HDU-5363 Key Set 【快速幂取模+递推】

Key Set

Time Limit:1000MS Memory Limit:131072KB 64bit IO Format:%I64d & %I64u

Submit

Status

Description

soda has a set S with n integers {1, 2, \dots, n}. A set is called key set if the sum of integers in the set is an even number. He wants to know how many nonempty subsets of S are key set.

Input

There are multiple test cases. The first line of input contains an integer T (1 \le T \le 10^5), indicating the number of test cases. For each test case:

The first line contains an integer n (1 \le n \le 10^9), the number of integers in the set.

Output

For each test case, output the number of key sets modulo 1000000007.

Sample Input

4

1

2

3

4

Sample Output

0

1

3

7

题意：给定一个集合{1,2,3……n},求此集合的非空子集中 key set的个数对1000000007的余数;key set：集合中所有的元素之和为偶数。

思路：递归思想：设集合{1 ,2,3……n}的key set个数为f(n),那么{1,2,,3……，n-1}的key set个数为f(n-1). 讨论： 1.n为偶数，f(n)=f(n-1)+f(n-1)+1； 2.n为奇数，f(n)=f(n-1)+2^(n-1)-1. 最后推导出公式f(n)=2^(n-1)-1; 在这里不再叙述，其实就是n这个元素加入到原来的·集合会产生多少·key set，没思路的话举例子也可以推出来（含n个元素的集合的幂集中的元素个数为2^n)

失误：刚开确实没思路，也不想举例子。

代码如下：

#include<cstdio>

__int64 quickpow(__int64 base ,__int64 m,__int64 mod)
{
__int64 ans=1;

while(m>0)
{
if(m&1)
{
ans=(ans*base)%mod;
--m;
}

base=(base*base)%mod;
m/=2;
}

return ans;
}

int main()
{
__int64 t,n,c;
scanf("%I64d",&t);

while(t--)
{
scanf("%I64d",&n);

c=quickpow(2,n-1,1000000007)-1;

printf("%I64d\n",c);
}
return 0;
}