牛客国庆集训派对Day1 C Utawarerumono 扩展欧几里德

#include <stdio.h>
#include <algorithm>
#include <climits>
#include <iostream>
#define LL long long
using namespace std;
const int maxn=1e3+10;
#define in(x) scanf("%lld",&x)
#define inf 9223372036854775807LL
LL ex_gcd(LL a,LL b,LL&x,LL &y)
{
    if(a==0&&b==0) return -1;
    if(b==0){x=1;y=0;return a;}
    LL d=ex_gcd(b,a%b,x,y);
    LL t=x;
    x=y;
    y=t-a/b*y;
    return d;
}

LL a,b,c;
LL p1,p2,q1,q2;
LL x_,y_,g,n,m;
 LL gcd(LL a,LL b)
{
    return b==0? a:gcd(b,a%b);
}
LL getans(LL x,LL y)
{
    LL r=x*(p2*x+p1)+y*(q2*y+q1);
    return r;
}
void getmeow(LL &x,LL &y,LL t)
{
    x=x_+m*t;y=y_-n*t;
}
int solve1()
{
    int lim=2e3;
    if(a==0&&b==0)
    {
        if(c!=0) {
            printf("%d\n",5/a);
            return 0;
        }
        LL mx=p1/(p2*-2),my=q1/(q2*-2); //0x+0y==0
        LL ans=inf;
//        cout<<mx-lim<<" "<<mx+lim<<endl;
//        cout<<getans(-4,-4);
        for(LL x=mx-lim;x<mx+lim;x++)
            for(LL y=my-lim;y<my+lim;y++)
                ans=min(ans,getans(x,y));
        printf("%lld\n",ans);
        return 0;
    }
    lim=1e5;
    if(a==0||b==0)
    {
        if(a==0)  {
            if(c%b!=0)
            {
                puts("Kuon");return 0;
            }
            LL mx=p1/(p2*-2);LL my=c/b; //0x+by=c b!=0 c>=0
            LL ans=inf;
            for(LL x=mx-lim;x<mx+lim;x++)
                    ans=min(ans,getans(x,my));
            printf("%lld\n",ans);
            return 0;
        }
        if(b==0) {
            if(c%a!=0) {puts("Kuon");return 0;}
            LL mx=c/a,my=q1/(q2*-2); //ax+0y=c
            LL ans=inf;
                for(LL y=my-lim;y<my+lim;y++)
                    ans=min(ans,getans(mx,y));
            printf("%lld\n",ans);
            return 0;
        }
    }
    return 1;
}
void bruce()
{
    long long ans = inf;
    bool flag = false;
    for(int i=-100000; i<=100000; i++) {
        long long y = c - (i*a);
        if(abs(y) % b == 0) {
            flag = true;
            y = y/b;
            long long nt = p2*i*i + p1*i +q2*y*y + q1*y;
            if(nt < ans){
                ans = nt;
                cout << ans << " " <<i<< " " << y <<endl;
            }

        } else continue;
    }
    if(!flag) cout << "Kuon" << endl;
    else cout << ans<< endl;
}
int main()
{
    #ifdef shuaishuai
    freopen("in.txt","r",stdin);
    #endif // shuaishuai
    in(a);in(b);in(c);
    in(p1);in(p2);in(q1);in(q2);
    // a>0 b>0 c>=0
    if(!solve1()) return 0;
    g=ex_gcd(a,b,x_,y_);
    m=b/g;n=a/g;
    if(c%g!=0)
    {
         {puts("Kuon");return 0;}
    }
    x_*=c/g;y_*=c/g;
//    printf("%lld %lld %lld\n",x_,y_,a*x_+y_*b);
    LL ans=inf;
//    printf("initans:%lld\n",ans);
    int lim=1e5;
    LL x=x_,y=y_;int i=0; //两个一元二次函数的加和，求偏导数会发现其极值存在于第三象限 注意求出的特解应化为最小的正整数解 这里我wa到死
    LL mx=-1,mi=1;
    int id=0;
    while(x>-lim) i--,getmeow(x,y,i);
    while(x<-lim) i++,getmeow(x,y,i);
    while(x<=lim)
    {
        getmeow(x,y,i);
        i++; 
        if(getans(x,y)<ans) ans=getans(x,y),id=i;
    }
    printf("%lld\n",ans);
//    printf("myans:%lld %lld %d\n",x,y,id);
}