POJ 2417 Discrete Logging(离散对数-小步大步算法)
Description
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)
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):
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 }