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;}