luogu2257 YY的GCD--莫比乌斯反演

标签：ret   int   \n   names   多少   return   clu   cpp   prim   

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给定N, M,求1<=x<=N, 1<=y<=M且gcd(x, y)为质数的(x, y)有多少对

多组数据T = 10000

N, M <= 10000000

推式子

\(\sum_{i=1}^n\sum_{j=1}^m[\gcd(i,j)=p]\)

\(=\sum_p\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}[\gcd(i,j)=1]\)

\(=\sum_p\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}\sum_{d|i,d|j}\mu(d)\)

\(=\sum_{d=1}^n\mu(d)\sum_p\lfloor\frac n{dp}\rfloor\lfloor\frac m{dp}\rfloor\)

令\(q=dp\)

\(=\sum_{q=1}^n(\sum_{p|q}\mu(\frac q p))\lfloor\frac nq\rfloor\lfloor\frac mq\rfloor\)

\(\mu\)线性筛

然后在对于质数枚举倍数求对于每个\(i\)的\(\sum_{p|i}\mu(\frac i p)\)

然后打数论分块就行了

#include <cstdio>
#include <functional>
using namespace std;

const int fuck = 10000000;
int prime[10000010], tot;
bool vis[10000010];
int mu[10000010], sum[10000010];

int main()
{
    mu[1] = 1;
    for (int i = 2; i <= fuck; i++)
    {
        if (vis[i] == false) prime[++tot] = i, mu[i] = -1;
        for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
        {
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0) break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= tot; i++)
        for (int j = 1; j * prime[i] <= fuck; j++)
            sum[j * prime[i]] += mu[j];
    for (int i = 1; i <= fuck; i++)
        sum[i] += sum[i - 1];
    int t; scanf("%d", &t);
    while (t --> 0)
    {
        int n, m;
        long long ans = 0; //别忘了初始化。。。
        scanf("%d%d", &n, &m);
        if (n > m) {int t = m; m = n; n = t; }
        for (int i = 1, j; i <= n; i = j + 1)
        {
            j = min(n / (n / i), m / (m / i));
            ans += (sum[j] - sum[i - 1]) * (long long)(n / i) * (m / i);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

luogu2257 YY的GCD--莫比乌斯反演

标签：ret   int   \n   names   多少   return   clu   cpp   prim   

原文地址：https://www.cnblogs.com/oier/p/10295755.html