Poj 3734 Blocks

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7381		Accepted: 3585

Description

Panda has received an assignment of painting a line of blocks. Since Panda is such an intelligent boy, he starts to think of a math problem of painting. Suppose there are N blocks in a line and each block can be paint red, blue, green or yellow. For some myterious reasons, Panda want both the number of red blocks and green blocks to be even numbers. Under such conditions, Panda wants to know the number of different ways to paint these blocks.

Input

The first line of the input contains an integer T(1≤T≤100), the number of test cases. Each of the next T lines contains an integer N(1≤N≤10^9) indicating the number of blocks.

Output

For each test cases, output the number of ways to paint the blocks in a single line. Since the answer may be quite large, you have to module it by 10007.

Sample Input

2
1
2

Sample Output

2
6

Source

PKU Campus 2009 (POJ Monthly Contest – 2009.05.17), Simon

 

 

答案的指数型生成函数显然是 e^2x + ((e^x+e^-x)/2)^2x 

我们拆开之后化简，得到 (e^4x + 2*e^2x +1)/4

因为n>=1，所以常数项对答案肯定没有贡献，直接忽略掉。

 

然后有一个定理是 e^kx 的指数型生成函数的x^n前的系数是 k^n/n!，所以直接替换就行了。

 

#include<cstdio>
#define ll long long
const int ha=10007;
int n,T;
inline int ksm(int x,int y){
	int an=1;
	for(;y;y>>=1,x=x*(ll)x%ha) if(y&1) an=an*(ll)x%ha;
	return an;
}
int main(){
	scanf("%d",&T);
	while(T--){
		scanf("%d",&n);
		printf("%d
",(ksm(4,n-1)+ksm(2,n-1))%ha);
	}
	return 0;
}