luogu3312 [SDOI2014]数表 (莫比乌斯反演+树状数组)

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

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

\(=\sum_{p=1}^ns(p)[s(p)\le a]\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}[\gcd(i,j)=1]\)

\(=\sum_{p=1}^ns(p)[s(p)\le a]\sum_{i=1}^{n/p}\sum_{j=1}^{m/p}\sum_{d|i,d|j}\mu(d)\)

\(=\sum_{p=1}^ns(p)[s(p)\le a]\sum_{d=1}^n\mu(d)\lfloor\frac n{pd}\rfloor\lfloor\frac m{pd}\rfloor\)

\(=\sum_{q=1}^n\sum_{p|q}s(p)[s(p)\le a]\mu(\frac q p)\lfloor\frac n{pd}\rfloor\lfloor\frac m{pd}\rfloor\)

离线，将询问按照\(a\)排序

由于前面最多只有nlogn个，可以线性筛之后都存一下，存一个三元组(p, s(p), 那一大坨子)，按照s(p)排序

离线处理询问，往树状数组里插值就行了，每次相当于在树状数组里查询前缀和之差，和普通的整除分块没什么太大的区别

#include <cstdio>
#include <algorithm>
#include <utility>
using namespace std;

struct query
{
    int n, m, a, id, ans;
} ask[20010];

struct info
{
    int val, id;
}  inf[100010];

int q;
bool vis[100010];
int d[100010], d1[100010], mu[100010], prime[100000], tot, fuck = 100000;

int c[100010];

void chenge(int x, int y)
{
    for (int i = x; i <= fuck; i += i & -i) c[i] += y;
}

int getsum(int x)
{
    int ans = 0;
    for (int i = x; i > 0; i -= i & -i) ans += c[i];
    return ans;
}

void add(int p)
{
    for (int q = p, dd = 1; q <= fuck; q += p, dd++)
        chenge(q, d[p] * mu[dd]);
}

int main()
{
    scanf("%d", &q);
    for (int i = 1; i <= q; i++)
    {
        scanf("%d%d%d", &ask[i].n, &ask[i].m, &ask[i].a);
        ask[i].id = i;
    }
    sort(ask + 1, ask + 1 + q, [](const query &a, const query &b) { return a.a < b.a; });
    
    mu[1] = d[1] = d1[1] = 1;
    for (int i = 2; i <= fuck; i++)
    {
        if (vis[i] == false) prime[++tot] = i, mu[i] = -1, d[i] = d1[i] = i + 1;
        for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
        {
            vis[i * prime[j]] = true;
            if (i % prime[j] == 0)
            {
                d1[i * prime[j]] = d1[i] * prime[j] + 1;
                d[i *prime[j]] = d[i] / d1[i] * d1[i * prime[j]];
                break;
            }
            d1[i * prime[j]] = prime[j] + 1;
            d[i * prime[j]] = d[i] * (prime[j] + 1);
            mu[i * prime[j]] = -mu[i];
        }
    }
    
    for (int i = 1; i <= fuck; i++)
        inf[i].id = i, inf[i].val = d[i];
    
    sort(inf + 1, inf + 1 + fuck, [](const info &a, const info &b) { return a.val < b.val; });
    
    for (int i = 1, j = 1; i <= q; i++)
    {
        while (j <= fuck && inf[j].val <= ask[i].a) { add(inf[j].id), j++; }
        int n = ask[i].n, m = ask[i].m; if (n > m) swap(n, m);
        int ans = 0;
        for (int i = 1, j; i <= n; i = j + 1)
        {
            j = min(n / (n / i), m / (m / i));
            ans += (getsum(j) - getsum(i - 1)) * (n / i) * (m / i);
        }
        ask[i].ans = ans;
    }
    
    sort(ask + 1, ask + 1 + q, [](const query &a, const query &b) { return a.id < b.id; });
    for (int i = 1; i <= q; i++) printf("%d\n", ask[i].ans & 2147483647);
    return 0;
}

题目要求对2^31取模，别忘了自然溢出最后对2147483647取一下and