poj 2417 Discrete Logging(baby-step算法)

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Description
给出P,B,N(P是素数)，解方程B^L = N (mod P)。
Input
多组输入，每组用例占一行包括三个整数P,B,N，以文件为结束输入。
Output
对于每组用例，如果方程有解则输出最小解，否则输出no solution。

参考资料：

http://www.cppblog.com/csu-yx-2013/archive/2012/07/29/185562.html?opt=admin

#include <stdio.h>#include <string.h>#include <math.h>#include <algorithm>using namespace std;#define MAX (1000000)long long nData[MAX];long long nKey[MAX];long long egcd(long long a, long long b, long long& x, long long& y){    if (b == 0)    {        x = 1;        y = 0;        return a;    }    long long ret = egcd(b, a % b, x, y);    long long t = x;    x = y;    y = t - (a / b) * y;    return ret;}long long GetPos(long long key){    return (key ^ 0xA5A5A5A5) % MAX;}void Add(long long key, long long data){    long long nPos = GetPos(key);    while (nData[nPos] != -1)    {        nPos = (nPos + 1) % MAX;    }    nData[nPos] = data;    nKey[nPos] = key;}int Query(int key){    int nPos = GetPos(key);    while (nData[nPos] != -1)    {        if (nKey[nPos] == key)        {            return nData[nPos];        }        nPos = (nPos + 1) % MAX;    }    return -1;}long long BabyStep(long long nA, long long nB, long long nP){    long long nM = ceil(sqrt((double)(nP - 1)));    long long x, y;    egcd(nP, nA, x, y);//y是nA%p的乘法逆    y = (y + nP) % nP;    long long nTemp = 1;    long long c = 1;//c是nA的—m次    memset(nData, -1, sizeof(nData));    memset(nKey, -1, sizeof(nKey));    for (long long j = 0; j < nM; ++j)    {        Add(nTemp, j);        nTemp = (nTemp * nA) % nP;        c = (c * y) % nP;    }    long long r = nB;    for (int i = 0; i < nM; ++i)    {        long long j = Query(r);        if (j != -1)        {            return i * nM + j;        }        r = (r * c) % nP;    }    return -1;}int main(){    long long nP, nB, nN;    while (scanf("%I64d%I64d%I64d", &nP, &nB, &nN) == 3)    {        long long nAns = BabyStep(nB, nN, nP);        if (nAns == -1)printf("no solution\n");        else printf("%I64d\n", nAns);    }    return 0;}

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