poj 2417 Discrete Logging (BSGS算法)

题意：

Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that BL == N (mod P)

思路：bsgs算法

详情见BSGS算法笔记

然后被map的count坑了一下？我想判断map中某个key是否存在，用count会TLE，find也会TLE,[]可以通过….不太懂，复杂度不都是log吗，差常数？还是有人会退化？

不过似乎[]比较安全就对了。

 1/* ***********************************************
 2Author :111qqz
 3Created Time :2017年07月23日 星期日 11时11分00秒
 4File Name :2417.cpp
 5************************************************ */
 6
 7#include <cstdio>
 8#include <cstring>
 9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define PB push_back
20#define fst first
21#define sec second
22#define lson l,m,rt<<1
23#define rson m+1,r,rt<<1|1
24#define ms(a,x) memset(a,x,sizeof(a))
25typedef long long LL;
26#define pi pair < int ,int >
27#define MP make_pair
28
29using namespace std;
30const double eps = 1E-8;
31const int dx4[4]={1,0,0,-1};
32const int dy4[4]={0,-1,1,0};
33const int inf = 0x3f3f3f3f;
34LL p,b,n;
35map<LL,LL>Hash;
36map<LL,LL>::iterator it;
37inline LL ksm(LL a,LL b,LL MOD)
38{
39    LL res = 1LL;
40    while (b)
41    {
42	if (b&1) res = (res*a)%MOD;
43	b = b >> 1;
44	a = (a*a)%MOD;
45    }
46    return res;
47}
48LL BSGS(LL a,LL b ,LL p) // a^x = b (mod p),求x
49{
50    a%=p;
51    b%=p;
52    if (!a&&!b) return 1;
53    if (!a) return -1;
54    Hash.clear();
55    LL m = ceil(sqrt(double(p)));
56    LL tmp = b;
57    for (LL j = 0 ; j <= m ; j++)
58    {
59	Hash[tmp]=j;
60	tmp = (tmp*a)%p;
61    }
62    tmp = ksm(a,m,p);
63    LL ret = 1;
64
65    for (LL i = 1  ; i <= m+1 ; i++)
66    {
67	ret = ret*tmp%p;
68	if (Hash[ret]) return i*m-Hash[ret]; //注意处理下%....虽然其实不处理也没关系...
69    }
70    return -1;
71
72}
73int main()
74{
75        #ifndef  ONLINE_JUDGE
76       // freopen("./in.txt","r",stdin);
77  #endif
78	while (~scanf("%lld%lld%lld",&p,&b,&n))   // B^L = n(mod p)
79	{
80	    LL ans = BSGS(b,n,p);
81	    if (ans==-1) printf("no solution\n");
82	    else printf("%lld\n",(ans%p+p)%p);
83	}
84
85
86  #ifndef ONLINE_JUDGE
87  fclose(stdin);
88  #endif
89    return 0;
90}

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