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  • bzoj1951

    CRT+LUCAS+费马小定理+拓展欧拉定理

    幂指数太大了怎么办?欧拉定理,n太大了怎么办?上lucas,模数太大了怎么办?上crt。然后就好了,唯一注意的是要用拓展欧拉定理,n%phi(p)+phi(p)

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    #include<iostream>
    using namespace std;
    typedef long long ll;
    const ll mod = 999911659ll, t[4] = {2, 3, 4679, 35617};
    ll n, g;
    ll a[5], fac[4][40010];
    ll power(ll x, ll t, ll mod)
    {
        ll ret = 1ll;
        for(; t; t >>= 1ll, x = x * x % mod) if(t & 1ll) ret = ret * x % mod;
        return ret;
    }
    ll inv(ll x, ll p)
    {
        return power(x, p - 2, p);
    }
    ll C(ll n, ll m, int id)
    {
        if(n < m) return 0;
        ll ret = fac[id][n] % t[id] * inv(fac[id][m], t[id]) % t[id] * inv(fac[id][n - m], t[id]) % t[id]; 
    //  printf("C(%lld %lld) = %lld mod = %lld
    ", n, m, ret, t[id]);
        return ret;
    }
    ll lucas(ll n, ll m, int id)
    {
        if(n < m) return 0;
        if(n < t[id] && m < t[id]) return C(n, m, id);
        return lucas(n % t[id], m % t[id], id) % t[id] * lucas(n / t[id], m / t[id], id) % t[id];
    }
    ll CRT()
    {
        ll M = mod - 1, ret = 0;
        for(int i = 0; i < 4; ++i) ret = (ret + a[i] * (M / t[i]) % M * inv(M / t[i], t[i]) % M) % M;
        return ret % M;
    }
    int main()
    {
        scanf("%lld%lld", &n, &g);
        for(int i = 0; i < 4; ++ i)
        {
            fac[i][0] = 1ll;
            for(int j = 1; j <= t[i]; ++j) fac[i][j] = fac[i][j - 1] * (ll)j % t[i];
            for(ll j = 1ll; j * j <= n; ++j) if(n % j == 0)
            {
                ll mul = lucas(n, j, i);
    //          printf("C(%lld %lld) = %lld
    ", n, j, mul);
                a[i] = (a[i] + mul) % t[i];
                if(j * j != n) 
                {
                    mul = lucas(n, n / j, i);
    //              printf("C(%lld %lld) = %lld
    ", n, n / j, mul);
                    a[i] = (a[i] + mul) % t[i];
                }
            }
        }
        printf("%lld
    ", power(g % mod, CRT() % (mod - 1) + mod - 1, mod) % mod);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/19992147orz/p/7481989.html
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