POJ 3243 Clever Y（离散对数-拓展小步大步算法）
Description
Little Y finds there is a very interesting formula in mathematics:
XY mod Z = K
Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?
Input
Input data consists of no more than 20 test cases. For each test case, there would be only one line containing 3 integers
X,
Z,
K (0 ≤
X,
Z,
K ≤ 10
9).
Input file ends with 3 zeros separated by spaces.
Input file ends with 3 zeros separated by spaces.
Output
For each test case output one line. Write "No Solution" (without quotes) if you cannot find a feasible
Y (0 ≤
Y <
Z). Otherwise output the minimum
Y you find.
题目大意：求离散对数。没有保证Z一定是素数。
PS：说一下上面没有提到的一个东西，最后一个O(sqrt(m))的循环中，逆元是没有必要每次都求的，可以预先求出来，让算法复杂度降至O(sqrt(m))。
在我的代码中，v = k^(-1) * a^(-i*m)，可以在前面先求出k^(-1)和a^(-m)，然后每次v都乘以a^(-m)，而不需要每次对k*a^(i*m)求逆元。
代码（63MS）：
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 void exgcd(LL a, LL b, LL &x, LL &y) { 38 if(!b) x = 1, y = 0; 39 else { 40 exgcd(b, a % b, y, x); 41 y -= x * (a / b); 42 } 43 } 44 45 LL inv(LL a, LL n) { 46 LL x, y; 47 exgcd(a, n, x, y); 48 return (x + n) % n; 49 } 50 51 LL pow_mod(LL x, LL p, LL n) { 52 LL ret = 1; 53 while(p) { 54 if(p & 1) ret = (ret * x) % n; 55 x = (x * x) % n; 56 p >>= 1; 57 } 58 return ret; 59 } 60 61 LL BabyStep_GiantStep(LL a, LL b, LL n) { 62 for(LL i = 0, e = 1; i <= 64; ++i) { 63 if(e == b) return i; 64 e = (e * a) % n; 65 } 66 LL k = 1, cnt = 0; 67 while(true) { 68 LL t = __gcd(a, n); 69 if(t == 1) break; 70 if(b % t != 0) return -1; 71 n /= t; b /= t; k = (k * a / t) % n; 72 ++cnt; 73 } 74 hashmap.init(); 75 hashmap.insert(1, 0); 76 LL e = 1, m = LL(ceil(sqrt(n + 0.5))); 77 for(int i = 1; i < m; ++i) { 78 e = (e * a) % n; 79 hashmap.insert(e, i); 80 } 81 LL p = inv(pow_mod(a, m, n), n), v = inv(k, n); 82 for(int i = 0; i < m; ++i) { 83 LL t = hashmap.find((b * v) % n); 84 if(t != -1) return i * m + t + cnt; 85 v = (v * p) % n; 86 } 87 return -1; 88 } 89 90 int main() { 91 LL x, z, k; 92 while(cin>>x>>z>>k) { 93 if(x == 0 && z == 0 && k == 0) break; 94 LL ans = BabyStep_GiantStep(x % z, k % z, z); 95 if(ans == -1) puts("No Solution"); 96 else cout<<ans<<endl; 97 } 98 }