[hdu6434]Problem I. Count

标签：题目   void   mit   while   bre   splay   arp   style   cpp   

题目大意：$T(T\leqslant 10^5)$组数据，每组数据给你$n(n\leqslant 2\times 10^7)$，求$\sum\limits_{i=1}^n\sum\limits_{j=1}^{i-1}[(i+j,i-j)==1]$

题解：
$$
\def\dsum{\displaystyle\sum\limits}
\begin{align*}
&\dsum_{i=1}^n\dsum_{j=1}^{i-1}[(i+j,i-j)=1]\\
&令k=i-j\\
=&\dsum_{i=1}^n\dsum_{k=1}^{i-1}[(2i-k,k)=1]\\
=&\dsum_{i=1}^n\dsum_{k=1}^{i-1}[(2i,k)=1]\\
\end{align*}
$$

$$
\therefore
(2i,k)=1\Rightarrow
\begin{cases}
(i,k)=1\\
(2,k)=1\\
\end{cases}\\
当i为偶数时：\\
(i,k)=1\\
\Rightarrow (2,k)=1\\
\therefore ans=\varphi(i)\\
当i为奇数时：\\
(2,k)=1\\
\Rightarrow k为奇数\\
\therefore k必为与i互质的数的奇数\\
\because (i,k)=1\Rightarrow(i,i-k)=1\\
\therefore 当i为奇数的时候，k奇偶各半\\
\therefore ans=\dfrac{\varphi(i)}2
$$

卡点：无

C++ Code：

#include <cstdio>
#define maxn 20000010
int Tim, n;
long long pre[maxn];
int plist[maxn << 3], pt, phi[maxn];
bool isp[maxn];
void sieve(int n) {
	phi[1] = 1;
	pre[1] = 0;
	isp[1] = true;
	for (int i = 2; i < n; i++) {
		if (!isp[i]) {
			plist[pt++] = i;
			phi[i] = i - 1;
		}
		for (int j = 0; j < pt && i * plist[j] < n; j++) {
			int tmp = i * plist[j];
			isp[tmp] = true;
			if (i % plist[j] == 0) {
				phi[tmp] = phi[i] * plist[j];
				break;
			}
			phi[tmp] = phi[i] * phi[plist[j]];
		}
		pre[i] = pre[i - 1] + ((i & 1) ? phi[i] / 1 : phi[i]);
	}
}
int main() {
	sieve(maxn);
	scanf("%d", &Tim);
	while (Tim --> 0) {
		scanf("%d", &n);
		printf("%lld\n", pre[n]);
	}
	return 0;
}

[hdu6434]Problem I. Count

标签：题目   void   mit   while   bre   splay   arp   style   cpp   

原文地址：https://www.cnblogs.com/Memory-of-winter/p/9671153.html