Codeforces Round #460 (Div. 2).E 费马小定理+中国剩余定理
3 seconds
256 megabytes
standard input
standard output
Given an integer x. Your task is to find out how many positive integers n (1 ≤ n ≤ x) satisfy
where a, b, p are all known constants.
The only line contains four integers a, b, p, x (2 ≤ p ≤ 106 + 3, 1 ≤ a, b < p, 1 ≤ x ≤ 1012). It is guaranteed that p is a prime.
Print a single integer: the number of possible answers n.
2 3 5 8
2
4 6 7 13
1
233 233 10007 1
1
In the first sample, we can see that n = 2 and n = 8 are possible answers.
题意:给出a, b, p, x,求有多少个n满足:n*a^n%p==b(n<=x)
思路:先要知道一个很简单的性质:a^n%p=(a%p)^(n%p-1)%p=a^(n%p-1),即a^n仅有p-1种结果。
①暴力枚举a^i%p (i从0到p-1)算出每个余数,可以得到式子n*a^i%p==b
②对于每个a^i,用逆元求出c,n%p = c= b*inv(a^i)%p
③则n%(p-1)==i 且n%p==c,最小的n可以用CRT求出,n的周期是P=p*(p-1).
代码:
//#include "bits/stdc++.h"
#include "cstdio"
#include "map"
#include "set"
#include "cmath"
#include "queue"
#include "vector"
#include "string"
#include "cstring"
#include "time.h"
#include "iostream"
#include "stdlib.h"
#include "algorithm"
#define db double
#define ll long long
//#define vec vector<ll>
#define Mt vector<vec>
#define ci(x) scanf("%d",&x)
#define cd(x) scanf("%lf",&x)
#define cl(x) scanf("%lld",&x)
#define pi(x) printf("%d\n",x)
#define pd(x) printf("%f\n",x)
#define pl(x) printf("%lld\n",x)
#define inf 0x3f3f3f3f
#define rep(i, x, y) for(int i=x;i<=y;i++)
const int N = 1e6 + ;
const int mod = 1e9 + ;
const int MOD = mod - ;
const db eps = 1e-;
const db PI = acos(-1.0);
using namespace std;
ll a,b,p,x;
ll m[],f[];
ll exgcd(ll a, ll b, ll &x, ll &y)
{
ll d;
//if (a == 0 && b == 0) return -1;
if (b == )
{
x = ;
y = ;
return a;
}
d = exgcd(b, a%b, y, x);
y -= a / b * x;
return d;
} ll inv(ll a, ll MOD)
{
ll x, y, d;
d = exgcd(a, MOD, x, y);
if (d == )
return (x % MOD + MOD) % MOD;
// else return -1;
}
ll china(ll *f, ll *m){
ll M = , ret = ;
for(int i = ; i < ; i ++) M *= m[i];
for(int i = ; i < ; i ++){
ll w = M / m[i];
ret = (ret + w * inv(w, m[i]) * f[i]) % M;
}
return (ret + M) % M;
}
ll qpow(ll x,ll n)
{
ll ans=;
x%=p;
while(n){
if(n&) ans=ans*x%p;
x=x*x%p;
n>>=;
}
return ans;
} int main()
{
cl(a),cl(b),cl(p),cl(x);
a%=p;
ll ans=,P=p*(p-);
m[]=p-,m[]=p;
for(int i=;i<p-;i++)
{
f[]=i;
f[]=inv(qpow(a,i),p)*b%p;
ll xx=china(f,m);
ans+=(x-xx+P)/P;
}
pl(ans);
return ;
}
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