洛谷 2257 - YY的GCD

莫比乌斯反演半模板题

很容易可以得到
\[Ans = \sum\limits_{p \in prime} \sum\limits_{d = 1}^{\min (\left\lfloor\frac{a}{p}\right\rfloor, \left\lfloor\frac{b}{p}\right\rfloor)} \mu(d) \left\lfloor\frac{a}{pd}\right\rfloor\left\lfloor\frac{b}{pd}\right\rfloor\]
那么现在由于想要进行整除分块，所以希望将 \(\sum\) 内部的向下取整部分移到外部，故令 \(T = dp\) ，则有
\[\begin{aligned} Ans &= \sum\limits_{T = 1}^{\min (a, b)} \sum\limits_{p | T, p \in prime} \mu(\left\lfloor\frac{T}{p}\right\rfloor) \left\lfloor\frac{a}{T}\right\rfloor\left\lfloor\frac{b}{T}\right\rfloor \\ &= \sum\limits_{T = 1}^{\min (a, b)} \left\lfloor\frac{a}{T}\right\rfloor\left\lfloor\frac{b}{T}\right\rfloor \left( \sum\limits_{p | T, p \in prime} \mu(\left\lfloor\frac{T}{p}\right\rfloor) \right) \end{aligned}\]
那么用筛法预处理一下 \(\mu\) 的那一部分就可以直接整除分块了

代码

#include 
#include 
#include 

using namespace std;

typedef long long LL;

const int MAXN = 1e07 + 10;

int prime[MAXN];
int vis[MAXN]= {0};
int pcnt = 0;
int mu[MAXN]= {0};
LL tsum[MAXN]= {0}, sum[MAXN]= {0};
const int MAX = 1e07;
void prime_Acqu () {
    mu[1] = 1;
    for (int i = 2; i <= MAX; i ++) {
        if (! vis[i]) {
            prime[++ pcnt] = i;
            mu[i] = - 1;
        }
        for (int j = 1; j <= pcnt && i * prime[j] <= MAX; j ++) {
            vis[i * prime[j]] = 1;
            if (! (i % prime[j]))
                break;
            mu[i * prime[j]] = - mu[i];
        }
    }
    for (int j = 1; j <= pcnt; j ++)
        for (int i = 1; i * prime[j] <= MAX; i ++)
            tsum[i * prime[j]] += mu[i];
    for (int i = 1; i <= MAX; i ++)
        sum[i] = sum[i - 1] + tsum[i];
}

LL Calc (int a, int b) {
    LL ans = 0;
    int limit = min (a, b);
    for (int l = 1, r; l <= limit; l = r + 1) {
        r = min (a / (a / l), b / (b / l));
        ans += (sum[r] - sum[l - 1]) * (a / l) * (b / l);
    }
    return ans;
}

int T;

int getnum () {
    int num = 0;
    char ch = getchar ();

    while (! isdigit (ch))
        ch = getchar ();
    while (isdigit (ch))
        num = (num << 3) + (num << 1) + ch - '0', ch = getchar ();

    return num;
}

int main () {
    prime_Acqu ();
    T = getnum ();
    for (int Case = 1; Case <= T; Case ++) {
        int a = getnum (), b = getnum ();
        LL ans = Calc (a, b);
        printf ("%lld\n", ans);
    }

    return 0;
}

/*
2
10 10
100 100
*/

转载于:https://www.cnblogs.com/Colythme/p/10268784.html