题目链接：点击传送
INTSUB - Interesting Subset
You are given a set X = {1, 2, 3, 4, … , 2n-1, 2n} where n is an integer. You have to find the number of interesting subsets of this set X.
A subset of set X is interesting if there are at least two integers a & b such that b is a multiple of a, i.e. remainder of b divides by a is zero and a is the smallest number in the set.
Input
The input file contains multiple test cases. The first line of the input is an integer T(<=30) denoting the number of test cases. Each of the next T lines contains an integer 'n' where 1<=n<=1000.
Output
For each test case, you have to output as the format below:
Case X: Y
Here X is the test case number and Y is the number of subsets. As the number Y can be very large, you need to output the number modulo 1000000007.
Example
Input:
3
1
2
3 Output:
Case 1: 1
Case 2: 9
Case 3: 47
题意：给你2*n个数，你最小需要选两个，使得这个子集中含有最小值的倍数；
思路：枚举最小值，对于其倍数最小取一个，其余随意取与不取；
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x) cout<<"bug"<<x<<endl;
const int N=1e5+,M=1e6+,inf=;
const ll INF=1e18+,mod=1e9+;
ll qpow(ll a,ll b,ll c)
{
ll ans=;
while(b)
{
if(b&)ans=(ans*a)%c;
b>>=;
a=(a*a)%c;
}
return ans;
}
int main()
{
int T,cas=;
scanf("%d",&T);
while(T--)
{
int n;
scanf("%d",&n);
ll ans=;
for(int i=;i<=n;i++)
{
int p=(*n-i);
int b=((*n)/i-);
ans=(ans+(qpow(,p-b,mod)*(qpow(,b,mod)+(mod-))%mod)%mod)%mod;
}
printf("Case %d: %lld\n",cas++,ans);
}
return ;
}