0){ x = 1, y = 0; return; }
	exgcd(b, a%b, x, y);
	int z = y; y = x - (a / b) * y; x = z;
}

int main(){
//	freopen("_.txt", "w", stdout);
	int asdf; scanf("%d", &asdf);
	for(int ghjk = 1; ghjk <= asdf; ++ ghjk){
		ll a, b, c, p; scanf("%lld%lld%lld%lld", &a, &b, &c, &p);
		ll x = 0, y = 0, d = gcd(a,b), ans = 0; exgcd(a, b, x, y); 
		for(ll z = 0; c * z <= p; ++ z){
			ll k = p - c * z, xx = x * k / d, yy = y * k / d;
			if(k%d) continue;
			ll l = ceil(-1.0*xx*d/b), r = floor(1.0*yy*d/a);
//			printf("%lld %lld %lld %lld %lld %lld %lld %lld\n", a, xx, b, yy, d, k, l, r);
			ans += r - l + 1;
		}
		printf("Case %d: %lld\n", ghjk, ans);
	}
	return 0;
}