Minimum’s Revenge
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 283 Accepted Submission(s): 219Problem DescriptionThere 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?InputThe 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.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 Input2 2 3Sample OutputCase #1: 2 Case #2: 5HintIn 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.SourceRecommendStatistic | Submit | Discuss | Note
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+1)/2-1); 59 } 60 return 0; 61 } 62 /* 63 // 64 65 // 66 */
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?
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.
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
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