poj 3734 Blocks【指数型生成函数】

指数型生成函数，推一推可得：

[(1+frac{x^1}{1!}+frac{x^2}{2!}+frac{x^3}{3!}+...)^2+(1+frac{x^2}{2!}+frac{x^4}{4!}+frac{x^6}{6!}+...)^2 ]

[=e^{2x}+(frac{e^x+2^-x}{2})^2 ]

[=e^{2x}+frac{e^{2x}+e^{-2x}+2}{4} ]

[=frac{e^{4x}+2e^{2x}+1}{4} ]

因为

[e^x=sum_{i=0}^{inf}frac{x^i}{i!},e^{4x}=sum_{i=0}^{inf}frac{(4x)^i}{i!}=sum_{i=0}^{inf}frac{4^ix^i}{i!} ]

所以展开可得

[=frac{1}{4}+frac{sum_{i=0}^{inf}frac{4^ix^i}{i!}+2*sum_{i=0}^{inf}frac{2^ix^i}{i!}}{4} ]

[=frac{1}{4}+frac{sum_{i=0}^{inf}(4^i+2^{i+1})*frac{x^i}{i!}}{4} ]

前面的常数不用管，这样取i个的答案也就是第i项的系数就是( 4ⁱ⁺²{i+1} )

#include<iostream>
#include<cstdio>
using namespace std;
const int mod=10007;
int T,n;
int ksm(int a,int b)
{
	int r=1;
	while(b)
	{
		if(b&1)
			r=r*a%mod;
		a=a*a%mod;
		b>>=1;
	}
	return r;
}
int main()
{
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		printf("%d
",(ksm(2,n-1)+ksm(4,n-1))%mod);
	}
	return 0;
}