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 = ;
 ll a, b, c, p, x, y, gcd_ab;

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

     if(b == ){//推理1，终止条件
         x = ;
         y = ;
         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 ;
     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 < ){
         k = -(xx / k1) + (xx % k1 != );
         xx += k * k1; yy -= k * k2;
     }
     if(yy < ){
         k = -(yy / k2) + (yy % k2 != );
         xx -= k * k1; yy += k * k2;
     }

     //cout << "aa = " << aa << " bb = " << bb << endl;
     ll x1, x2, y1, y2;
     if(xx <  || yy < ) return ;
     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 + ;
 }

 void solve()
 {

     int T, _case = ;
     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 = ;
         for(ll i = ; p - c * i >= ; ++i){
             ways += fun(p - c * i);
             //cout << "ways = " << ways << endl;
         }

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

 int main()
 {

     solve();

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

     return ;
 }