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

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。

//
//by coolxxx
//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<iomanip>
#include<map>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<memory.h>
#include<time.h>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
//#include<stdbool.h>
#include<math.h>
#pragma comment(linker,"/STACK:1024000000,1024000000")
#define min(a,b) ((a)<(b)?(a):(b))
#define max(a,b) ((a)>(b)?(a):(b))
#define abs(a) ((a)>0?(a):(-(a)))
#define lowbit(a) (a&(-a))
#define sqr(a) ((a)*(a))
#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
#define mem(a,b) memset(a,b,sizeof(a))
#define eps (1e-10)
#define J 10000
#define mod 1000000007
#define MAX 0x7f7f7f7f
#define PI 3.14159265358979323
#define N 100004
using namespace std;
typedef long long LL;
double anss;
LL aans;
int cas,cass;
LL n,m,lll,ans;

int main()
{
    #ifndef ONLINE_JUDGEW
//    freopen("1.txt","r",stdin);
//    freopen("2.txt","w",stdout);
    #endif
    int i,j,k;
    int x,y,z;
//    init();
//    for(scanf("%d",&cass);cass;cass--)
    for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
//    while(~scanf("%s",s))
//    while(~scanf("%d%d",&n,&m))
    {
        printf("Case #%d: ",cass);
        scanf("%lld",&n);
        if(n==2)puts("2");
        else printf("%lld
",n*(n+1)/2-1);
    }
    return 0;
}
/*
//

//
*/

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

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5932 5931 5930 5929 5928 

 

Statistic | Submit | Discuss | Note