[POJ 3243]Clever Y

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 ≤ 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

题解

扩展BSGS：

当模数 $c$ 不是质数的时候，显然不能直接使用 $BSGS$ 了，考虑它的扩展算法。

前提：同余性质。

令 $d = gcd(a, c)$ ， $A = a \cdot d,B = b \cdot d, C = c \cdot d$

则 $a \cdot d \equiv b \cdot d \pmod{c \cdot d}$

等价于 $a \equiv b \pmod{c}$

因此我们可以先消除因子。

对于现在的问题 $(A \cdot d)^x \equiv B \cdot d \pmod{C \cdot d}$ 当我们提出 $d = gcd(a, c)$ （$d \neq 1$）后，原式化为 $A \cdot (A \cdot d)^{x-1} \equiv B \pmod{C}$ 。

即求 $D \cdot A^{x-cnt} \equiv B \pmod{C}$ ，令 $x = i \cdot r-j+cnt$ 。之后的做法就和 $BSGS$ 一样了。

值得注意的是因为这样求出来的解 $x \geq cnt$ 的，但有可能存在解 $x < cnt$ ，所以一开始需要特判。

 

  1 //It is made by Awson on 2018.1.15
  2 #include <set>
  3 #include <map>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <queue>
  7 #include <stack>
  8 #include <cstdio>
  9 #include <string>
 10 #include <vector>
 11 #include <cstdlib>
 12 #include <cstring>
 13 #include <iostream>
 14 #include <algorithm>
 15 #define LL long long
 16 #define Abs(a) ((a) < 0 ? (-(a)) : (a))
 17 #define Max(a, b) ((a) > (b) ? (a) : (b))
 18 #define Min(a, b) ((a) < (b) ? (a) : (b))
 19 #define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
 20 using namespace std;
 21 const LL MOD = 233333;
 22 void read(LL &x) {
 23     char ch; bool flag = 0;
 24     for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
 25     for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
 26     x *= 1-2*flag;
 27 }
 28 void write(LL x) {
 29     if (x > 9) write(x/10);
 30     putchar(x%10+48);
 31 }
 32 
 33 LL a, b, c, ans;
 34 struct MAP {
 35     LL ha[MOD+5]; int id[MOD+5];
 36     void clear() {for (int i = 0; i < MOD; i++) ha[i] = id[i] = -1; }
 37     int count(LL x) {
 38     LL pos = x%MOD;
 39     while (true) {
 40         if (ha[pos] == -1) return 0;
 41         if (ha[pos] == x) return 1;
 42         ++pos; if (pos >= MOD) pos -= MOD;
 43     }
 44     }
 45     void insert(LL x, int idex) {
 46     LL pos = x%MOD;
 47     while (true) {
 48         if (ha[pos] == -1 || ha[pos] == x) {ha[pos] = x, id[pos] = idex; return; }
 49         ++pos; if (pos >= MOD) pos -= MOD;
 50     }
 51     }
 52     int query(LL x) {
 53     LL pos = x%MOD;
 54     while (true) {
 55         if (ha[pos] == x) return id[pos];
 56         ++pos; if (pos >= MOD) pos -= MOD;
 57     }
 58     }
 59 }mp;
 60 
 61 LL quick_pow(LL a, LL b, LL c) {
 62     LL ans = 1;
 63     while (b) {
 64     if (b&1) ans = ans*a%c;
 65     a = a*a%c, b >>= 1;
 66     }
 67     return ans;
 68 }
 69 LL gcd(LL a, LL b) {return b ? gcd(b, a%b) : a; }
 70 LL exBSGS(LL a, LL b, LL c) {
 71     if (b == 1) return 0;
 72     LL cnt = 0, d = 1, t;
 73     while ((t = gcd(a, c)) != 1) {
 74     if (b%t) return -1;
 75     ++cnt, b /= t, c /= t, d = d*(a/t)%c;
 76     if (d == b) return cnt;
 77     }
 78     mp.clear();
 79     LL tim = ceil(sqrt(c)), tmp = b%c;
 80     for (int i = 0; i <= tim; i++) {
 81     mp.insert(tmp, i); tmp = tmp*a%c;
 82     }
 83     t = tmp = quick_pow(a, tim, c); tmp = (tmp*d)%c;
 84     for (int i = 1; i <= tim; i++) {
 85     if (mp.count(tmp)) return tim*i-mp.query(tmp)+cnt;
 86     tmp = tmp*t%c;
 87     }
 88     return -1;
 89 }
 90 void work() {
 91     while ((~scanf("%lld%lld%lld", &a, &c, &b))) {
 92     if (c == 0) return;
 93     if ((ans = exBSGS(a%c, b%c, c)) == -1) printf("No Solution\n");
 94     else write(ans), putchar('\n');
 95     }
 96 }
 97 int main() {
 98     work();
 99     return 0;
100 }

 

posted @ 2018-01-16 07:54  NaVi_Awson  阅读(362)  评论(2)  编辑  收藏  举报