hdu 5922/Minimum’s Revenge
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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
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
:如果不是1和所以点相连的话,那么其他点如果相连那个最小公倍数一定大于他们本身,所以所以点都与1相连才是最短。所以答案就是:2 + 3 + 4 ....+n;
<span style="font-size:14px;color:#ff0000;background-color: rgb(255, 255, 255);">#include <iostream>#include<stdio.h>using namespace std;int main(){ int T; cin>>T; for(int t=1;t<=T;t++) { __int64 n; scanf("%I64d",&n); printf("Case #%d: %I64d\n",t,n*(n+1)/2-1); } return 0;}</span>
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