标签：highlight space ++ 答案 uva algorithm span prime can
思路
将题意转化为$\sum_{i = 1}^{n} \sum_{j = 1}^{i - 1}gcd(i, j)$，考虑每个最大公因数的值$k$对答案的影响。假设
$gcd(A,B) = k$
那么肯定可以表示成
$gcd(ak,bk) = k$
$gcd(a, b) = 1$
假设$a>b$那么$b$有$\varphi (a)$种选法且$ak \leq n$等价于$a \leq \left \lfloor \frac{n}{k} \right \rfloor$，因此对于一种$k$，它对答案的贡献为
$\sum_{a = 2}^{\left \lfloor \frac{n}{k} \right \rfloor} k*\varphi (a)$
$k \sum_{a = 1}^{\left \lfloor \frac{n}{k} \right \rfloor} \varphi (a) - k$
$k (\sum_{a = 1}^{\left \lfloor \frac{n}{k} \right \rfloor} \varphi (a) - 1)$
所以
$ans = \sum_{k = 1}^{n} k (\sum_{a = 1}^{\left \lfloor \frac{n}{k} \right \rfloor} \varphi (a) - 1)$
维护欧拉函数前缀和之后整除分块，时间复杂度$O(n+\sqrt{n})$
代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define DBG(x) cerr << #x << " = " << x << endl
using namespace std;
typedef long long LL;
const int N = 1e7 + 5;
int n;
int tag[N], pri[N], cnt;
LL phi[N];
LL pre[N];
void getPrime() {
phi[1] = 1;
for(int i = 2; i < N; i++) {
if(!tag[i]) {
pri[cnt++] = i;
phi[i] = i - 1;
}
for(int j = 0; j < cnt && i * pri[j] < N; j++) {
tag[i * pri[j]] = 1;
if(i % pri[j] == 0) {
phi[i * pri[j]] = phi[i] * pri[j];
break;
}
phi[i * pri[j]] = phi[i] * (pri[j] - 1);
}
}
for(int i = 1; i < N; i++) pre[i] = pre[i - 1] + phi[i];
}
int main() {
getPrime();
while(~scanf("%d", &n) && n != 0) {
LL ans = 0;
for(LL l = 1, r; l <= n; l = r + 1) {
r = 1LL * n / (1LL * n / l);
LL tmp1 = ((r + l) * (r - l + 1)) / 2LL;
LL tmp2 = pre[n / l] - 1LL;
ans += tmp1 * tmp2;
}
printf("%lld\n", ans);
}
return 0;
}
/*
10
100
200000
0
*/
UVA - 11426 GCD - Extreme (II) (欧拉函数)
标签：highlight space ++ 答案 uva algorithm span prime can
原文地址：https://www.cnblogs.com/DuskOB/p/10725591.html
