Codeforces Round #460 (Div. 2).E 费马小定理+中国剩余定理

E. Congruence Equation

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

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.

Input

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.

Output

Print a single integer: the number of possible answers n.

Examples

input

2 3 5 8

output

2

input

4 6 7 13

output

1

input

233 233 10007 1

output

1

Note

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
",x)
#define pd(x) printf("%f
",x)
#define pl(x) printf("%lld
",x)
#define inf 0x3f3f3f3f
#define rep(i, x, y) for(int i=x;i<=y;i++)
const int N   = 1e6 + 5;
const int mod = 1e9 + 7;
const int MOD = mod - 1;
const db  eps = 1e-10;
const db  PI  = acos(-1.0);
using namespace std;
ll a,b,p,x;
ll m[2],f[2];
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    ll d;
    //if (a == 0 && b == 0) return -1;
    if (b == 0)
    {
        x = 1;
        y = 0;
        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 == 1)
        return (x % MOD + MOD) % MOD;
    // else   return -1;
}
ll china(ll *f, ll *m){
    ll M = 1, ret = 0;
    for(int i = 0; i < 2; i ++) M *= m[i];
    for(int i = 0; i < 2; 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=1;
    x%=p;
    while(n){
        if(n&1) ans=ans*x%p;
        x=x*x%p;
        n>>=1;
    }
    return  ans;
}

int main()
{
    cl(a),cl(b),cl(p),cl(x);
    a%=p;
    ll ans=0,P=p*(p-1);
    m[0]=p-1,m[1]=p;
    for(int i=0;i<p-1;i++)
    {
        f[0]=i;
        f[1]=inv(qpow(a,i),p)*b%p;
        ll xx=china(f,m);
        ans+=(x-xx+P)/P;
    }
    pl(ans);
    return 0;
}