Inequality - 2019.9.13 balance

Face questions

【Problem Description】

You have a balance (balance of the left and right sides can be put weight) the weight of \ (A, B \) ( \ (. 1 \ Le A, B \ 10000 Le \) ) Two weight. Let you find out the weight of a scheme called \ (c \) ( \ (1 \ Le c \ Le 50000 \) ) items, if a variety of programs, please output two kinds of weight requires a minimum sum of the number of Program.

[Input Format]

There are several lines of three numbers, \ (A, B, C \) .

Represented by 000 at the end.

[Output format]

Several lines of two numbers, representing each interrogation \ (A \) number and \ (B \) number

If no solution output no solution

[Sample input]

700 300 200
500 200 300
500 200 500
275 110 330
275 110 385
648 375 4002
3 1 10000
0 0 0

[Sample Output]

solution

Containing a maximum value of the absolute value function parameters.

program

#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
 
#define INF 0x3f3f3f3f
 
ll exgcd(ll a, ll b, ll& x, ll& y) {
    if (b) {
        ll xp, yp;
        ll d = exgcd(b, a % b, xp, yp);
        x = yp; y = xp - (a / b) * yp;
        return d;
    } else {
        x = 1; y = 0;
        return a;
    }
}

ll f(ll k, ll x, ll b, ll d, ll y, ll a) { 
    return abs(x + b * k / d) + abs(y - a * k / d);
}
 
int main() {
    ll a, b, c;
    while (cin >> a >> b >> c && a && b && c) {
        ll d, x, y;
        d = exgcd(a, b, x, y);
        if (c % d) {
            cout << "no solution" << endl;
            continue;
        }
        x *= c / d; y *= c / d;
        // assert(a * x + b * y == c);
        vector<ll> cand;
        cand.push_back(-d * x / b);
        cand.push_back(-d * x / b + 1);
        cand.push_back(-d * x / b - 1);
        cand.push_back(d * y / a);
        cand.push_back(d * y / a + 1);
        cand.push_back(d * y / a - 1);
        ll bval = INF, xf, yf;
        for (const int& k : cand) {
            if (f(k, x, b, d, y, a) < bval) {
                bval = f(k, x, b, d, y, a);
                xf = x + b * k / d;
                yf = y - a * k / d;
            }
        }
        cout << abs(xf) << " " << abs(yf) << endl;
    }
    return 0;
}