P3911 最小公倍数之和
题目描述:
对于A1,A2…AN,求
∑ i=1N∑ j=1Nlcm(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
代码:
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define maxn 50005
#define rgt register
int N, M, cnt[maxn], mu[maxn], p[maxn], tot, v[maxn];
ll s[maxn];
ll ans=0;
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mu
if ( !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;
}