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SPOJ : Interesting Subset 想法題

2019-11-08 02:23:22
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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}這個集合里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小于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。最后相乘,全部加起來后 mod 1000000007 就是我們所要求的答案。

#include<cstdio>#define mod 1000000007typedef 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; }
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