P3327 [SDOI2015]约数个数和

Posted 2021-02-06 olinr

\(\color{#0066ff}{ 题目描述 }\)

设\(d(x)\)为\(x\)的约数个数，给定\(N、M\)，求?\(\sum^N_{i=1}\sum^M_{j=1}d(ij)\)

\(\color{#0066ff}{输入格式}\)

输入文件包含多组测试数据。第一行，一个整数T，表示测试数据的组数。接下来的T行，每行两个整数N、M。

\(\color{#0066ff}{输出格式}\)

T行，每行一个整数，表示你所求的答案。

\(\color{#0066ff}{输入样例}\)

2
7 4
5 6

\(\color{#0066ff}{输出样例}\)

110
121

\(\color{#0066ff}{数据范围与提示}\)

1<=N, M<=50000

1<=T<=50000

\(\color{#0066ff}{ 题解 }\)

题目要求

\[ \sum_{i=1}^n\sum_{j=1}^md(ij) \]

看似不好求，然而有结论。。。

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

然后就是套路了

\[ \sum_{i=1}^n\sum_{j=1}^m\sum_{x|i}\sum_{y|j} \sum_{d|gcd(x, y)} \mu(d) \]

枚举x和y，注意还要是i和j的约数

\[ \sum_{d = 1}^{min(n,m)}\sum_{i=1}^n\sum_{j=1}^m\sum_{x=1}^{\lfloor\frac i d \rfloor}\sum_{y=1}^{\lfloor\frac j d \rfloor}[xd|i\&\&yd|j] \mu(d) \]

能提前的全部提前

\[ \sum_{d = 1}^{min(n,m)}\mu(d)\sum_{i=1}^n\sum_{j=1}^m\sum_{x=1}^{\lfloor\frac i d \rfloor}\sum_{y=1}^{\lfloor\frac j d \rfloor}[xd|i\&\&yd|j] \]

\[ \sum_{d = 1}^{min(n,m)}\mu(d)\sum_{i=1}^n\sum_{x=1}^{\lfloor\frac i d \rfloor}[xd|i]\sum_{j=1}^m\sum_{y=1}^{\lfloor\frac j d \rfloor}[yd|j] \]

最后就成这东西了

\[ \sum_{d = 1}^{min(n,m)}\mu(d)\sum_{x=1}^{\lfloor\frac n d \rfloor}\lfloor\frac{n}{xd}\rfloor\sum_{y=1}^{\lfloor\frac m d \rfloor}\lfloor\frac{m}{yd}\rfloor \]

把\(\lfloor\frac{n}{d}\rfloor\)当成一个整体，那么其实就是\(\begin{aligned}\sum_{x=1}^n \lfloor\frac{n}{x}\rfloor\end{aligned}\)

直接数列分块即可

#include<bits/stdc++.h>
#define LL long long
LL in() {
    char ch; LL x = 0, f = 1;
    while(!isdigit(ch = getchar()))(ch == '-') && (f = -f);
    for(x = ch ^ 48; isdigit(ch = getchar()); x = (x << 1) + (x << 3) + (ch ^ 48));
    return x * f;
}
const int maxn = 5e4 + 100;
LL pri[maxn], mu[maxn], cnt[maxn], ans[maxn], tot;
bool vis[maxn];
void predoit() {
    mu[1] = 1;
    for(int i = 2; i < maxn; i++) {
        if(!vis[i]) pri[++tot] = i, mu[i] = -1;
        for(int j = 1; j <= tot && (LL)i * pri[j] < maxn; j++) {
            vis[i * pri[j]] = true;
            if(i % pri[j] == 0) break;
            else mu[i * pri[j]] = -mu[i];
        } 
    }
    for(int i = 1; i < maxn; i++) {
        mu[i] += mu[i - 1];
        for(LL l = 1, r; l <= i; l = r + 1) {
            r = i / (i / l);
            ans[i] += (r - l + 1) * (i / l);
        }
    }
}
LL work(LL n, LL m) {
    LL res = 0;
    for(LL l = 1, r; l <= std::min(n, m); l = r + 1) {
        r = std::min(n / (n / l), m / (m / l));
        res += (mu[r] - mu[l - 1]) * ans[n / l] * ans[m / l];
    }
    return res;
}
int main() {
    predoit();
    for(int T = in(); T --> 0;) printf("%lld\n", work(in(), in()));
    return 0;
}