UVA10820 交表 Send a Table

\(\Large\textbf{Description:} \large{输入n，求有多少个二元组(x,y)满足：1\leqslant x,y\leqslant n，且x和y互素。}\)

\(\Large\textbf{Solution:} \large{不难发现除了(1,1)外，其他二元组的x，y都不相等。设满足x < y的二元组有g(n)个，那么答案为2\times g(n) + 1。}\)

\(\large{其中g(n)=\sum_{i=2}^n\Phi(i)}。\)

\(\Large\textbf{Code:}\)

#include <cmath>
#include <cstdio>
#include <algorithm>
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for (int i = (a); i <= (b); ++i)
using namespace std;
const int N = 5e4 + 5;
int n, cnt, prime[N], phi[N];
LL sum[N];
bool vis[N];

inline int read() {
	char ch = gc();
	int ans = 0;
	while (ch > '9' || ch < '0') ch = gc();
	while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
	return ans;
}

inline void euler() {
	phi[1] = 1;
	rep(i, 2, N) {
		if (!vis[i]) prime[++cnt] = i, phi[i] = i - 1;
		for (int j = 1; j <= cnt && i * prime[j] <= N; ++j) {
			vis[prime[j] * i] = true;
			if (i % prime[j] == 0) {
				phi[i * prime[j]] = phi[i] * prime[j];
				break;
			}
			else phi[i * prime[j]] = phi[i] * (prime[j] - 1);
		}
	}
}

int main() {
	euler();
	rep(i, 1, N) sum[i] = sum[i - 1] + phi[i];
	n = read();
	while (n) {
		printf("%lld\n", 2 * sum[n] - 1);
		n = read();
	}
	return 0;
}

\(\large\color{pink}{by\quad Miraclys}\)