1 #include <stdio.h>
2 #include <string.h>
3 #include"vector"
4 #include"iostream"
5 #include <algorithm>
6 #include"bits/stdc++.h"
7 using namespace std;
8 #include"map"
9 #define ll long long
10 #define int ll
11
12 int a,b,p;
13 int k,g;
14 vector<int >v;
15 int phi;
16 vector<int > ans;
17
18 int ksm(int a,int b,int mod)
19 {
20 int ans = 1;
21 for(;b;b>>=1,a*=a,a%=mod)if(b&1)ans*=a,ans%=mod;
22 return ans;
23 }
24
25 void fenjie(int x)
26 {
27 for(int i=2;i*i<=x;i++)
28 {
29 if(x%i==0)v.push_back(i);
30 while(x%i==0)x/=i;
31 }
32 if(x>1) v.push_back(x);
33 }
34
35 int find_root()
36 {
37 for(int i=2;i;i++)
38 {
39 int f=1;
40 for(auto j:v)
41 {
42 if(ksm(i,phi/j,p)==1)
43 {
44 f=0;break;
45 }
46 }
47 if(f==1)return i;
48 }
49 }
50
51 int bsgs(int a,int b,int p)
52 {
53 b%=p;
54 int m=ceil(sqrt(p));
55 map<int ,int >mp;
56 int k=b;
57 for(int i=0;i<=m;i++)
58 {
59 mp[k]=i;
60 k=k*a%p;
61 }
62 int kt=ksm(a,m,p) ; int now = 1;
63
64 for(int i=1;i<=m;i++) // 从1开始，否则会出现负数解
65 {
66 now=now*kt%p;
67
68 if(mp[now])
69 return i*m-mp[now];
70 }
71 return -1;
72 }
73 int exgcd(int a,int b,int &x,int &y)
74 {
75 if(!b)x=1,y=0;
76 else
77 {
78 exgcd(b,a%b,y,x),y-=a/b*x;
79 }
80 }
81
82 void solve(int a,int b,int p)
83 {
84 // cout<<a<<" "<<b<<" " <<p<<endl;
85 int x,y;
86 int d=__gcd(a,p);
87 if(b%d!=0)return ;
88
89 int t=b/d;
90 exgcd(a,p,x,y);
91 x%=p,x+=p,x%=p;
92 x*=t;
93 //cout<<x<<" "<<b/d<<" "<<p/d<<" "<<d<<endl;
94 int delta=p/d;
95 while(x<p)ans.push_back(ksm(g,x,p+1)),x=(x+delta); // 这里要mod p+1， 因为这个函数里的p为phi（p）
96
97
98 }
99 signed main()
100 {
101 cin>>p>>k>>b; phi=p-1;
102 fenjie(phi);
103 g=find_root();
104 int gb=bsgs(g,b,p);
105
106 solve(k,gb,phi);
107
108
109 sort(ans.begin(),ans.end());
110 ans.erase(unique(ans.begin(),ans.end()),ans.end());
111 cout<<ans.size()<<endl;
112 for(auto i:ans)cout<<i<<" ";
113
114
115
116
117 }
In this problem, we try to compute discrete kth root modulo n; given n, k, a; find all the solutions for x such that xk = a (mod n) and x is coprime with n.
Input
For each input file, there are 3 space seperated integers n, k, a.
n = pe for some odd prime p, integer e > 0; 0 <= a < n <= 109, 0 <= k < phi(n), where phi is Euler's totient function; the numbers n, a are coprimes.
Output
The first line of the output contains a single integer m, the number of solutions in the range [0, n - 1] that are coprimes with n, followed by m lines that contain the m solutions in ascending order. It is guranteed that m <= 104.
Example
Input:
5 1 3
Output:
1
3
这题不错，综合了很多的知识
题解网上一大堆，我就不写了 嘿嘿