题解P4454 [CQOI2018]破解D-H协议

题面比较长,但其实就是一道板子题

根据题意, \(g^a\equiv A(mod~p),g^b\equiv B(mod~p)\),可以直接通过BSGS求出 \(a,b\) 的值,然后直接用快速幂算出来即可

关于BSGS,可以看看我的博客

代码

#include<iostream>
#include<cmath>
#include<unordered_map>
using namespace std;
typedef long long ll;

int ksm(int a, int b, int p)
{
    int res = 1;
    while (b)
    {
        if (b & 1)
            res = (ll)res * a % p;
        a = (ll)a * a % p;
        b >>= 1;
    }
    return res;
}

int BSGS(int a, int b, int p)
{
    if (1 % p == b % p)
        return 0;
    unordered_map<int, int> hash;
    int k = sqrt(p) + 1;
    for (int i = 0, j = b % p; i < k; i++, j = (ll)j * a % p)
        hash[j] = i;
    int ak = ksm(a, k, p);
    for (int i = 1, j = ak; i <= k; i++, j = (ll)j * ak % p)
        if (hash.count(j))
            return i * k - hash[j];
    return -1;
}

int main()
{
    int g, p;
    scanf("%d%d", &g, &p);
    int n;
    scanf("%d", &n);
    while (n--)
    {
        int A, B;
        scanf("%d%d", &A, &B);
        int a = BSGS(g, A, p);
        int b = BSGS(g, B, p);
        printf("%d\n", ksm(g, (ll)a * b % (p - 1), p));
    }
    return 0;
}

标签：P4454,int,题解,ll,CQOI2018,res,include,ak,BSGS
来源： https://www.cnblogs.com/A2484337545/p/14695479.html