POJ 3243：Clever Y BSGS

Clever Y

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 7782 Accepted: 1939

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 58 332 4 30 0 0

Sample Output

9No Solution

m=sqrt(z)，将b^0一直到b^(m-1)次方插入到哈希表中，然后不断枚举i*m，查找是否有x^(i*m)=k*x^(j) mod z。

代码：

#pragma warning(disable:4996)#include <iostream>#include <functional>#include <algorithm>#include <cstring>#include <vector>#include <string>#include <cstdio>#include <cmath>#include <queue>#include <stack>#include <deque>#include <set>#include <map>using namespace std;typedef long long ll;#define INF 0x333f3f3f#define repp(i, n, m) for (int i = n; i <= m; i++)#define rep(i, n, m) for (int i = n; i < m; i++)#define sa(n) scanf("%d", &(n))const int maxn = 1e6 + 7;const double PI = acos(-1.0);const int mod = 1e9 + 7;struct ed{ll next;ll val;ll id;}edge[maxn];int num;int head[maxn];void insert(ll x, ll y){ll k = x%maxn;num++;edge[num].val = x;edge[num].id = y;edge[num].next = head[k];head[k] = num;}ll find(ll x){int k = x%maxn;for (int i = head[k]; i != -1; i = edge[i].next)if (edge[i].val == x)return edge[i].id;return -1;}ll x, z, k;void solve(){ll i, j, m;num = 0;memset(edge, -1, sizeof(edge));memset(head, -1, sizeof(head));m = sqrt(1.0*z);if (k == 1){printf("0\n");return;}ll p = 1;for (i = 0; i < m; i++, p = p*x%z){insert(p*k%z, i);}ll xx = 1;for (i = m;; i += m){ll pos = find(xx = p*xx%z);if (pos != -1){printf("%lld\n", i - pos);return;}if (i > z)break;}printf("No Solution\n");}int main(){#ifndef ONLINE_JUDGE  freopen("i.txt", "r", stdin);freopen("o.txt", "w", stdout);#endifwhile (scanf("%lld%lld%lld", &x, &z, &k) != EOF){if (x == 0 && z == 0 && k == 0)break;solve();}return 0;}