hdu 5922/Minimum’s Revenge

最新推荐文章于 2019-02-26 18:46:14 发布

原创最新推荐文章于 2019-02-26 18:46:14 发布 · 371 阅读

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本文探讨了一个关于图论中最小生成树的问题。在一个由n个顶点构成的图中，任意两个不同顶点间的边权为其索引的最小公倍数。文章提供了计算最小生成树总权重的有效算法，并附带了示例输入输出。

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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

：如果不是1和所以点相连的话，那么其他点如果相连那个最小公倍数一定大于他们本身，所以所以点都与1相连才是最短。所以答案就是：2 + 3 + 4 ....+n；

<span style="font-size:14px;color:#ff0000;background-color: rgb(255, 255, 255);">#include <iostream>
#include<stdio.h>
using namespace std;

int main()
{
    int  T;
    cin>>T;
    for(int t=1;t<=T;t++)
    {
        __int64 n;
        scanf("%I64d",&n);
        printf("Case #%d: %I64d\n",t,n*(n+1)/2-1);
    }
    return 0;
}</span>