A - Minimum’s Revenge

最新推荐文章于 2020-01-27 10:07:07 发布

原创最新推荐文章于 2020-01-27 10:07:07 发布 · 244 阅读

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本文介绍了一种计算特定图中最小生成树总权重的方法。给定由1到n编号的顶点组成的图，任意两个不同顶点之间的边权为其编号的最小公倍数。文章通过示例展示了如何求解最小生成树的总权重，并提供了一段C++代码实现。

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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?

InputThe first line contains only one integer T (T≤100T≤100), which indicates the number of test cases.

For each test case, the first line contains only one integer n (2≤n≤1092≤n≤109), indicating the number of vertices.
OutputFor 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.

#include <iostream>
#include <stdio.h>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int i,t,k;
    long long x,y,n;
    long long ans=0;
    cin>>t;
    for(k=1;k<=t;k++){
        cin>>n;
        x=2+n;
        y=n-1;
        if(n%2==0)
            x/=2;
        else
            y/=2;
        ans = x*y;
        printf("Case #%d: %lld\n",k,ans);
    }

    return 0;
}