Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

扩展 BSGS 模板

用于快速求方程 $a ^ x = b$ 的最小非负整数解。

查看代码
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#include <cstdio>
#include <cmath>
#include <unordered_map>
using namespace std;
typedef long long LL;
const int inf = 1e8;
int exgcd(int a, int b, int &x, int &y)
{
if (!b)
{
x = 1, y = 0;
return a;
}
int d = exgcd(b, a % b, y, x);
y -= a / b * x;
return d;
}
int bsgs(int a, int b, int p)
{
if (1 % p == b % p)
return 0;
int k = sqrt(p) + 1;
unordered_map<int, int> hash;
for (int i = 0, j = b % p; i < k; i++)
{
hash[j] = i;
j = (LL)j * a % p;
}
int t = 1;
for (int i = 0; i < k; i++)
t = (LL)t * a % p;
for (int i = 1, j = t; i <= k; i++)
{
if (hash.count(j))
return i * k - hash[j];
j = (LL)j * t % p;
}
return -inf;
}
int exbsgs(int a, int b, int p)
{
b = (b % p + p) % p;
if (1 % p == b % p)
return 0;
int x, y;
int d = exgcd(a, p, x, y);
if (d > 1)
{
if (b % d)
return -inf;
exgcd(a / d, p / d, x, y);
return exbsgs(a, (LL)b / d * x % (p / d), p / d) + 1;
}
return bsgs(a, b, p);
}
int main()
{
int a, p, b;
while (~scanf("%d%d%d", &a, &p, &b) && a && p && b)
{
int res = exbsgs(a, b, p);
if (res < 0)
puts("No Solution");
else
printf("%d\n", res);
}
return 0;
}