BZOJ1951 [Sdoi2010]古代猪文

这道题是各种数论搞在一起的题目。。。

首先由burnside引理可以知道答案是ans = (G^sigma(C(n, d))) % MOD

然后由费马小定理，ans =  (G^（sigma(C(n, d)) % (MOD - 1)）) % MOD

之后把MOD - 1分解为2 * 3 * 4679 * 35617，用Lucas定理分别求出sigma(C(n, d)) % m，其中m = 2, 3, 4679, 35617

最后再用中国剩余定理合起来快速幂一下就好辣！

/**************************************************************
    Problem: 1951
    User: rausen
    Language: C++
    Result: Accepted
    Time:76 ms
    Memory:2772 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
 
using namespace std;
typedef long long LL;
const LL p = 999911659;
const int x[4] = {2, 3, 4679, 35617};
 
LL mod[4][50000], a[50000], b[5], cnt, n, g, y, z;
 
inline LL pow(LL a, LL b, LL p){
    LL res = 1;
    while (b){
        if (b & 1) res *= a, res %= p;
        a *= a, a %= p;
        b >>= 1;
    }
    return res;
}
 
LL extend_gcd(int a, int b, LL &x, LL &y){
    if (!b){
        x = 1, y = 0;
        return a;
    }
    LL res = extend_gcd(b, a % b, y, x);
    y -= a / b * x;
    return res;
}
 
inline LL GCD(int a, int p){
    LL x, y, d = extend_gcd(a, p, x, y);
    while (x < 0) x += p;
    return x;
}
 
inline LL C(int i, int n, int k, int p){
    return mod[i][n] * GCD(mod[i][n - k] * mod[i][k] % p, p) % p;
}
 
inline LL Lucas(int i, int n, int k, int p){
    LL res = 1;
    while (n && k){
        res *= C(i, n % p, k % p, p), res %= p;
        if (!res) return 0;
        n /= p, k /= p;
    }
    return res;
}
 
inline LL Chinese(){
    LL M = 1, res = 0;
    for (int i = 0; i < 4; ++i) M *= x[i];
    for (int i = 0; i < 4; ++i){
        LL w = M / x[i], x1, y1, d = extend_gcd(w, x[i], x1, y1);
        res += x1 * w * b[i], res %= M;
    }
    return (res + M) % M;
}
 
int main(){
    scanf("%lld%lld", &n, &g);
    g %= p;
    if (!g){
        printf("0
");
        return 0;
    }
    for (int i = 0; i < 4; ++i){
        mod[i][0] = 1;
        for (int j = 1; j <= x[i]; ++j)
            mod[i][j] = (mod[i][j - 1] * j) % x[i];
    }
    int maxi = (int) sqrt(n);
    for (int i = 1; i <= maxi; ++i)
        if (!(n % i)){
            a[++cnt] = i;
            if (i * i != n) a[++cnt] = n / i;
        }
         
    for (int i = 1; i <= cnt; ++i)
        for (int j = 0; j < 4; ++j){
            b[j] += Lucas(j, n, a[i], x[j]);
            if (b[j] > x[j]) b[j] -= x[j];
        }
    int z = Chinese();
    printf("%lld
", pow(g, z, p));
    return 0;
}