Victor, a top coder, loves the Internet. Every day he registers on many new websites, and every
time he is asked to think of a new password. Victor wrote a password generator, which produces
a new password when given a prime number. All he has to do now is to learn how to generate
various prime numbers.
Victor has a favorite number q. To obtain the prime numbers, he decided to use prime divisors
of q
n − 1, increasing the exponent n daily. However, to get rid of duplicate prime numbers, he
chooses on the n-th day only those numbers which he has not encountered earlier, i.e. those that
were not the prime divisors of q
m − 1 for each m < n.
Help our coder to write a program that finds acceptable prime divisors for the given q and n.
Since Victor’s password generator only accepts relatively small numbers, he needs prime numbers
not greater than 107
.
Input
The only line of the input file contain two integers q and n (2 ≤ q ≤ 109
, 1 ≤ n ≤ 109
).
Output
The first line of the output file must contain the number of the desired divisors of the number q
n−1.
The second line must contain these divisors in ascending order, separated by space characters. All
these numbers must be prime, not greater than 107 and must not occur for the lower values of n.
Example
input.txt output.txt
3 4 1
5
2 6 0
Example explanation
In the first example q
n − 1 = 80 = 24
· 5. The numbers 2 and 5 are thus prime divisors. However,
2 divides 3
1 − 1, and 5 does not divide the numbers 3
1 − 1, 3
2 − 1 and 3
3 − 1, so Victor can take
only one prime divisor — the number 5. In the second example, 2
6 − 1 = 63 = 32
· 7. No prime
divisor fits, because 3 divides 2
2 − 1, and 7 divides 2
3 − 1.
思路： 题目说 素数 <= 1e7 ，那么我们枚举每一个素数，看是否满足 q^n mod p ==1, 如果满足，那么我们反解这个同余方程
由引理：
若 a,n 互质，那么满足 a^x ≡ 1 ( mod n ) 最小正整数 x0 为 φ ( n ) 的因数，
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 20000005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const ll mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-5
#define pll pair<ll,ll>
#define lson 2*x
#define rson 2*x+1
//const ll maxn= 1005;
long long qupower(ll a, ll b,ll mod) {
long long ans = 1;
while (b) {
if (b & 1)ans = ans * a%mod;
b >>= 1;
a = a * a%mod;
}
return ans;
}
inline int read() {
int an = 0, x = 1; char c = getchar();
while (c > '9' || c < '0') {
if (c == '-') {
x = -1;
}
c = getchar();
}
while (c >= '0'&&c <= '9') {
an = an * 10 + c - '0'; c = getchar();
}
return an * x;
}
const int N = 10000000;
int v[maxn], prime[maxn];
int res[maxn];
int tot;
int m;
int n, q;
bool judge(int x) {
if (x <= n)return false;
if (qupower(q, n, x) != 1)return false;
int cur = x - 1;
int tmp = x - 1;
while (tmp > 1) {
int sy = v[tmp];
if (qupower(q, cur / sy, x) == 1)cur = cur / sy;
if (n > cur)return false;
tmp = tmp / sy;
}
return true;
}
int main() {
ios::sync_with_stdio(false);
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
cin >> q >> n;
for (int i = 2; i < N; i++) {
if (!v[i]) {
v[i] = i;
prime[tot++] = i;
}
for (int j = 0; j < tot&&i*prime[j] < N; j++) {
v[i*prime[j]] = prime[j];
if (i%prime[j] == 0)break;
}
}
int cnt = 0;
for (int i = 0; i < tot; i++) {
if (judge(prime[i]))res[++cnt] = prime[i];
}
cout << cnt << endl;
for (int i = 1; i <= cnt; i++)cout << res[i] << ' ';
}