能量采集题解-优快云博客

本文链接：https://blog.youkuaiyun.com/C2022lihan/article/details/119894135

这篇博客探讨了在二维平面上，从(0,0)到任意点(x,y)之间挡了多少个点的问题。通过欧几里得算法求最大公约数(gcd)，并利用Euler函数预处理gcd的区间和，博主展示了如何计算gcd的总和，并给出了C++代码实现，该代码使用分块方法高效地计算了所有点对的gcd之和。

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$(0, 0) 和 (x, y) 之间挡了 g c d (x, y) - 1 个点$

$f (x, y) = 2 * (g c d (x, y) - 1) + 1 = 2 * g c d (x, y) - 1$

$_{x=1}^n∑_{y=1}^m(2*gcd(x,y)-1)$

$2*∑_{x=1}^n∑_{y=1}^m gcd(x,y) -n*m$

$目标∑_{x=1}^n∑_{y=1}^m gcd(x,y)$

$=∑x=1n∑y=1m∑d∣gcd(x,y)ϕ(d)=∑_{x=1}^n∑_{y=1}^m∑_{d|gcd(x,y)}\phi(d)$

$=∑(ϕ(d)∗⌊n/d⌋∗⌊m/d⌋)=∑(\phi(d)*\lfloor n/d \rfloor *\lfloor m/d \rfloor)$

$整除分块，预处理ϕ(x)区间和整除分块，预处理\phi(x)区间和$

#include <cstdio>
#include <iostream>
using namespace std;
#define LL long long
#define ULL unsigned long long

template <typename T> void read (T &x) { x = 0; T f = 1;char tem = getchar ();while (tem < '0' || tem > '9') {if (tem == '-') f = -1;tem = getchar ();}while (tem >= '0' && tem <= '9') {x = (x << 1) + (x << 3) + tem - '0';tem = getchar ();}x *= f; return; }
template <typename T> void write (T x) { if (x < 0) {x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0'); }
template <typename T> void print (T x, char ch) { write (x); putchar (ch); }
template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }

const int Maxn = 1e5;

int cnt, primes[Maxn + 5];
int phi[Maxn + 5]; LL pre[Maxn + 5];
bool vis[Maxn + 5];
void Euler () {
	for (int i = 2; i <= Maxn; i++) {
		if (vis[i] == 0) {
			primes[++cnt] = i;
			phi[i] = i - 1;
		}
		for (int j = 1; j <= cnt; j++) {
			if (primes[j] > Maxn / i) break;
			vis[primes[j] * i] = 1;
			if (i % primes[j] == 0) {
				phi[i * primes[j]] = phi[i] * primes[j];
				break;
			}
			phi[i * primes[j]] = phi[i] * (primes[j] - 1);
		}
	}
	phi[1] = 1;
	for (int i = 1; i <= Maxn; i++)
		pre[i] = pre[i - 1] + phi[i];
}

int main () {
	//answer = 2 * ∑∑{gcd (i, j)} - n * m
	Euler ();
	int n, m;
	read (n); read (m);
	
	int l = 1, r;
	LL res = 0;
	while (l <= Min (n, m)) {
		r = Min (n / (n / l), m / (m / l));
		res += (pre[r] - pre[l - 1]) * (n / l) * (m / l);
		l = r + 1;
	}
	write (res * 2 - (LL)n * m);
	return 0;
}