拓展欧几里得求 ax + by = c的通解（a >=0, b >= 0）

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>

using namespace std;

#define ll long long

// 题目：给定三种物品的价格A,B,C和拥有的钱P(C / gcd(A, B, C) >= 200)
// 求解 AX + BY + CZ = P的解个数(case = 100)
// A, B, C, P ∈ [0, 100000000]

// 解：
// AX + BY = P - CZ (C >= 200 -> Z <= 1e8 / 200 = 5e5)
// 复杂度O(case * 5e5 * log(1e8))

const int INF = 1e9;
const ll MOD = 19940417;
ll a, b, c, p, x, y, gcd_ab;

ll exgcd(ll a, ll b, ll &x, ll &y)
{

    if(b == 0){//推理1，终止条件
        x = 1;
        y = 0;
        return a;
    }
    ll r = exgcd(b, a%b, x, y);
    //先得到更底层的x2,y2,再根据计算好的x2,y2计算x1，y1。
    //推理2，递推关系
    ll t = y;
    y = x - (a/b) * y;
    x = t;
    return r;
}

ll fun(ll n)
{
    if(n % gcd_ab) return 0;
    ll tim = n / gcd_ab;
    ll xx, yy;
    // a * (tim * x) + b * (tim * y) = gcd(a, b) * tim = n;
    xx = x * tim;
    yy = y * tim;
    ll k, k1, k2;
    //a * (xx + F) + b * (yy + G) = n
    // Fa + Gb = 0
    // F = lcm(a, b) / a * t
    // G = lcm(a, b) / g * t
    k1 = b / gcd_ab;
    k2 = a / gcd_ab;

    if(xx < 0){
        k = -(xx / k1) + (xx % k1 != 0);
        xx += k * k1; yy -= k * k2;
    }
    if(yy < 0){
        k = -(yy / k2) + (yy % k2 != 0);
        xx -= k * k1; yy += k * k2;
    }

    //cout << "aa = " << aa << " bb = " << bb << endl;
    ll x1, x2, y1, y2;
    if(xx < 0 || yy < 0) return 0;
    k = xx / k1;
    xx -= k * k1;
    yy += k * k2;
    x1 = xx, y1 = yy;
    k = yy / k2;
    xx += k * k1;
    yy -= k * k2;
    x2 = xx, y2 = yy;
    y2 = y2 - k * k2;
    //x1 + (x2 - x1) / k1 = x2
    return (x2 - x1) / k1 + 1;
}

void solve()
{

    int T, _case = 0;
    cin >> T;
    while(T--){
        
        cin >> a >> b >> c >> p;
        //ax + by = gcd(a, b)
        gcd_ab = exgcd(a, b, x, y);
        //cout << x << " " << y << endl;
        ll ways = 0;
        for(ll i = 0; p - c * i >= 0; ++i){
            ways += fun(p - c * i);
            //cout << "ways = " << ways << endl;
        }

        cout << "Case " << ++_case << ": " << ways << endl;
    }
}

int main()
{

    solve();

    //cout << "ok" << endl;

    return 0;
}