Codeforces-gym/101853G：Hard Equation（BSGS）

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G. Hard Equation
time limit per test 10.0 s
memory limit per test 256 MB
inputs tandard input
outputs tandard output
Consider the following equation $a^x\equiv b\quad mod\quad m$Given a, b and m, your task is to find a value x that satisfy the equation for the given values. Can you?

Input
The first line contains an integer $T (1 ≤ T ≤ 500)$, in which T is the number of test cases.

Each test case consists of a line containing three integers a, b and m $(0 ≤ a, b < m ≤ 10^9)$.

Output
For each test case, print a single line containing an integer $x (0 ≤ x ≤ 10^{17})$ that satisfy the equation , for the given a, b and m.

If there are multiple solutions, print any of them. It is guaranteed that an answer always exist for the given input.

Example
input
3
3 9 11
2 3 5
2 1 5
output
2
3
4

思路：BSGS算法。 $a^x\equiv b\quad mod\quad m$其中 $x=i*\sqrt{m}-j，(j<\sqrt{m})$，转化一下即变为 $a^{i*\sqrt{m}}\equiv b*a^j\quad mod\quad m$我们只需枚举 $b*a^j$并将其与 $j$保存在map中，然后遍历 $a^{i*\sqrt{m}}$判断一下map里是否有对应的值即可。

#include<bits/stdc++.h>
using namespace std;
const int MAX=5e5+10;
typedef long long ll;
ll BSGS(ll a,ll b,ll m)
{
    a%=m;
    b%=m;
    if(b==1)return 0;
    unordered_map<ll,ll>ma;
    ll n=sqrt(2*m)+1;
    ll e=1;
    for(int i=0;i<n;i++)
    {
        if(ma.count(b*e%m)==0)ma[b*e%m]=i;//很奇怪去掉ma.count(b*e%m)==0就会WA
        e=e*a%m;
    }
    ll t=1;
    for(int i=1;i<=n+1;i++)
    {
        t=t*e%m;
        if(ma.count(t))return i*n-ma[t];
    }
    return -1;
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        ll a,b,m;
        scanf("%lld%lld%lld",&a,&b,&m);
        printf("%lld\n",BSGS(a,b,m));
    }
    return 0;
}