Minimum’s Revenge

                                                                      Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                      Total Submission(s): 1020    Accepted Submission(s): 544

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 theleast 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

223

 

Sample Output

Case #1: 2Case #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.
 

 简要分析：

有n个节点的图，顶点编号1-n，任意连接两个不同的顶点的边的权值为这两个顶点编号的最小公倍数，求该图最小生成树的权值和。

注意到任何顶点与1号顶点相连的边的权值为最小，那么该题就转化为2+3+4+...+n求和。

#include <iostream>#include <stdio.h>#include <algorithm>#include <math.h>#include <queue>#include <string.h>#include <stack>#include <vector>#define LL long longint main(){    int mycase=1;    int T;    scanf("%d",&T);    while(T--)    {        LL n;        scanf("%d",&n);        LL ans=(2+n)*(n-1)/2;  //数据规模过大，用long long型        printf("Case #%d: %lld\n",mycase,ans);        mycase++;    }    return 0;}