bzoj3929 Discrete Logging 大步小步算法

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
ll p,a,b;
ll ksm(ll x,ll y) 
{
    ll res=1;
    while(y) {
        if(y&1)res=res*x%p;
        y>>=1;
        x=x*x%p;
    }
    return res;
}
map<ll,ll> h;
ll solve(ll a,ll b,ll p) 
//设x=im-c,则x+c=im
//a^x * a^c =a^im
{
    ll m=ceil(sqrt(p));//向上取整 
    h.clear();
    for(ll i=0; i<m; i++) 
    //求出b*a,b*a^2,b*a^3...即等式左边那一段，放到map中 
	{
        if(!h.count(b))
		   h[b]=i;
        b=b*a%p;
    }
    ll now=1,base=ksm(a,m);
    for(ll i=1; i<=m+1; i++) 
    //枚举i,算a^im的结果，然后到hash表中去找 
	{
        now=now*base%p;
        if(h.count(now))return i*m-h[now];
    }
    return -1;
}
int main() 
{
    while(scanf("%lld%lld%lld",&p,&a,&b)!=EOF) 
	{
        ll ans=solve(a,b,p);
        if(ans==-1)printf("no solution
");
        else printf("%lld
",ans);
    }
    return 0;
}


#include<bits/stdc++.h>
using namespace std;
pair<long long, long long> x[1000001];
long long p, b, n;
long long pow(long long x, long long y, long long p)
{
    long long ans = 1;
    while(y)
	{
        if(y & 1) ans *= x, ans %= p;
        y >>= 1;
        x *= x;
        x %= p;
    }
    return ans;
}
 
long long find(long long l, long long r, long long v)
{
    while(l <= r)
	{
        long long mid = (l + r) / 2;
        if(x[mid].first < v)
		{
            l = mid + 1;
        }
        else{
            r= mid-1;
        }
    }
    return (x[l].first == v) ? l : -1;
}
 
int main()
{
    cin >> p >> b >> n;
	//b^x =N % p 
   long long sq = (double)(sqrt(p - 2) + 0.5);

    x[0].first = 1;
    x[0].second = 0;
    for(long long i = 1; i < sq; ++i)
    //求出b的若干次方Mod P的结果，放到一个表里 
	{
        x[i].first = x[i - 1].first * b;
        x[i].first %= p;
        x[i].second = i;
    }
    sort(x, x + sq - 1);//排序 
    for(long long i = 0; i <= sq; ++i)
	{
        long long v = n * pow(b, p - 1 - i * sq, p);
        v %= p;
        long long j = find(0, sq - 1, v);//到表里去找 
        if(j != -1)
		{
            cout << i * sq + x[j].second << endl;
            return 0;
        }
    }
    cout << "no solution" << endl;
    return 0;
}