POJ 2417 Discrete Logging（离散对数-小步大步算法）

Description

Given a prime P, 2 <= P < 2 ³¹, 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 

    B

^L

 == N (mod P)

Input

Read several lines of input, each containing P,B,N separated by a space.

Output

For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print "no solution".

 

题目大意：求离散对数，B^L = N (mod P)，其中P是素数，求L。

思路：大步小步算法的模板题。

PS：白书上的模板少了个ceil向上取整……

PS：用map来哈希TLE了。速度和手写哈希原来差距这么大……

 

代码（47MS）：

 1 #include <cstdio>

 2 #include <iostream>

 3 #include <algorithm>

 4 #include <cstring>

 5 #include <cmath>

 6 using namespace std;

 7 typedef long long LL;

 8 

 9 const int SIZEH = 65537;

10 

11 struct hash_map {

12     int head[SIZEH], size;

13     int next[SIZEH];

14     LL state[SIZEH], val[SIZEH];

15 

16     void init() {

17         memset(head, -1, sizeof(head));

18         size = 0;

19     }

20 

21     void insert(LL st, LL sv) {

22         LL h = st % SIZEH;

23         for(int p = head[h]; ~p; p = next[p])

24             if(state[p] == st) return ;

25         state[size] = st; val[size] = sv;

26         next[size] = head[h]; head[h] = size++;

27     }

28 

29     LL find(LL st) {

30         LL h = st % SIZEH;

31         for(int p = head[h]; ~p; p = next[p])

32             if(state[p] == st) return val[p];

33         return -1;

34     }

35 } hashmap;

36 

37 LL inv(LL a, LL n) {

38     if(a == 1) return 1;

39     return ((n - n / a) * inv(n % a, n)) % n;

40 }

41 

42 LL pow_mod(LL x, LL p, LL n) {

43     LL ret = 1;

44     while(p) {

45         if(p & 1) ret = (ret * x) % n;

46         x = (x * x) % n;

47         p >>= 1;

48     }

49     return ret;

50 }

51 

52 LL BabyStep(LL a, LL b, LL n) {

53     LL e = 1, m = LL(ceil(sqrt(n + 0.5)));

54     hashmap.init();

55     hashmap.insert(1, 0);

56     for(int i = 1; i < m; ++i) {

57         e = (e * a) % n;

58         hashmap.insert(e, i);

59     }

60     LL v = inv(pow_mod(a, m, n), n);

61     for(int i = 0; i < m; ++i) {

62         LL t = hashmap.find(b);

63         if(t != -1) return i * m + t;

64         b = (b * v) % n;

65     }

66     return -1;

67 }

68 

69 int main() {

70     LL p, b, n;

71     while(cin>>p>>b>>n) {

72         LL ans = BabyStep(b, n, p);

73         if(ans == -1) puts("no solution");

74         else cout<<ans<<endl;

75     }

76 }

View Code