BZOJ 2818 GCD - NekoMio's Blog

Description#

给定整数N，求1<=x,y<=N且Gcd(x,y)为素数的数对(x,y)有多少对.

Input#

一个整数N

Output#

如题

Sample Input#

4

Sample Output#

4

题解#

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

$= \sum_{d}{\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}{\sum_{j=1}^{\lfloor\frac{n}{d}\rfloor}{[gcd(i,j)=1]}}}$

$= \sum_{T}{\lfloor\frac{n}{T}\rfloor \lfloor\frac{n}{T}\rfloor\sum_{d|T}{\mu(\frac{n}{T})}}$
令 $f[i]=\sum_{d|T}^{} \mu(\frac{T}{d})$

然后线性筛

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#include <algorithm>
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#include <cstdio>
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#include <cstring>
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using namespace std;
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const int N = 10000000;
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int mu[N + 5], prime[N + 5], g[N + 5];
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long long sum[N + 5], cnt;
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bool isnprime[N + 5];
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void get_g()
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{
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    mu[1] = 1;
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    for (int i = 2; i <= N; i++)
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    {
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        if (!isnprime[i])
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        {
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            prime[++cnt] = i;
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            mu[i] = -1;
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            g[i] = 1;
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        }
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        for (int j = 1; j <= cnt && i * prime[j] <= N; j++)
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        {
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            isnprime[i * prime[j]] = 1;
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            if (i % prime[j])
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                mu[i * prime[j]] = -mu[i], g[i * prime[j]] = mu[i] - g[i];
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            else
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            {
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                mu[i * prime[j]] = 0;
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                g[i * prime[j]] = mu[i];
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                break;
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            }
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        }
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    }
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    for (int i = 1; i <= N; i++)
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    {
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        sum[i] = sum[i - 1] + g[i];
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    }
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}
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int main()
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{
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    get_g();
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    int n;
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    scanf("%d", &n);
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    long long ans = 0, last;
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    for (int i = 1; i <= n; i = last + 1)
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    {
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        last = n / (n / i);
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        ans += (long long)(sum[last] - sum[i - 1]) * (n / i) * (n / i);
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    }
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    printf("%lld\n", ans);
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}