2021“MINIEYE杯”中国大学生算法设计超级联赛（1）1005. Minimum spanning tree（min25筛）

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);
	}
}