bzoj 1467: Pku3243 clever Y 扩展BSGS

1467: Pku3243 clever Y

Time Limit: 4 Sec  Memory Limit: 64 MB
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Description

小Y发现，数学中有一个很有趣的式子： X^Y mod Z = K 给出X、Y、Z，我们都知道如何很快的计算K。但是如果给出X、Z、K，你是否知道如何快速的计算Y呢？

Input

本题由多组数据（不超过20组），每组测试数据包含一行三个整数X、Z、K（0 <= X, Z, K <= 109）。 输入文件一行由三个空格隔开的0结尾。

Output

对于每组数据：如果无解则输出一行No Solution，否则输出一行一个整数Y(0 <= Y < Z)，使得其满足XY mod Z = K，如果有多个解输出最小的一个Y。

Sample Input

5 58 33
2 4 3
0 0 0

Sample Output

9
No Solution

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=2e5+,M=4e6+,inf=1e9+,mod=1e9+;
const int  MAXN=  ;
struct LINK{
    ll data;
    ll j;
    ll next;
}HASH_LINK[];
ll ad, head[MAXN];ll Gcd(ll a, ll b){
return b ? Gcd(b, a % b) : a;
}ll Ext_Gcd(ll a, ll b, ll &x, ll &y){
    if(!b){
       x = ; y = ;
       return a;
    }
    ll r = Ext_Gcd(b, a % b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return r;
}ll POWER(ll a, ll b, ll c){
    ll ans = ;
    while(b){
       if(b & ) ans = ans * a % c;
       a = a * a % c;
       b >>= ;
    }
    return ans;
}void init(){
    memset(head, -, sizeof(head));
    ad = ;
}ll Hash(ll a){
    return a % MAXN;
}void INSERT_HASH(ll i, ll buf){
    ll hs = Hash(buf), tail;
    for(tail = head[hs]; ~tail; tail = HASH_LINK[tail]. next)
       if(buf == HASH_LINK[tail]. data) return;
    HASH_LINK[ad]. data = buf;
    HASH_LINK[ad]. j    = i;
    HASH_LINK[ad]. next = head[hs];
    head[hs] = ad ++;
}ll BSGS(ll a, ll b, ll c){
    ll i, buf, m, temp, g, D, x, y, n = ;
    for(i = , buf = ; i < ; i ++, buf = buf * a % c)
       if(buf == b) return i;
    D = ;
    while((g = Gcd(a, c)) != ){
       if(b % g) return -; // g | b 不满足，则说明无解
       b /= g;
       c /= g;
       D = D * a / g % c;
       ++ n;
    }
    init();
    m = ceil(sqrt((long double) c));
    for(i = , buf = ; i <= m; buf = buf * a % c, i ++) INSERT_HASH(i, buf);
    for(i = , temp = POWER(a, m, c), buf = D; i <= m; i ++, buf = temp * buf % c){
       Ext_Gcd(buf, c, x, y);
       x = ((x * b) % c + c) % c;
       for(ll tail = head[Hash(x)]; ~tail; tail = HASH_LINK[tail].next)
           if(HASH_LINK[tail]. data == x) return HASH_LINK[tail].j + n + i * m;
    }
    return -;
}
int main()
{
    ll a,b,n;
    while(~scanf("%lld%lld%lld",&a,&n,&b))
    {
        if(a==&&b==&&n==)break;
        if(n<b)
        {
            printf("No Solution\n");
            continue;
        }
        ll ans=BSGS(a,b,n);
        if(ans==-)
        printf("No Solution\n");
        else
        printf("%lld\n",ans%mod);
    }
    return ;
}