JZOJ 5026.Number

首先有一个结论:当 \(n\) 有平方质因子时,除非 \(n=4\),否则有无数解。

一个并不严谨的证明:
\(n = \prod\limits_{i=1}^m p_i^{c_i},x = k\prod\limits_{i=1}^m p_i^{c_i+c'_i}\),其中 \(\gcd(n,k)=1\)
\(\sigma_0(x) = \sigma_0(k)\prod\limits_{i=1}(c_i+c'_i+1)\)
对于 \(1 \le i \le m\),考虑关于 \(c'_i\) 的方程 \(p_i^{c_i} = c_i + c'_i + 1\)
对于其解 \(c'_i = x_0\),设 \(d = \frac{c_i + x_0 + 1}{p_i}\)。注意到除非 \(c_i=1\)\(p_i=c_i=2\),一定仍存在 \(x'_0\) 满足 \(c_0 + x'_0 + 1 = d\),此时便可将一个 \(p_i\) 的贡献加入到 \(k\) 中,并且易知这样的 \(k\) 有无数种。
而若 \(p_i = c_i = 2\),易证除非 \(m=1\),否则亦有无数解。

首先 Pollard-Rho 分解质因子,然后容易发现 \(m \le 15\),考虑用一个简单的状压 DP 统计答案。

代码:

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#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <tr1/unordered_set>
using namespace std;
using namespace tr1;
const int CNT = 1e7;
const int M = 15;
unordered_set<long long> vis;
inline long long fmul(long long a,long long b,long long mod)
{
return (a * b - (long long)((long double)a / mod * b) * mod + mod) % mod;
}
inline long long fpow(long long a,long long b,long long mod)
{
long long ret = 1;
for(;b;b >>= 1)
(b & 1) && (ret = fmul(ret,a,mod)),a = fmul(a,a,mod);
return ret;
}
inline bool witness(long long x,long long p)
{
if(fpow(x,p - 1,p) ^ 1)
return 0;
long long k = p - 1;
while(!(k & 1))
{
k >>= 1;
long long t = fpow(x,k,p);
if((t ^ 1) && (t ^ p - 1))
return 0;
if(t == p - 1)
return 1;
}
return 1;
}
inline bool Miller_Rabin(long long n)
{
if(n == 1)
return 0;
const int base[] = {2,3,7,61,24251,19260817};
for(register int i = 0;i < 6;++i)
if(n == base[i])
return 1;
else if(!witness(base[i],n))
return 0;
return 1;
}
bool Pollard_Rho(long long n)
{
if(Miller_Rabin(n))
{
if(vis.count(n))
return 0;
vis.insert(n);
return 1;
}
const long long c = 1e9 + 9;
const int k = 200;
long long t1 = 1e9 + 7;
long long t2 = (fmul(t1,t1,n) + c) % n;
long long p = 1;
int i = 0;
while(t1 ^ t2)
{
++i,p = fmul(p,abs(t1 - t2),n);
if(!p)
{
long long d = __gcd(abs(t1 - t2),n);
return Pollard_Rho(d) && Pollard_Rho(n / d);
}
if(!(i % k))
{
long long d = __gcd(p,n);
p = 1;
if((d ^ 1) && (d ^ n))
return Pollard_Rho(d) && Pollard_Rho(n / d);
}
t1 = (fmul(t1,t1,n) + c) % n,t2 = (fmul(t2,t2,n) + c) % n,t2 = (fmul(t2,t2,n) + c) % n;
}
long long d = __gcd(p,n);
if((d ^ 1) && (d ^ n))
return Pollard_Rho(d) && Pollard_Rho(n / d);
}
int T;
long long n,mod;
int prime[CNT + 5],cnt,bz[CNT + 5];
int m,full;
long long p[M + 5],f[(1 << M) + 5];
int main()
{
freopen("number.in","r",stdin),freopen("number.out","w",stdout);
for(register int i = 2;i <= CNT;++i)
{
if(!bz[i])
prime[++cnt] = i;
for(register int j = 1;j <= cnt && i * prime[j] <= CNT;++j)
{
bz[i * prime[j]] = 1;
if(!(i % prime[j]))
break;
}
}
scanf("%d",&T);
for(int flag;T;--T)
{
scanf("%lld%lld",&n,&mod);
if(n == 4)
{
printf("%lld\n",8 % mod);
continue;
}
vis.clear(),flag = 0;
for(register int i = 1;i <= cnt && prime[i] <= n;++i)
for(;!(n % prime[i]);n /= prime[i])
if(vis.count(prime[i]))
flag |= 1;
else
vis.insert(prime[i]);
if(flag || (n > 1 && !Pollard_Rho(n)))
{
puts("-1");
continue;
}
memset(f,0,sizeof f),m = 0,f[0] = 1;
for(register unordered_set<long long>::iterator it = vis.begin();it != vis.end();++it)
p[++m] = *it;
full = (1 << m) - 1;
for(register int i = 0;i <= full;++i)
for(register int j = 1;j <= m;++j)
if(!(i & (1 << j - 1)))
f[i | (1 << j - 1)] = (f[i | (1 << j - 1)] + fmul(f[i],fpow(p[__builtin_popcount(i) + 1] % mod,p[j] - 1,mod),mod)) % mod;
printf("%lld\n",f[full]);
}
}