Blog of RuSun

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

O(n) / O(1) GCD

用于 $O(n)$ 预处理 $O(1)$ 查询GCD。

查看代码
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#include <cstdio>
#include <cmath>
using namespace std;
const int N = 1e6 + 10, M = 1e3 + 10;
int n, m, q;
int cnt, p[N], v[N], div[N][3], g[M][M];
void init()
{
v[1] = 1;
for (int i = 2; i < N; i++)
{
if (!v[i])
{
p[++cnt] = i;
v[i] = i;
}
for (int j = 1; j <= cnt && i * p[j] < N; j++)
{
v[i * p[j]] = p[j];
if (i % p[j])
break;
}
}
div[1][0] = div[1][1] = div[1][2] = 1;
for (int i = 1; i < N; i++)
{
for (int j = 0; j < 3; j++)
div[i][j] = div[i / v[i]][j];
for (int j = 0; j < 3; j++)
if (div[i][j] * v[i] <= m)
{
div[i][j] *= v[i];
break;
}
}
for (int i = 1; i <= m; i++)
for (int j = 1; j <= i; j++)
g[i][j] = g[j][i] = g[j][i % j];
}
int main()
{
scanf("%d%d", &n, &q);
m = sqrt(n);
init();
for (int a, b; q; q--)
{
scanf("%d%d", &a, &b);
int res = 1, d;
for (int i = 0; i < 3; i++)
if (div[a][i] <= m)
d = g[div[a][i]][b % div[a][i]];
else
d = b % div[a][i] == 0 ? div[a][i] : 1;
res *= d;
b /= d;
printf("%d\n", res);
}
return 0;
}