洛谷 P1829 [国家集训队]Crash的数字表格 / JZPTAB

题意

求

$$\sum _{i=1}^n\sum _{j=1}^m\text{lcm}(i,j)(\mathrm{m}\mathrm{o}\mathrm{d}20101009)$$

思路

容易想到原式等价于

$$\sum _{i=1}^n\sum _{j=1}^m\frac{i\ast j}{\gcd (i,j)}$$

枚举$i,j$的最大公约数$d$，显然$\gcd (\frac{i}{d},\frac{j}{d})=1$，即$\frac{i}{d}$和$\frac{j}{d}$互质

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{d|i,d|j,\gcd (\frac{i}{d},\frac{j}{d})=1}\frac{i\ast j}{d}$$

变换求和顺序

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

记$sum(n,m)=\sum _{i=1}^n\sum _{j=1}^m[\gcd (i,j)=1]i\ast j$

对其进行化简，用$\epsilon (\gcd (i,j))$替换$[\gcd (i,j)=1]$

$$\sum _{i=1}^n\sum _{j=1}^m\sum _{d|\gcd (i,j)}\mu (d)\ast i\ast j$$

转化为首先枚举约数

$$\sum _{d=1}^{\min (n,m)}\sum _{d|i}^n\sum _{d|j}^m\mu (d)\ast i\ast j$$

设$i=i^{\prime }\ast d,j=j^{\prime }\ast d$，则可以进一步转化

$$\sum _{d=1}^{\min (n,m)}\mu (d)\ast d^2\ast \sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{m}{d}\rfloor }i\ast j$$

前半段可以处理前缀和，后半段可以$O(1)$求，设

$$Q(n,m)=\sum _{i=1}^n\sum _{j=1}^{m}i\ast j=\frac{n\ast (n+1)}{2}\ast \frac{m\ast (m+1)}{2}$$

显然可以$O(1)$求解

到现在

$$sum(n,m)=\sum _{d=1}^{\min (n,m)}\mu (d)\ast d^2\ast Q(\lfloor \frac{n}{d}\rfloor ,\lfloor \frac{m}{d}\rfloor )$$

可以用数论分块求解

回带到原式中

$$\sum _{d=1}^{\min (n,m)}d\ast sum(\lfloor \frac{n}{d}\rfloor ,\lfloor \frac{m}{d}\rfloor )$$

又可以数论分块求解了

然后就做完啦

代码

/*
Author:loceaner
*/
#include 
#include 
#include 
#include 
#include 
#define int long long
using namespace std;

const int A = 1e7 + 11;
const int B = 1e6 + 11;
const int mod = 20101009;
const int inf = 0x3f3f3f3f;

inline int read() {
	char c = getchar(); int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = x * 10 + (c ^ 48);
	return x * f;
}

bool vis[A];
int n, m, mu[A], p[B], sum[A], cnt;

void getmu() {
    mu[1] = 1;
    int k = min(n, m);
    for (int i = 2; i <= k; i++) {
        if (!vis[i]) p[++cnt] = i, mu[i] = -1;
        for (int j = 1; j <= cnt && i * p[j] <= k; ++j) {
            vis[i * p[j]] = 1;
            if (i % p[j] == 0) break;
            mu[i * p[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= k; i++) sum[i] = (sum[i - 1] + i * i % mod * mu[i]) % mod;
}

int Sum(int x, int y) { 
	return (x * (x + 1) / 2 % mod) * (y * (y + 1) / 2 % mod) % mod; 
}

int solve2(int x, int y) {
    int res = 0;
    for (int i = 1, j; i <= min(x, y); i = j + 1) {
        j = min(x / (x / i), y / (y / i));
        res = (res + 1LL * (sum[j] - sum[i - 1] + mod) * Sum(x / i, y / i) % mod) % mod;
    }
    return res;
}

int solve(int x, int y) {
    int res = 0;
    for (int i = 1, j; i <= min(x, y); i = j + 1) {
        j = min(x / (x / i), y / (y / i));
        res = (res + 1LL * (j - i + 1) * (i + j) / 2 % mod * solve2(x / i, y / i) % mod) % mod;
    }
    return res;
}

signed main() {
    n = read(), m = read();
    getmu();
    cout << solve(n, m) << '\n';
}