hdu 5922 Minimum’s Revenge【贪心】水题

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探讨了给定N个顶点的图中，每两点间边的权值为其点权的最小公倍数时，如何快速计算最小生成树的总权值。此问题可通过直接利用点1与其他点构成生成树并计算权值和的方法解决。

Minimum’s Revenge

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 237 Accepted Submission(s): 186

Problem Description

There is a graph of n vertices which are indexed from 1 to n. For any pair of different vertices, the weight of the edge between them is the least common multiple of their indexes.

Mr. Frog is wondering about the total weight of the minimum spanning tree. Can you help him?

Input

The first line contains only one integer T (

T≤100), which indicates the number of test cases.

For each test case, the first line contains only one integer n (

2≤n≤109), indicating the number of vertices.

Output

For each test case, output one line "Case #x：y",where x is the case number (starting from 1) and y is the total weight of the minimum spanning tree.

Sample Input

2
2
3

Sample Output

Case #1: 2
Case #2: 5
Hint
In the second sample, the graph contains 3 edges which are (1, 2, 2), (1, 3, 3) and (2, 3, 6). Thus the answer is 5.

题目大意:

给你一个数N，表示一共有N个顶点，其中每个顶点的权值是从1-N,其中两点间有一条无向边，其权值为两个点的点权的lcm，问最小生成树的权值和。

思路：

很明显直接拿点1和其他点连线构成生成树的权值最小，值为从2加到n、

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
#define ll __int64
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        ll n;
        scanf("%I64d",&n);
        ll ans;
        ans=(2+n)*(n-1)/2;
        printf("Case #%d: ",++kase);
        printf("%I64d\n",ans);
    }
}