HDU - 2815 Mod Tree 【高次不定方程 + 扩展 BSGS + 板子】

Mod Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7143 Accepted Submission(s): 1750

Problem Description

这里写图片描述
The picture indicates a tree, every node has 2 children.
The depth of the nodes whose color is blue is 3; the depth of the node whose color is pink is 0.
Now out problem is so easy, give you a tree that every nodes have K children, you are expected to calculate the minimize depth D so that the number of nodes whose depth is D equals to N after mod P.

Input

The input consists of several test cases.
Every cases have only three integers indicating K, P, N. (1<=K, P, N<=10^9)

Output

The minimize D.
If you can’t find such D, just output “Orz,I can’t find D!”

Sample Input

3 78992 453
4 1314520 65536
5 1234 67

Sample Output

Orz,I can’t find D!
8
20

题意：一个树上，每个节点有k个孩子，给你一个p和N $求 a^x \% p == N$ ,且不保证p为素数

分析：这是一个板子题，如果p为素数的时候，是BSGS，这里要用到扩展BSGS，原理看下这里吧
就当做个板子吧

参考代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <cmath>
#include <unordered_map>

using namespace std;

typedef long long ll;

unordered_map<ll,ll> ha;

ll MOD;
ll gcd(ll a, ll b) {
    return !b ? a : gcd(b, a % b);
}

#define mod(x) ((x) % MOD)

ll qpow(ll a,ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) {
            res = mod(res * a);
        }
        a = mod(a * a);
        b >>= 1;
    }
    return res;
}

ll ebsgs(ll a, ll b, ll p) {
    if (b == 1) return 0;
    ll t,d = 1,k = 0;
    while ((t = gcd(a, p)) != 1) {
        if (b % t) return -1;
        k++,b /= t,p /= t, d = (d*(a / t)) % p;
        if (b == d) return k;
    }
    ha.clear();
    ll m = ceil(sqrt(p+0.5));
    ll a_m = qpow(a,m);
    ll mul = b;
    for (ll j = 1; j <= m; j++) {
        mul = (mul * a) % p;
        ha[mul] = j;
    }
    for (ll i = 1; i <= m; i++) {
        d = (d * a_m) % p;
        if (ha[d]) return i * m - ha[d] + k;
    }
    return -1;
}

int main() {
    ll a,b;
    while (~scanf("%lld%lld%lld", &a, &MOD, &b)) {
        if (b >= MOD) {
            puts("Orz,I can’t find D!");continue;
        }
        if (ebsgs(mod(a), mod(b),MOD) == -1) puts("Orz,I can’t find D!");
        else printf("%lld\n", ebsgs(mod(a), mod(b),MOD));
    }
    return 0;
}