SPOJ interesting subset

最新推荐文章于 2018-04-17 20:25:22 发布

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本文探讨了一种算法问题，即计算特定整数集合中有多少个子集满足特定条件：存在两个元素a和b使得b是a的倍数，并且a是子集中最小的数。通过分析和使用快速幂运算来解决此问题。

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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

//思维题

题意：给出n，求{1,2,3,4,5…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b%a==0且a最小

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组合（减一为减去都不在子集的情况)，对不能整除它的也进行组合，相乘。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define MOD 1000000007
long long quickPow(long long a,long long b){
	long long re=1,te=a;
	while(b){
		if(b&1){
			re=(re*te)%MOD;
		}
		te=(te*te)%MOD;
		b=b>>1;
	}
	return re;
}
int main(){
	int n;
	long long num,ans;
	cin>>n;
	for(int j=1;j<=n;j++){
		scanf("%lld",&num);
		ans=0;
		for(long long i=1;i<=num;i++){
			ans=(ans+((quickPow(2,2*num/i-1)-1)*quickPow(2,2*num-i-(2*num/i-1)))%MOD)%MOD;
		}
		printf("Case %d: %lld\n",j,ans);
	}
	return 0;
}