东莞企业网站后缀,如何快速推广自己的品牌,公司名字大全英文,做网站后台要学最小公倍数之和
题目描述#xff1a;
对于A1#xff0c;A2…AN#xff0c;求 ∑i1N∑i1Nlcm(Ai,Aj)\sum_{i1}^{N}\sum_{i1}^{N} lcm(Ai,Aj)∑i1N∑i1Nlcm(Ai,Aj)
题解#xff1a;
莫比乌斯反演#xff0c;直接强推一波 推导过程我也是一知半解#xff0c;大体如图…最小公倍数之和
题目描述
对于A1A2…AN求 ∑i1N∑i1Nlcm(Ai,Aj)\sum_{i1}^{N}\sum_{i1}^{N} lcm(Ai,Aj)∑i1N∑i1Nlcm(Ai,Aj)
题解
莫比乌斯反演直接强推一波 推导过程我也是一知半解大体如图 然后预处理f(T)即可 2021/8/21时隔一年再更
代码
#includebits/stdc.h
using namespace std;
#define ll long long
#define maxn 50005
#define rgt registerint N, M, cnt[maxn], mu[maxn], p[maxn], tot, v[maxn];
ll s[maxn];
ll ans0;int main(){scanf( %d, N ); for ( rgt int i 1, x; i N; i ) scanf( %d, x ), cnt[x], M max( M, x );N M, mu[1] 1;for ( rgt int i 2; i N; i ){//线性筛出muif ( !v[i] ) p[tot] i, mu[i] -1;for ( rgt int j 1; j tot i * p[j] N; j ){v[i * p[j]] 1;if ( i % p[j] 0 ){ mu[i * p[j]] 0; break; }else mu[i * p[j]] -mu[i];}}for ( rgt int i 1; i N; i )for ( rgt int j i; j N; j i )s[j] 1ll * mu[i] * i;//预处理提到过的那玩意for ( rgt int T 1; T N; T ){rgt ll cur(0);for ( rgt int i 1, I N / T; i I; i ) cur 1ll * cnt[i * T] * i;//暴力求解ans T * cur * cur * s[T];} printf( %lld\n, ans );return 0;
}
