poj 3734 指数型生成函数

最新推荐文章于 2021-03-23 22:45:27 发布

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#指数型生成函数

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本文介绍了一种解决特定计数问题的方法，该问题是求解在一定条件下，将多个方块涂成红色、蓝色、绿色或黄色的不同方式的数量。通过构造指数生成函数并使用快速幂运算来高效地计算答案。

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这道题就是和hdu 2065 解法一样

g(x)=(1+x^2/2!+x^4/4!......)^2*(1+x+x^2/2!+x^3/3!.....)^2=([(e^(-x)+e^x)/2]^2)*e^(2*x);

需要自己算出其指数生成函数中的系数为4^(n-1)+2^(n-1)，然后用快速幂计算结果即可。

#include <iostream>
using namespace std;
typedef long long ll;
ll n;
const int mod=10007;
int quickmod(int a,ll b)
{
    int ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%mod;
        b/=2;
        a=a*a%mod;
    }
    return ans;
}
int main(int argc, const char * argv[])
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n;
        cout<<(quickmod(4,n-1)+quickmod(2,n-1))%mod<<endl;
    }
    return 0;
}

Blocks

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6311		Accepted: 3030

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