方程的解

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<iostream>
 4 #include<cmath>
 5 #define xian 65535
 6 #define ll long long
 7 using namespace std;
 8 inline ll read(){
 9     ll s=0,w=1;char ch=getchar();
10     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)w=-1;ch=getchar();}
11     while(ch>=‘0‘&&ch<=‘9‘)s=s*10+ch-‘0‘,ch=getchar();
12     return s*w;
13 }
14 
15 ll T;
16 ll a,b,c;
17 ll ltmp;
18 ll exgcd(ll a,ll b,ll &x,ll &y){
19     if(!b){
20         x=1;y=0;
21         return a;
22     }
23     ll lgcd=exgcd(b,a%b,x,y);
24     ltmp=x;
25     x=y;
26     y=ltmp-a/b*y;
27     return lgcd;
28 }
29 
30 int main(){
31 //    freopen("da.in","r",stdin);
32 //    freopen("te.out","w",stdout);
33     T=read();
34     while(T){
35         --T;
36         a=read();b=read();c=read();
37         //cout<<c<<endl;
38 
39         if(a==0&&b==0){
40             if(!c){puts("ZenMeZheMeDuo");continue; }
41             else{puts("0");continue; }
42         }
43         if(!a){
44             if(c*b<0){puts("0");continue; }
45             else{
46                 if(!c){puts("0");continue; }
47                 if(c%b){puts("0");continue; }
48                 else{puts("ZenMeZheMeDuo");continue; }
49             }
50         }
51         if(!b){
52             if(c*a<0){puts("0");continue; }
53             else{
54                 if(!c){puts("0");continue; }
55                 if(c%a){puts("0");continue; }
56                 else{puts("ZenMeZheMeDuo");continue; }
57             }
58         }
59 
60         if(a+b==c){puts("1");continue; }
61         if(a<0)a*=-1,b*=-1,c*=-1;
62         if(b>0&&c<0){puts("0");continue; }
63         if(b>0&&c>0){
64             if(a==1&&b==1){
65                 if(c-1>xian)puts("ZenMeZheMeDuo");
66                 else printf("%lld\n",(c-1));
67                 continue;
68             }
69         }
70 
71         ll x,y;
72         ll lin=exgcd(a,b,x,y);
73 
74         if(c%lin){puts("0");continue; }
75         if(a*b<0){puts("ZenMeZheMeDuo");continue; }
76 
77         ll xi=c/lin;
78         x*=xi;y*=xi;
79 
80         if(x==0&&y==0){puts("0");continue; }
81         if(x<0&&y<0){puts("0");continue; }
82 
83         ll jx=b/lin,jy=a/lin;
84         ll ans=0;
85 
86         if(x<=0)ans=ceil(y*1.0/jy)-1-(-1*x/jx);
87         else if(y>=0)ans=ceil(x*1.0/jx)-1-(-1*y/jy);
88         else ans=ceil(x*1.0/jx)-1+ceil(1.0*y/jy);
89 
90         if(ans<=0){puts("0");continue; }
91         if(ans>xian){puts("ZenMeZheMeDuo");continue; }
92 
93         printf("%lld\n",ans);
94     }
95     return 0;
96 }