题意
求
\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\text{lcm}(i,j)(\bmod 20101009) \]
思路
容易想到原式等价于
\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\frac{i* j}{\gcd(i,j)} \]
枚举\(i,j\)的最大公约数\(d\)，显然\(\gcd(\frac id,\frac jd)=1\)，即\(\frac id\)和\(\frac jd\)互质
\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^m\sum\limits_{d|i,d|j,\gcd(\frac id,\frac jd)=1}\frac{i*j}d \]
变换求和顺序
\[\sum\limits_{d=1}^{n}d\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}[\gcd(i,j)=1]i*j \]
记\(sum(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}[\gcd(i,j)=1]i*j\)
对其进行化简，用\(\varepsilon(\gcd(i,j))\)替换\([\gcd(i,j)=1]\)
\[\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}\sum\limits_{d|\gcd(i,j)}\mu(d)*i*j \]
转化为首先枚举约数
\[\sum\limits_{d=1}^{\min(n,m)}\sum\limits_{d|i}^{n}\sum\limits_{d|j}^{m}\mu(d)*i*j \]
设\(i=i'*d,j=j'*d\)，则可以进一步转化
\[\sum\limits_{d=1}^{\min(n,m)}\mu(d)*d^2*\sum\limits_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{d}\rfloor}i*j \]
前半段可以处理前缀和，后半段可以\(O(1)\)求，设
\[Q(n,m)=\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m}i*j=\frac{n*(n+1)}{2}*\frac{m*(m+1)}{2} \]
显然可以\(O(1)\)求解
到现在
\[sum(n,m)=\sum\limits_{d=1}^{\min(n,m)}\mu(d)*d^2*Q(\lfloor\frac nd \rfloor,\lfloor\frac md\rfloor) \]
可以用数论分块求解
回带到原式中
\[\sum\limits_{d=1}^{\min(n, m)}d*sum(\lfloor\frac nd \rfloor,\lfloor\frac md\rfloor) \]
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';
}