Problem Description
Given n-1 points, numbered from 2 to n, the edge weight between the two points a and b is lcm(a, b). Please find the minimum spanning tree formed by them.
A minimum spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. That is, it is a spanning tree whose sum of edge weights is as small as possible.
lcm(a, b) is the smallest positive integer that is divisible by both a and b.
Input
The first line contains a single integer t (t<=100) representing the number of test cases in the input. Then t test cases follow.
The only line of each test case contains one integers n (2<=n<=10000000) as mentioned above.
Output
For each test case, print one integer in one line, which is the minimum spanning tree edge weight sum.
Sample Input
2
2
6
Sample Output
0
26
手玩样例很容易就能发现所有合数一定和质因子相连，所以边权为本身，质数一定和2相连，因此边权是这个质数 * 2。所以总的答案就是2到n求和再加上n以内的质数和再减去2 * 2。n以内的质数和可以用min25筛求。
#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef long long ll;
const int N = 1000010;
struct Min25 {
ll prime[N], id1[N], id2[N], flag[N], ncnt, m;
ll g[N], sum[N], a[N], T, n;
inline int ID(ll x) {
return x <= T ? id1[x] : id2[n / x];
}
inline ll calc(ll x) {
return x * (x + 1) / 2 - 1;
}
inline ll f(ll x) {
return x;
}
inline void Init() {
memset(prime, 0, sizeof(prime));
memset(id1, 0, sizeof(id1));
memset(id2, 0, sizeof(id2));
memset(flag, 0, sizeof(flag));
memset(g, 0, sizeof(g));
memset(sum, 0, sizeof(sum));
memset(a, 0, sizeof(a));
ncnt = m = T = n = 0;
}
inline void init() {
T = sqrt(n + 0.5);
for (int i = 2; i <= T; i++) {
if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) {
flag[i * prime[j]] = 1;
if (i % prime[j] == 0) break;
}
}
for (ll l = 1; l <= n; l = n / (n / l) + 1) {
a[++m] = n / l;
if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
g[m] = calc(a[m]);
}
for (int i = 1; i <= ncnt; i++)
for (int j = 1; j <= m && (ll)prime[i] * prime[i] <= a[j]; j++)
g[j] = g[j] - (ll)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
}
inline ll solve(ll x) {
if (x <= 1) return x;
return n = x, init(), g[ID(n)];
}
}a;
signed main() {
int t;
cin >> t;
while(t--) {
long long n;
cin >> n;
a.Init();
printf("%lld
", (2 + n) * (n - 1) / 2 +a.solve(n) - 4);
}
}