# 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\n");
else printf("%lld\n",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;
}
```

bzoj3929 Discrete Logging 大步小步算法

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