思路：題意轉化為求 (ax+by=dis) || (ax+cy=dis) || (bx+cy=dis) 三個式子有解時的最小|x| + |y|。顯然求解特解x，y直接用擴展歐幾裏得，那麽怎麽求|x| + |y|？xy關系為一條直線，那麽|x| + |y|應該是在x取0或者y取0的時候，但是要整數，所以只能在周圍取幾個點。我們知道x=x1+b/gcd*t，那麽x1+b/gcd*t = 0可以解得 t = -x1 * gcd / b。然後在附近取幾個點。

代碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include
 
<cmath>
#include<algorithm>
#define LS(n) node[(n)].ch[0]
#define RS(n) node[(n)].ch[1]
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int maxn = 32767 + 10;

ll ex_gcd(ll a, ll b, ll &x, ll &y){
    ll d, t;
    if(b == 0){
        x = 1;
        y = 0;
        
 
return a;
    }
    d = ex_gcd(b, a%b, x, y);
    t = x-a/b*y;
    x = y;
    y = t;
    return d;
}

ll dis;
//求|x|+|y|最小
ll solve(ll a, ll b){
    ll x, y, d = ex_gcd(a, b, x, y);
    if(dis % d != 0) return INF;
    x = x * dis / d;
    y = y * dis / d;
    a /= d, b /= d;
    ll ans = abs(x) + abs(y);

    ll k;
    k 
 
= -x / b - 5;
    for(int i = 0; i <= 10; i++){
        ans = min(ans, abs(x + b * (k + i)) + abs(y - a * (k + i)));
    }

    k = y / a - 5;
    for(int i = 0; i <= 10; i++){
        ans = min(ans, abs(x + b * (k + i)) + abs(y - a * (k + i)));
    }

    return ans;
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        ll a, b, A, B;
        scanf("%lld%lld%lld%lld", &A, &B, &a, &b);
        dis = abs(A - B);
        ll ans;
        ans = min(solve(a, a + b), min(solve(a, b), solve(b, a + b)));
        printf("%lld\n", ans == INF? -1 : ans);
    }
    return 0;
}

ZOJ 3593 One Person Game（ExGcd + 最優解）題解