Minimum’s Revenge

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 283    Accepted Submission(s): 219

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 multipleof 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

 

 

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题目链接：

　　http://acm.hdu.edu.cn/showproblem.php?pid=5922

题目大意：

　　N个点1~N(N<=10⁹)，x与y之间的边的权为x与y的最小公倍数，求最小生成树的边权和。

题目思路：

　　【模拟】【构造】

　　这样构造数：质数与1相连，其余与任意一个因子相连，这样边权最小，所以当N=2时答案为2，否则为2+3+...+N。

 1 //
 2 //by coolxxx
 3 //#include<bits/stdc++.h>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<string>
 7 #include<iomanip>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-10)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 100004
34 using namespace std;
35 typedef long long LL;
36 double anss;
37 LL aans;
38 int cas,cass;
39 LL n,m,lll,ans;
40 
41 int main()
42 {
43     #ifndef ONLINE_JUDGEW
44 //    freopen("1.txt","r",stdin);
45 //    freopen("2.txt","w",stdout);
46     #endif
47     int i,j,k;
48     int x,y,z;
49 //    init();
50 //    for(scanf("%d",&cass);cass;cass--)
51     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
52 //    while(~scanf("%s",s))
53 //    while(~scanf("%d%d",&n,&m))
54     {
55         printf("Case #%d: ",cass);
56         scanf("%lld",&n);
57         if(n==2)puts("2");
58         else printf("%lld\n",n*(n+1)/2-1);
59     }
60     return 0;
61 }
62 /*
63 //
64 
65 //
66 */

View Code

Minimum’s Revenge

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 283    Accepted Submission(s): 219

HDU 5922 Minimum’s Revenge 【模拟】 (2016CCPC东北地区大学生程序设计竞赛)