Topic encryption

First of all, we can think of using exgcd when we see this problem, but how to use it? We know that as soon as this equation has a bunch of solutions, then the limit of this problem is the number of solutions. We think that if we simplify the equation, we can get $y=\frac{c}{b}-\frac{a}{b}*x $, and then we will use the knowledge of compulsory education. When a and b are different, they must be the same. If exgcd has solutions, they must be the same. It's an infinite group. It's a direct verdict. If we consider a b's identity sign, then we can get a decreasing straight line that must pass through the first quadrant. We do the following operations on the equation: $a(x-b)+b(y+a)=c $. Obviously, it is also true. At this time, ab can be transformed into a positive sign. If x decreases, then y increases. We can get the maximum solution of Y by moduling x to b, then get the minimum solution of Y by moduling y to a, and then the transformation of y must be a sequence of equal difference numbers. If the tolerance is a, then the range of y can be found and the proposition can be solved.

And then I'll give you A special judgment. Read it and remember to type A minus sign.

#include<iostream>
#include<cstdio>
#define ll long long
using namespace std;
ll rd()
{
    ll s=0,w=1;
    char cc=getchar();
    while(cc<'0'||cc>'9') {if(cc=='-') w=-1;cc=getchar();}
    while(cc>='0'&&cc<='9') s=(s<<3)+(s<<1)+cc-'0',cc=getchar();
    return s*w;
}
ll exgcd(ll a,ll b,ll &x,ll &y,ll c)
{
    if(b==0){x=c/a;y=0;return a;}
    ll d=exgcd(b,a%b,y,x,c);
    y-=a/b*x;
    return d;
}
int main()
{
//    freopen("data.in","r",stdin);
    //freopen("data.ans","w",stdout);
    ll T=rd(),a,b,c,x,y,g,ans;
    while(T--)
    {
        a=rd(),b=rd(),c=rd();
        //cout<<a<<" "<<b<<endl;
        if(!a&&!b)
        {
            if(!c)printf("ZenMeZheMeDuo\n");
            else printf("0\n");
            continue;
        }
        if(a<0&&b<0) a=-a,b=-b,c=-c;
        g=exgcd(a,b,x,y,c);
        if(c%g){puts("0");continue;}
        if(a==0)
        {
            if((b<=0&&c>=0)||(b>=0&&c<=0))printf("0\n");
            else printf("ZenMeZheMeDuo\n");
            continue;
        }
        if(b==0)
        {
            if((a<=0&&c>=0)||(a>=0&&c<=0))printf("0\n");
            else printf("ZenMeZheMeDuo\n");
            continue;
        }
        if(a*b<0)
        {
            printf("ZenMeZheMeDuo\n");
            continue;
        }
        if(a<0&&b<0)a=-a,b=-b,c=-c;
        a/=g;b/=g;c/=g;x%=b;
        while(x<=0)x+=b;
        y=(c-a*x)/b;
        ll ymin=y%a;while(ymin<=0)ymin+=a;
        if(ymin>y)ans=0;
        else ans=(y-ymin)/a+1;
        if(ans>65535)printf("ZenMeZheMeDuo\n");
        else printf("%lld\n",ans);
    }
}
/*
g++ 6.cpp -o 6
./6
4
1 -1 3
1 1 65536
1 1 65537
2 3 24
*/