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SPOJ : Interesting Subset 想法题

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

题目链接

题意:给你一个n,让你求{1,2,3…2n-1,2n}这个集合里面有多少个子集满足:至少有两个元素a,b满足a%b==0且a小于b

解题思路:对每一个1-n中的数,我们总能找到一个数能整除它,因此我们只要对能整除它的进行排列组合且至少要有一个,对不能整除它的也进行排列组合,可以没有。最后相乘,全部加起来后 mod 1000000007 就是我们所要求的答案。

#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; } PRintf("Case %lld: %lld/n",tt,ans); } return 0;}
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