[国家集训队]Crash的数字表格

题意

求 (sumlimits_{i = 1}^N sumlimits_{j = 1}^M lcm (i, j))

Solution

易知，原式

[sumlimits_{i = 1}^N sumlimits_{j = 1}^M frac{ij}{gcd (i, j)} ]

枚举 (gcd (i, j)) ，且将 (d) 提出来得

[sumlimits_{d = 1}^{min (N, M)} d sumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor} sumlimits_{j = 1}^{leftlfloorfrac{M}{d} ight floor} ij[(i, j) = 1] ]

将公式 (sumlimits_{k | n} mu(k) = [n = 1]) 代入，得

[sumlimits_{d = 1}^{min (N, M)} d sumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor} sumlimits_{j = 1}^{leftlfloorfrac{M}{d} ight floor} ij sumlimits_{k | (i, j)} mu(k) ]

套路枚举 (k) ，得

[sumlimits_{d = 1}^{min (N, M)} d sumlimits_{k = 1}^{min (leftlfloorfrac{N}{d} ight floor, leftlfloorfrac{M}{d} ight floor)} mu(k) sumlimits_{i = 1}^{leftlfloorfrac{N}{d} ight floor} sumlimits_{j = 1}^{leftlfloorfrac{M}{d} ight floor} ij [k | (i, j)] ]

那么 (ij) 存在贡献时其必定是 (k) 的倍数，故

[sumlimits_{d = 1}^{min (N, M)} d sumlimits_{k = 1}^{min (leftlfloorfrac{N}{d} ight floor, leftlfloorfrac{M}{d} ight floor)} mu(k) sumlimits_{ki = 1}^{leftlfloorfrac{N}{d} ight floor} sumlimits_{kj = 1}^{leftlfloorfrac{M}{d} ight floor} k^2 ij ]

将 (k) 提出，得

[sumlimits_{d = 1}^{min (N, M)} d sumlimits_{k = 1}^{min (leftlfloorfrac{N}{d} ight floor, leftlfloorfrac{M}{d} ight floor)} k^2 mu(k) ( sumlimits_{i = 1}^{leftlfloorfrac{N}{kd} ight floor} i) (sumlimits_{j = 1}^{leftlfloorfrac{M}{kd} ight floor} j) ]

那么就可以预处理 (sumlimits_{k = 1}^n k^2 mu(k)) ，后面的用整除分块就好了

Code

#include <iostream>
#include <cstdio>
#include <cstring>

#define MOD 20101009

using namespace std;

typedef long long LL;

const int MAXN = 1e07 + 10;

int prime[MAXN];
int vis[MAXN]= {0};
int pcnt = 0;
int mu[MAXN]= {0};
LL sum[MAXN]= {0};
const int MAX = 1e07;
void prime_Acqu () {
	mu[1] = 1;
	for (int i = 2; i <= MAX; i ++) {
		if (! vis[i]) {
			prime[++ pcnt] = i;
			mu[i] = - 1;
		}
		for (int j = 1; j <= pcnt && i * prime[j] <= MAX; j ++) {
			vis[i * prime[j]] = 1;
			if (! (i % prime[j]))
				break;
			mu[i * prime[j]] = - mu[i];
		}
	}
	for (int i = 1; i <= MAX; i ++)
		sum[i] = (sum[i - 1] + 1ll * i * 1ll * i % MOD * mu[i] % MOD) % MOD;
}

int N, M;
inline LL calc (int n) {
	LL fn = (LL) n;
	return (fn * (fn + 1) >> 1) % MOD;
}
LL Solve () {
	LL ans = 0;
	int limit = min (N, M);
	for (int d = 1; d <= limit; d ++) {
		LL total = 0;
		int minlim = min (N / d, M / d);
		for (int l = 1, r; l <= minlim; l = r + 1) {
			r = min ((N / d) / ((N / d) / l), (M / d) / ((M / d) / l));
			total = (total + (sum[r] - sum[l - 1] + MOD) % MOD * calc (N / d / l) % MOD * calc (M / d / l) % MOD) % MOD;
		}
		ans = (ans + (LL) (d) * total % MOD) % MOD;
	}
	return ans;
}

int main () {
	prime_Acqu ();
	scanf ("%d%d", & N, & M);
	LL ans = Solve ();
	cout << ans << endl;

	return 0;
}

/*
4 5
*/