题解 P2834 【能力测验】

[ans=sum_{i=1}^nsum_{j=1}^m(n;mod;i) imes(m;mod;j),i eq j ]

我们假设

[nleq m ]

[egin{aligned} ans & =sum_{i=1}^nsum_{j=1}^m(n;mod;i) imes(m;mod;j),i eq j \ & =sum_{i=1}^{n}sum_{j=1}^m(n;mod;i) imes(m;mod;j)-sum_{i=1}^n(n;mod;i) imes(m;mod;i) \ & =sum_{i=1}^{n}sum_{j=1}^m(n-leftlfloordfrac{n}{i} ight floorcdot i) imes(m-leftlfloordfrac{m}{j} ight floorcdot j)-sum_{i=1}^n(n-leftlfloordfrac{n}{i} ight floorcdot i) imes(m-leftlfloordfrac{m}{i} ight floorcdot i) \ & =sum_{i=1}^{n}(n-leftlfloordfrac{n}{i} ight floorcdot i)sum_{j=1}^m(m-leftlfloordfrac{m}{j} ight floorcdot j)-sum_{i=1}^n(n-leftlfloordfrac{n}{i} ight floorcdot i) imes(m-leftlfloordfrac{m}{i} ight floorcdot i) \ & =sum_{i=1}^{n}(n-leftlfloordfrac{n}{i} ight floorcdot i)sum_{j=1}^m(m-leftlfloordfrac{m}{j} ight floorcdot j)-sum_{i=1}^n(ncdot m - icdot(leftlfloordfrac{n}{i} ight floorcdot m+leftlfloordfrac{m}{i} ight floorcdot n)+leftlfloordfrac{n}{i} ight floorleftlfloordfrac{m}{i} ight floor 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{icdot(i+1)cdot(2cdot 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;	
}