高次同余方程（BSGS算法模板）

最新推荐文章于 2021-10-09 09:18:08 发布

原创最新推荐文章于 2021-10-09 09:18:08 发布 · 263 阅读

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本文深入探讨了Baby-step Giant-step算法，一种用于求解离散对数问题的有效方法。该算法尤其适用于当模数为素数的情况，通过将问题分解成两部分，即Baby-step和Giant-step，从而显著降低了计算复杂度。文章提供了详细的算法实现代码，并扩展讨论了当模数不是素数时的解决方案。

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裸题（n是素数）

POJ 2417

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//baby_step giant_step
// a^x = b (mod n) n为素数，a,b < n
// 求解上式 0<=x < n的解
#define MOD 76543
int hs[MOD],head[MOD],Next[MOD],id[MOD],top;
void insert(int x,int y)
{
    int k = x%MOD;
    hs[top] = x, id[top] = y, Next[top] = head[k], head[k] = top++;
}
int find(int x)
{
    int k = x%MOD;
    for(int i = head[k]; i != -1; i = Next[i])
        if(hs[i] == x)
            return id[i];
    return -1;
}
int BSGS(int a,int b,int n)
{
    memset(head,-1,sizeof(head));
    top = 1;
    if(b == 1)return 0;
    int m = sqrt(n*1.0), j;
    long long x = 1, p = 1;
    for(int i = 0; i < m; ++i, p = p*a%n)insert(p*b%n,i);
    for(long long i = m; ;i += m)
    {
        if( (j = find(x = x*p%n)) != -1 )return i-j;
        if(i > n)break;
    }
    return -1;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int a,b,n;
    while(scanf("%d%d%d",&n,&a,&b) == 3)
    {
        int ans = BSGS(a,b,n);
        if(ans == -1)printf("no solution\n");
        else printf("%d\n",ans);
    }
    return 0;
}

扩展（n不一定是素数）

POJ 3243

#include <cstdio>
#include <stack>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <map>
#include <vector>
#include <queue>
#include <set>
#define eps 1e-8
typedef long long ll;
const double PI = acos(-1.0);
const int maxn = 1e6;
const int INF = 0x3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9+7;
using namespace std;
/*
 *  baby_step giant _step
 *  a^x = b(mod n) n不要求是素数
 *  求解上式0 ≤ x < n的解
 */
#define MOD maxn
ll hs[MOD];
int head[MOD];
ll _next[MOD];
ll id[MOD];
int top;

void insert(ll x, ll y)
{
    ll k = x % MOD;
    hs[top] = x;
    id[top] = y;
    _next[top] = head[k];
    head[k] = top++;
    return ;
}

ll find(ll x)
{
    ll k = x % MOD;
    for (int i = head[k]; i != -1; i = _next[i])
    {
        if (hs[i] == x)
        {
            return id[i];
        }
    }
    return -1;
}

ll BSGS(ll a, ll b, ll n)
{
    memset(head, -1, sizeof(head));
    top = 1;
    if (b == 1)
    {
        return 0;
    }
    ll m = sqrt(n * 1.0), j;
    long long x = 1, p = 1;
    for (int i = 0; i < m; i++, p = p * a % n)
    {
        insert(p * b % n, i);
    }
    for (long long i = m; ; i+=m)
    {
        if ((j = find(x = x * p % n)) != -1)
        {
            return i - j;
        }
        if (i > n)
        {
            break;
        }
    }

    return -1;
}
int main()
{
    ios::sync_with_stdio(false);
    ll X,Z,K;
    while(cin>>X>>Z>>K && X+Z+K>0)
    {
        ll ans = BSGS(X,K,Z);
        if(ans == -1)
        {
            cout<<"No Solution"<<endl;
            continue;
        }
        cout<<ans<<endl;
    }

    return 0;
}