SPOJ : Interesting Subset 想法題

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

題目鏈接

題意：給你一個n，讓你求{1,2,3…2n-1,2n}這個集合裏面有多少個子集滿足：至少有兩個元素a,b滿足a%b==0且a小於b

解題思路：對每一個1-n中的數，我們總能找到一個數能整除它，因此我們只要對能整除它的進行排列組合且至少要有一個，對不能整除它的也進行排列組合，可以沒有。最後相乘，全部加起來後 mod 1000000007 就是我們所要求的答案。

#include<cstdio>
#define mod 1000000007
typedef long long ll;
ll t,n,ans;
ll quickpower(ll a,ll b,ll c){
    ll ss=1;
    while(b){
        if(b&1){
            ss=ss*a%c;
        }
        a=a*a%c;
        b/=2;
    }
    return ss;
}
int main(){
    scanf("%lld",&t);
    for(ll tt=1;tt<=t;tt++){
        ans=0;
        scanf("%lld",&n);
        ll k=2*n;
        for(ll i=1;i<=n;i++){
            ll a1=k/i-1;
            ll a2=k-i-a1;
            ans+=((quickpower(2,a1,mod)-1)*quickpower(2,a2,mod)%mod);
            ans=ans%mod;
        }
        printf("Case %lld: %lld\n",tt,ans);
    }
    return 0;
}