Minimum’s Revenge

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1409 Accepted Submission(s): 749

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.

Source

2016CCPC东北地区大学生程序设计竞赛
- 重现赛

思路：水题，注意变量类型设为long long

#include <iostream>

#include<stdio.h>

using namespace std;

int main()

{

    int t;

    scanf("%d",&t);

    long long n;

    int path=1;

    while(t--)

    {

        scanf("%lld",&n);

        long long  sum=0;

            sum=((2+n)*(n-1))/2;

            printf("Case #%d: %lld\n",path++,sum);

    }

    return 0;

}

巴特西

hdu5922Minimum’s Revenge

Minimum’s Revenge

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