POJ2417 Discrete Logging （离散对数取模）

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原创最新推荐文章于 2026-04-21 11:58:18 发布 461 阅读 AI 写同款· GEO 优化 ›

2月26日 Discrete logging(哈希表+离散对数+BSGS算法) 开学第一天，有点困，不过天气不错，手脚能放开了。今天学习了离散对数的几类问题，也是开始涉及密码学的核心：对于离散对数：$A^x\equiv B \mod p$如果要求B，那就是幂模问题，用快速幂求模。如果未知x，就是一个重要的离散对数/密码学问题，直观的我们想：是不是可以从1枚举到p-1（欧拉函数）？不过这个开销比较大，有一种算法（BSGS，大步小步/北上广深（!?）算法）可以解决这类问题。将x写... 阅读详情

Discrete Logging

Given a prime P, 2 <= P < 2 31, an integer B, 2 <= B < P, and an integer N, 1 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that

    BL == N (mod P)

Input
Read several lines of input, each containing P,B,N separated by a space.
Output
For each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print “no solution”.
Sample Input

5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111

Sample Output

0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587

Hint
The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat’s theorem that states

   B(P-1) == 1 (mod P)


for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat's theorem is that for any m

   B(-m) == B(P-1-m) (mod P) .

直接套模板

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;
typedef long long ll;
ll q_pow(ll a, ll b, ll mod){
    ll ans = 1;
    while(b){
        if(b & 1) ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return ans;
}

ll gcd(ll a,ll b){
    return b ? gcd(b,a%b) : a;
}

void ex_gcd(ll a,ll b,ll &x,ll &y){
    if(!b){
        x = 1;
        y = 0;
        return;
    }
    ex_gcd(b,a%b,y,x);
    y -= a / b * x;
    return;
}

ll BSGS(ll a, ll b, ll p) {
    a %= p; b %= p;
    map<ll, ll> h;
    map<ll,bool> vis;//这道题必须多加一个map作为标记，否则直接查找会超时！！
    ll m = ceil(sqrt(p)), x, y, d, t = 1, v = 1;
    for(ll i = 0; i < m; ++i) {
        if(h.count(t)) h[t] = min(h[t], i);
        else h[t] = i,vis[t] = 1;
        t = (t*a) % p;
    }
    ex_gcd(t,p,x,y);
    ll inv = x > 0 ? x % p : x % p + p;
    for(ll i = 0; i < m; ++i) {
        if(vis[b]) return i*m + h[b];
        b = b * inv % p;
    }
    return -1;
}
int main(){
    ll b,p,n;
    while(~scanf("%lld%lld%lld",&p,&b,&n)){
        if(n >= p){
            puts("no solution");
            continue;
        }
        ll ans = BSGS(b,n,p);
        if(ans == -1) puts("no solution");
        else printf("%lld\n",ans);
    }
    return 0;
}