题解 P2834 【能力测验】

\[ans=\sum_{i=1}^n\sum_{j=1}^m(n\;mod\;i)\times(m\;mod\;j),i\neq j \]

我们假设

\[n\leq m \]

\[\begin{aligned} ans & =\sum_{i=1}^n\sum_{j=1}^m(n\;mod\;i)\times(m\;mod\;j),i\neq j \\ & =\sum_{i=1}^{n}\sum_{j=1}^m(n\;mod\;i)\times(m\;mod\;j)-\sum_{i=1}^n(n\;mod\;i)\times(m\;mod\;i) \\ & =\sum_{i=1}^{n}\sum_{j=1}^m(n-\left\lfloor\dfrac{n}{i}\right\rfloor\cdot i)\times(m-\left\lfloor\dfrac{m}{j}\right\rfloor\cdot j)-\sum_{i=1}^n(n-\left\lfloor\dfrac{n}{i}\right\rfloor\cdot i)\times(m-\left\lfloor\dfrac{m}{i}\right\rfloor\cdot i) \\ & =\sum_{i=1}^{n}(n-\left\lfloor\dfrac{n}{i}\right\rfloor\cdot i)\sum_{j=1}^m(m-\left\lfloor\dfrac{m}{j}\right\rfloor\cdot j)-\sum_{i=1}^n(n-\left\lfloor\dfrac{n}{i}\right\rfloor\cdot i)\times(m-\left\lfloor\dfrac{m}{i}\right\rfloor\cdot i) \\ & =\sum_{i=1}^{n}(n-\left\lfloor\dfrac{n}{i}\right\rfloor\cdot i)\sum_{j=1}^m(m-\left\lfloor\dfrac{m}{j}\right\rfloor\cdot j)-\sum_{i=1}^n(n\cdot m - i\cdot(\left\lfloor\dfrac{n}{i}\right\rfloor\cdot m+\left\lfloor\dfrac{m}{i}\right\rfloor\cdot n)+\left\lfloor\dfrac{n}{i}\right\rfloor\left\lfloor\dfrac{m}{i}\right\rfloor i^2) \end{aligned} \]

然后

前面的等式先用数论分块算出j循环的值，然后第二次数论分块分i的值，从而将前面的等式算出来。

后面的等式直接数论分块

r = min(n / (n / l), m / (m / l));

换个样子就可以分了。

然后要用平方和公式，所以得先把2,6的逆元算出来

\[\sum_{i=1}^ni^2=\frac{i\cdot(i+1)\cdot(2\cdot i +1)}{6} \]

剩下的就没什么了

#include<cstdio>
#define Starseven main
#define ll long long
const ll mod = 19940417;

ll Add(ll l, ll r) {
	ll re = 0, ny2 = 9970209;
	re = (1 + r) * r % mod * ny2 % mod;
	re %= mod;
	l--;
	re = (re - (l + 1) * l % mod * ny2 % mod) % mod;
	re = (re + mod) % mod;
	return re;
}

ll All(ll l, ll r) {
	ll ny6 = 3323403;
	ll re = 0;
	re = r * (r + 1) % mod * ((2 * r + 1) % mod) % mod * ny6 % mod;
	l--;
	re = (re - (l * (l + 1) % mod *((2 * l + 1) % mod) % mod * ny6 % mod)) % mod;
	re = (re + mod) % mod;
	return re; 
}

int Starseven(void) {
	ll n, m;
	read(n);
	read(m);
	if(n > m) swap(n, m);
	ll ans = 0, judge = 0;
	for (ll l = 1, r; l <= m; l = r + 1) {
		r = m / (m / l);
		ll hack = (r - l + 1) * m % mod;
		hack = (hack - (m / l) * Add(l, r) % mod) % mod;
		hack = (hack + mod) % mod;
		judge = (judge + hack) % mod;
	} 
	for (ll l = 1, r; l <= n; l = r + 1) {
		r = n / (n / l);
		ll hack = (r - l + 1) * n % mod;
		hack = (hack - (n / l) * Add(l, r) % mod) % mod;
		hack = (hack + mod) % mod;
		hack = hack * judge % mod;
		ans = (ans + hack) % mod;
	}
	for (ll l = 1, r; l <= n; l = r + 1) {
		r = min(n / (n / l), m / (m / l));
		ll hack = n * m % mod * (r - l + 1) % mod;
		hack += (n / l) * (m / l) % mod * All(l, r) % mod;
		hack %= mod;
		ll a = m * (n / l) % mod, b = n * (m / l) % mod;
		hack = (hack - Add(l, r) * (a + b) % mod) % mod;
		hack = (hack + mod) % mod;
		ans = (ans - hack) % mod;
		ans = (ans + mod) % mod;
	}
	write(ans);
	puts("");
	return 0;	
}