【POJ3243】【拓展BSGS】Clever Y

Description

Little Y finds there is a very interesting formula in mathematics:

X^Y 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 ≤ 10⁹).
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.

Sample Input

5 58 33
2 4 3
0 0 0

Sample Output

9
No Solution

Source

POJ Monthly--2007.07.08, Guo, Huayang

【分析】

时间卡得真紧TAT,被T出翔。

后面用HASH + 链表才过...

 /*
 五代李煜
 《浪淘沙令·帘外雨潺潺》
 帘外雨潺潺，春意阑珊。罗衾不耐五更寒。梦里不知身是客，一晌贪欢。
 独自莫凭栏，无限江山，别时容易见时难。流水落花春去也，天上人间。
 */
 #include <iostream>
 #include <cstdio>
 #include <ctime>
 #include <cmath>
 #include <algorithm>
 #include <cstring>
 #include <string>
 #include <map>
 #include <set>
 #include <vector>
 #define LOCAL
 const int MAXN =  + ;
 const int maxn = ;//用来HASH
 const int INF = 0x7fffffff;
 using namespace std;
 typedef long long ll;
 struct Hash{
        ll a,b,next;
 }Hash[maxn << ];
 ll flg[maxn + ];
 ll top,idx;
 void ins(ll a,ll b){
      ll k = b & maxn;
      if (flg[k] != idx){
         flg[k] = idx;
         Hash[k].next = -;
         Hash[k].a = a;
         Hash[k].b = b;
         return ;
      }
      while (Hash[k].next != -){
            if(Hash[k].b == b) return ;
            k = Hash[k].next;
      }
      Hash[k].next = ++ top;
      Hash[top].next = -;
      Hash[top].a = a;
      Hash[top].b = b;
 }
 ll find(ll b){
     ll k = b & maxn;
     if (flg[k] != idx) return -;
     while (k != -){
           if(Hash[k].b == b) return Hash[k].a;
           k = Hash[k].next;
     }
     return -;
 }
 ll gcd(ll a,ll b) {return b == ? a: gcd(b, a % b);}
 ll exgcd(ll a, ll b, ll &x, ll &y){
     if (b == ){x = ; y = ; return a;}
     ll tmp = exgcd(b, a % b, y, x);
     y -= x * (a / b);
     return tmp;
 }
 //求解线性同余方程 形如Ax = B(mod C)
 ll solve(ll a, ll b, ll c){
     ll x, y, Ans;
     ll tmp = exgcd(a, c, x, y);
     Ans = (ll)(x * b) % c;
     return Ans >=  ? Ans : Ans + c;
 }
 ll pow(ll a, ll b, ll c){
     if (b == ) return a % c;
     ll tmp = pow(a, b / , c);
     if (b %  == ) return (tmp * tmp) % c;
     else return (((tmp * tmp) % c) * a) % c;
 }
 ll BSGS(ll A, ll B, ll C){
     top = maxn;
     ++idx;
     ll buf =  % C, D = buf, K, tmp;
     for (ll i = ; i <= ; i++){
         if (buf == B) return i;
         buf = (buf * A) % C;
     }
     ll d = ;
     while ((tmp = gcd(A, C)) != ){
           if (B % tmp != ) return -;
           d++;
           B /= tmp;
           C /= tmp;
           D = D * A / tmp % C;
     }
     //hash表记录1-sqrt(c)的值
     ll M = (ll)ceil(sqrt(C * 1.0));
     buf =  % C;
     for (ll i = ; i <= M; i++){
         ins(i, buf);
         buf = (buf * A) % C;
     }
     K = pow(A, M, C);
     for (ll i = ; i <= M; i++){
         tmp = solve(D, B, C);
         ll w;
         if (tmp >=  && (w = find(tmp)) != -) return i * M + w + d;
         D = (D * K) % C;
     }
     return -;
 }

 int main(){

     ll A, B, C;
     while (scanf("%lld%lld%lld", &A, &C, &B) != EOF){
           if (A ==  && C ==  && B == ) break;
           B %= C;
           ll tmp = BSGS(A, B, C);
           if (tmp >= ) printf("%lld\n", tmp);
           else printf("No Solution\n");
     }
     return ;
 }