SPOJ - INTSUB 数学

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INTSUB - Interesting Subset

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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

Submit solution!

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 ;
}

Interesting Subset

SPOJ - INTSUB