题解[SDOI2014]数表

P3312 [SDOI2014]数表

令 \(g(i)\) 表示 \(i\) 的所有约数和

题目要求的是 \(\sum\limits_{i=1}^n\sum\limits_{j=1}^mg(gcd(i,j)),g(gcd(i,j))\le a\)

令 \(f(i)=\sum\limits_{x=1}^n\sum\limits_{y=1}^m[gcd(x,y)=i],F(i)=\sum\limits_{x=1}^n\sum\limits_{y=1}^m[i|gcd(x,y)]\)

那么 \(F(i)=\sum\limits_{i|d}f(d)\)

由莫比乌斯反演得 \(f(i)=\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)F(d)=\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)\sum\limits_{x=1}^n\sum\limits_{y=1}^m{d|gcd(x,y)}\\=\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\)

先不考虑 \(a\),枚举所有的约数

那么 \(\sum\limits_{i=1}^n\sum\limits_{j=1}^mg(gcd(i,j))=\sum\limits_{d=1}^{min(n,m)}g(i)f(i)\\=\sum\limits_{i=1}^{min(n,m)}g(i)\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor=\sum\limits_{d=1}^{min(n,m)}\lfloor\frac{n}{d}\rfloor\lfloor\frac{m}{d}\rfloor\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)g(i)\)

我们可以现预处理出来所有的约数和,离线处理询问,按照 \(a\) 排序,当约数和小于当前寻问的 \(a\) 时,考虑它对答案的影响,若当前为 \(i\),那么他会对 \(2i,3i...ki\) 位置的数造成影响,考虑用树状数组来维护 \(\sum\limits_{i|d}\mu(\lfloor\frac{d}{i}\rfloor)g(i)\)每一次询问前更新树状数组,然后查询前缀和即可

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#define lowbit(x) (-(x) & (x))
using namespace std;
const int N = 1e5 + 5;
typedef long long ll;
const ll MOD = (ll)1 << 31;

bool st[N];
ll d[N], tr[N], ans[N];
int prime[N], mu[N], tot;

void add(int x, int k)
{
    while (x < N)
    {
        tr[x] += k;
        x += lowbit(x);
    }
}

ll query(int x)
{
    ll res = 0;
    while (x)
    {
        res += tr[x];
        x -= lowbit(x);
    }
    return res;
}

void add2(int x)
{
    for (int i = 1; i * x < N; i++)
        add(i * x, mu[i] * d[x]);
}

struct que
{
    int n, m, a, id;
    void init(int i)
    {
        scanf("%d%d%d", &n, &m, &a);
        id = i;
    }
    bool operator < (const que temp) const
    {
        return a < temp.a;
    }
} Q[N], di[N];

void init()
{
    mu[1] = 1;
    for (int i = 2; i < N; i++)
    {
        if (!st[i])
            prime[++tot] = i, mu[i] = -1;
        for (int j = 1; i * prime[j] < N; j++)
        {
            st[i * prime[j]] = true;
            if (i % prime[j] == 0)
                break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i = 1; i < N; i++)
        for (int j = i; j < N; j += i)
            d[j] += i;
    for (int i = 1; i < N; i++)
        di[i].id = i, di[i].a = d[i];
    sort(di + 1, di + N);
}

int main()
{   
    init();
    int q;
    scanf("%d", &q);
    for (int i = 1; i <= q; i++)
        Q[i].init(i);
    sort(Q + 1, Q + 1 + q);
    int t = 0;
    for (int i = 1; i <= q; i++)
    {
        while (di[t + 1].a <= Q[i].a && t + 1 < N)
            t++, add2(di[t].id);
        int n = Q[i].n, m = Q[i].m;
        int k = min(n, m);
        for (int l = 1, r; l <= k; l = r + 1)
        {
            r = min(k, min(n / (n / l), m / (m / l)));
            ans[Q[i].id] += (ll)(n / l) * (m / l) * (query(r) - query(l - 1));
        }
    }
    for (int i = 1; i <= q; i++)
        printf("%lld\n", ans[i] % MOD);
    return 0;
}

标签：lfloor,frac,gcd,limits,题解,sum,rfloor,SDOI2014,数表
来源： https://www.cnblogs.com/A2484337545/p/14679501.html