答案相当于全部减去互质的对数,再减去$x,y$存在倍数关系的对数。
互质部分莫反搞一搞,倍数部分就是$\sum_{i=l}^r2*(int)(r/i)-1$
by _jimmywang_ @ 2021-06-19 21:53:36
problem b
by SSerxhs @ 2021-06-19 21:53:41
啊真实
by _jimmywang_ @ 2021-06-19 21:54:54
P2522
by _jimmywang_ @ 2021-06-19 21:55:17
借楼问 F 怎么做,不会 $\text{Grundy number}$ ![e](https://xn--9zr.tk/kel)
by Mobius127 @ 2021-06-19 21:58:05
借楼问我 D 这个贪心怎么 hack
```cpp
#include <iostream>
#define MAXN 200000
#define QWQ cout << "QWQ" << endl;
using namespace std;
int a[MAXN + 10];
int cnt[MAXN + 10];
int ys[MAXN + 10];
int main() {
int n, ans = 0;
cin >> n;
for(int p = 1; p <= n; p++)
cin >> a[p], cnt[a[p]]++, ys[a[p]] = a[p];
for(int p = 1; p <= n; p++) {
if(ys[a[p]] != ys[a[n - p + 1]]) {
ans++;
if(cnt[ys[a[p]]] > cnt[ys[a[n - p + 1]]]) ys[a[n - p + 1]] = ys[a[p]], cnt[ys[a[n - p + 1]]] += cnt[ys[a[p]]];
else ys[a[p]] = ys[a[n - p + 1]], cnt[ys[a[p]]] += cnt[ys[a[n - p + 1]]];
}
}
cout << ans << endl;
}
```
能帮忙 hack 一下吗/kel
还有 E 怎么去重,口胡了一种枚举 l ~ r 的所有数字乘上一个互质数对的做法,用欧拉函数随便搞搞,但是布吉岛怎么去重嘤嘤嘤
by SIXIANG32 @ 2021-06-19 21:59:37
最后二十分钟去洗澡了,没来得及完善 E,500pts 啊/ll/ll/ll
by SIXIANG32 @ 2021-06-19 22:00:31
@[wangjinbo](/user/228843) 没看,但想问ABC相当于CF的什么水平?
by Morgen_Kornblume @ 2021-06-19 22:00:54
@[Lucky_Yukikaze](/user/93701) Div 2.5
by SSerxhs @ 2021-06-19 22:03:49
@[SSerxhs](/user/29826) 那么ARC和AGC呢
by Morgen_Kornblume @ 2021-06-19 22:07:51