【BZOJ2242】【SDoi2011】计算器 快速幂+EXGCD+BSGS

 1  
 2 #include <iostream>
 3 #include <cstdio>
 4 #include <cmath>
 5 #include <map>
 6 #define ll long long 
 7 using namespace std;
 8 ll y,z,p;
 9  
10 ll fast_pow(ll y,ll z,ll p)
11 {
12     ll ans=1;
13     while (z)
14     {
15         if (z&1)    ans=ans*y%p;
16         y=y*y%p;
17         z>>=1;
18     }
19     return ans;
20 }
21  
22 ll gcd(ll a,ll b)
23 {return b==0?a:gcd(b,a%b);}
24  
25 void ex_gcd(ll a,ll b,ll &x,ll &y)
26 {
27     if (!b) {x=1;y=0;return;}
28     ex_gcd(b,a%b,x,y);
29     ll t=x; x=y; y=t-a/b*y; 
30 }
31  
32 void solve1()
33 {
34     printf("%lld\n",fast_pow(y,z,p));
35 }
36  
37 void solve2()
38 {
39     ll d=gcd(y,p);
40     if (z%d)    {printf("Orz, I cannot find x!\n");return;}
41     y/=d; z/=d;
42     ll a,b;
43     ex_gcd(y,p,a,b);
44     a=a*z%p;
45     while (a<0)  a+=p;
46     printf("%lld\n",a);
47 }
48  
49 void solve3()
50 {
51     y%=p,z%=p;
52     if (!y && !z)   {puts("1"); return;}
53     if (!y) {printf("Orz, I cannot find x!\n");return;}
54     map<ll,ll> mp;
55     mp.clear();
56     ll m=ceil(sqrt(p));
57     ll t=fast_pow(y,m,p),k=z%p;//直接m即可 
58     for (int i=0;i<m;i++)
59     {
60         if (!mp[k]) if (i) mp[k]=i;//注意变量，及时订正。 
61                     else mp[k]=-1;
62         k=k*y%p;
63     }
64     k=1;
65     for (int i=0;i<m;i++)
66     {
67         if (mp[k])//注意！mp与mp的判断 
68         {
69             if (mp[k]==-1)  mp[k]=0;
70             printf("%lld\n",i*m-mp[k]);
71             return;
72         }
73         k=k*t%p;
74     }
75     printf("Orz, I cannot find x!\n");//各种傻 
76 }
77  
78 int main()
79 {
80     int T,t;
81     scanf("%d%d",&T,&t);
82     while (T--)
83     {
84         scanf("%lld%lld%lld",&y,&z,&p);
85         if (t==1)   solve1();
86         if (t==2)   solve2();
87         if (t==3)   solve3();
88     }
89 }