HDU Minimum’s Revenge 2016CCPC东北地区大学生程序设计竞赛 - 重现赛

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本文探讨了一个关于图论中最小生成树的问题，具体为给定一个由n个顶点构成的图，每对不同的顶点间边的权重为其索引的最小公倍数。文章提供了解决方案，即如何计算该图最小生成树的总权重。

Minimum’s Revenge
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0


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.

边权=u,v的最小公倍数
u和任意点之间的边权>=u
所以对u>1 都选择与1相连得到的就是最小生成树
ans=2+3+….+n=(2+n)(n-1)/2

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>
//#define CHECK_TIME

int main()
{
    //freopen("/home/lu/文档/r.txt","r",stdin);
    //freopen("/home/lu/文档/w.txt","w",stdout);

    int T;
    ll n;
    scanf("%d",&T);
    for(int t=1;t<=T;++t){
        cin>>n;
        ll ans;
        if((n+2)%2==0)
            ans=((n+2)/2)*(n-1);
        else
            ans=((n-1)/2)*(n+2);
        printf("Case #%d: %lld\n",t,ans);
    }
    return 0;
}