记录编号 |
600467 |
评测结果 |
AAAAAAAAAA |
题目名称 |
[BZOJ 2818]GCD |
最终得分 |
100 |
用户昵称 |
LikableP |
是否通过 |
通过 |
代码语言 |
C++ |
运行时间 |
0.714 s |
提交时间 |
2025-05-04 16:51:58 |
内存使用 |
38.97 MiB |
显示代码纯文本
#include <cstdio>
#include <vector>
typedef long long ll;
using ::std::vector;
const int MAXN = 1e7 + 10;
bool vis[MAXN];
vector <ll> primes;
ll phi[MAXN], pre[MAXN];
void solve(int n) {
phi[0] = 0, phi[1] = 1;
for (int i = 2; i <= n; ++i) {
if (!vis[i]) primes.push_back(i), phi[i] = i - 1;
for (auto p : primes) {
if (i * (long long)p > n) break;
vis[i * p] = 1;
if (i % p == 0) {
phi[i * p] = phi[i] * p;
break;
} else {
phi[i * p] = phi[i] * (p - 1);
}
}
}
for (int i = 1; i <= n; ++i) {
pre[i] = pre[i - 1] + phi[i];
}
}
int n;
ll ans;
int main() {
freopen("gcd_prime.in", "r", stdin);
freopen("gcd_prime.out", "w", stdout);
scanf("%d", &n);
solve(n);
for (auto p : primes) {
ans += 2 * pre[n / p] - 1;
}
printf("%lld", ans);
return 0;
}