7.20 第一场 Minimum spanning tree

Minimum spanning tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 81 Accepted Submission(s): 43

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

大概题意

求n个节点的最小生成树，两两节点之间边的权值为两数的最小公倍数。

思路

惯性思维考虑了克鲁斯卡尔的复杂度。。。。。发现这数据量有点费劲

冷静过后梳理样例发现规律

若定点为素数，则与2相连时权值最小

若定点不为素数，则与其因数相连时权值最小

据此编写代码

代码

#include<iostream>
#include<cstdio>

using namespace std;
const int N = 10000010;

int primes[N], cnt;     // primes[]存储所有素数
bool st[N];         // st[x]存储x是否被筛掉

void get_primes(int n) {
    for (int i = 2; i <= n; i++) {
        if (!st[i]) primes[cnt++] = i;
        for (int j = 0; primes[j] <= n / i; j++) {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main() {
    int t;
    long long n;
    cin >> t;
    get_primes(10000010);
    while (t--) {
        cin >> n;
        long long res = 0;
        for (int i = 3; i <= n; ++i) {
            if (!st[i])res += i * 2;
            else res += i;
        }
        cout << res << endl;
    }
    return 0;
}
//6 4 10 6