hdu 5363 Key Set 快速幂

[align=left]Problem Description[/align]
soda has a set S with n integers {1,2,…,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.

[align=left]Input[/align]
There are multiple test cases. The first line of input contains an integer T (1≤T≤105), indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤109), the number of integers in the set.

[align=left]Output[/align]
For each test case, output the number of key sets modulo 1000000007.

[align=left]Sample Input[/align]

4
1
2
3
4

[align=left]Sample Output[/align]

0
1
3
7

题意：给出一个数n，表示一个集合中有n个数1-n;找出这个集合的非空子集的个数，并且这些非空子集中所有元素相加为偶数。
大家都知道n个数的非空子集有2的n次方减1个，那么给原集合中1剔除来，剩下的集合就有2的n-1次方-1个。那么剩下的集合中所有元素相加之和为奇数的加上1就为偶数了，剩下为偶数的就不加1.所以结果就是2的n次方-1。由于这个数据非常大，就要用快速幂。
快速幂的作用：减少计算时间。在计算中就可以取摸。取摸原理假设 m=x*y;那么 m%n=((x%n)*(y%n))%n;

#include<cstdio>
#include<cstring>
using namespace std;
long long t,n,ans,y;
int f()
{
ans=1;
y=2;;
n-=1;
while (n)
{
if (n&1) ans=(ans*y)%1000000007;
y=(y*y)%1000000007;
n>>=1;
}
return ans-1;
}
int main()
{
scanf("%lld",&t);
while (t--)
{
scanf("%lld",&n);
printf("%lld\n",f());
}
return 0;
}