POJ2417 Baby-Step-Gaint-Step 算法

考虑一个问题:A^x%p=B,给定A,B,p，求x的最小非负整数解。

在p是质数的情况下，这个问题比较简单。

A^x=B(mod P) (P is a Prime, A,B<P)
Let m = floor(sqrt(P))
Put A^0,A^1,...A^(m-1) into HashSet(You Can Also Use Map in STL),for Example M[A^i]=i.

if A^i=A^j(i<j), M[A^i=A^j]=i.

Enumerate i, Let x=i*m+j, A^(i*m)*A^j=B(mod C)
Let E=A^(i*m),F=A^j,E*F=B(mod P) because (E,C)=1,(E,C)|B,we can use EXTgcd to get F.

If F is in the HashSet,then the minimum answer x=i*m+M[F].

Because P is a Prime, and A,B<P, then A^(P-1)=1(mod P). 
Then if a answer exists, the minimum answer must less then P.

So the range of i is [0,P/m].

If for all i we cannot find a answer, then no solution.

我亲手胡乱写的东西，能看懂才怪！

再附上代码吧：（我的小数据范围是暴力的）

#include <cmath>
#include <cstdio>
#include <cctype>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;
inline void Exgcd(LL a, LL b, LL &d, LL &x, LL &y) {
	if (!b) { d = a, x = 1, y = 0; }
	else { Exgcd(b, a % b, d, y, x), y -= x * (a / b); }
}
inline LL gcd(LL a, LL b) {
	return (!b) ? a : gcd(b, a % b);
}
inline LL Solve(LL a, LL b, LL c) { // ax=b(mod c)
	LL d, x, y;
	Exgcd(a, c, d, x, y);
	return (x + c) % c * b % c;
}
LL ksm(LL x, LL y, LL p) {
	LL res = 1, t = x;
	for(; y; y >>= 1) {
		if (y & 1) res = res * t % p;
		t = t * t % p;
	}
	return res;
}

const int mod = 13131;
struct Hashset {
	int head[mod], next[40010], f[40010], v[40010], ind;
	void reset() {
		ind = 0;
		memset(head, -1, sizeof head);
	}
	void insert(int x, int _v) {
		int ins = x % mod;
		for(int j = head[ins]; j != -1; j = next[j])
			if (f[j] == x) {
				v[j] = min(v[j], _v);
				return;
			}
		f[ind] = x, v[ind] = _v;
		next[ind] = head[ins], head[ins] = ind++;
	}
	int operator [] (const int &x) const {
		int ins = x % mod;
		for(int j = head[ins]; j != -1; j = next[j])
			if (f[j] == x)
				return v[j];
		return -1;
	}
}S;

int main() {
	LL A, B, C;
	LL i;
	while(scanf("%I64d%I64d%I64d", &C, &A, &B) == 3) {
		if (C <= 100) {
			LL d = 1;
			bool find = 0;
			for(i = 0; i < C; ++i) {
				if (d == B) {
					find = 1;
					printf("%I64d\n", i);
					break;
				}
				d = d * A % C;
			}
			if (!find)
				puts("no solution");
		}
		else {
			int m = (int)sqrt(C);
			S.reset();
			LL d = 1;
			for(i = 0; i < m; ++i) {
				S.insert(d, i);
				d = d * A % C;
			}
			bool find = 0;
			int ins;
			for(i = 0; i * m < C; ++i) {
				LL t = Solve(ksm(A, i * m, C), B, C);
				if ((ins = S[t]) != -1) {
					printf("%I64d\n", i * m + ins);
					find = 1;
					break;
				}
			}
			if (!find)
				puts("no solution");
		}
	}
	
	return 0;
}