BZOJ 2219 数论之神 (CRT推论+BSGS+原根指标)

看了Po神的题解一下子就懂了A了!

不过Po神的代码出锅了…solve中"d-temp"并没有什么用QwQQwQQwQ…应该把模数除以p^temp次方才行.

来自BZOJ讨论板的hack数据

hack data

1 5 3125 7812

正确输出应该是625， 但是很多人输出3125…

CODE

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const LL INF = 1e15;
inline LL qpow(LL a, LL b, LL c) {
    LL re = 1;
    while(b) {
        if(b&1) re = re * a % c;
        a = a * a % c; b >>= 1;
    }
    return re;
}
LL gcd(LL a, LL b) { return b ? gcd(b, a%b) : a; }
void exgcd(LL a, LL b, LL &x, LL &y) {
    if(!b) { x = 1, y = 0; return; }
    exgcd(b, a%b, y, x); y -= x*(a/b);
}
int prime[100], cnt;
inline void Factor(int N) {
    cnt = 0;
    for(int i = 2; i*i <= N; ++i)
        if(N % i == 0) {
            prime[++cnt] = i;
            while(N % i == 0) N /= i;
        }
    if(N > 1) prime[++cnt] = N;
}
inline int Get_g(int p, int phi) { //找原根
    Factor(phi);
    for(int g = 2; ; ++g) {
        bool flg = true;
        for(int i = 1; i <= cnt; ++i)
            if(qpow(g, phi/prime[i], p) == 1)
                { flg = 0; break; }
        if(flg) return g;
    }
}
map<int, int>myhash;
inline LL Baby_Step_Giant_Step(LL a, LL b, LL p) {
    myhash.clear(); int m = int(sqrt(p) + 1);
    LL base = b;
    for(int i = 0; i < m; ++i) {
        myhash[base] = i;
        base = base * a % p;
    }
    base = qpow(a, m, p); LL tmp = 1;
    for(int i = 1; i <= m+1; ++i) {
        tmp = tmp * base % p;
        if(myhash.count(tmp))
            return i*m - myhash[tmp];
    }
    return -1;
}
inline LL solve(LL a, LL b, LL p, LL d, LL p_d) {
    b %= p_d;
    if(!b) return qpow(p, d-((d-1)/a+1), INF);
    LL temp = 0;
    while(b % p == 0) b /= p, ++temp, p_d /= p;
    if(temp % a) return 0;
    LL phi = p_d - p_d/p, g = Get_g(p_d, phi);
    LL ind = Baby_Step_Giant_Step(g, b, p_d);
    LL re = gcd(a, phi);
    if(ind % re) return 0;
    return re * qpow(p, temp-temp/a, INF);
}
int main() {
    int T, a, b, k;
    for(scanf("%d", &T); T; T--) {
        scanf("%d%d%d", &a, &b, &k); k = k<<1|1;
        LL ans = 1;
        for(int i = 2; i*i <= k && ans; ++i)
            if(k % i == 0) {
                int d = 0, p_d = 1;
                while(k % i == 0) k /= i, ++d, p_d *= i;
                ans *= solve(a, b, i, d, p_d);
            }
        if(k > 1 && ans) ans *= solve(a, b, k, 1, k);
        printf("%lld\n", ans);
    }
}