【POJ 3734】【用母函数推公式 或者 矩阵幂】 Blocks【现要给n块砖染红、蓝、绿、黄四种颜色。要求被染成红色和绿色的砖块数量必须为偶数，问染色方案数】

传送门：http://poj.org/problem?id=3734

描述：

Blocks

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6290 Accepted: 3019

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

212

Sample Output

26

Source

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

题意：

有一排砖，数量为N。现要将砖全部染上色，有红、蓝、绿、黄四种颜色。要求被染成红色和绿色的砖块数量必须为偶数，问一共有多少种染色方案。（由于答案较大，模10007） 

思路一：

挑战书上P202

分别列出三个递推式就好了

代码一：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <vector>using  namespace std;typedef  long  long  ll;typedef  vector<int>vec;typedef  vector<vec>mat;const  int  M=10007;//计算A*Bmat  mul(mat& A,mat& B){    mat C(A.size(),vec(B[0].size()));    for(int i=0 ; i<A.size() ; i++){        for(int k=0 ; k<B.size() ; k++){            for(int  j=0 ; j<B[0].size() ; j++){                C[i][j]=(C[i][j]+A[i][k]*B[k][j])%M;            }        }    }    return  C;}//计算A^nmat pow(mat A , ll n){    mat B(A.size(),vec(A.size()));    for(int i=0 ; i<A.size() ; i++){        B[i][i]=1;    }    while(n>0){        if(n&1)B=mul(B,A);        A=mul(A,A);        n>>=1;    }    return B;}ll  n;void solve(){    mat A(3,vec(3));    A[0][0]=2;A[0][1]=1;A[0][2]=0;    A[1][0]=2;A[1][1]=2;A[1][2]=2;    A[2][0]=0;A[2][1]=1;A[2][2]=2;    A=pow(A,n);    printf("%d\n",A[0][0]);}int main(){    int T;    scanf("%d",&T);    while(T--){        scanf("%d",&n);        solve();    }    return 0;}

代码二：

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;#define mod 10007int n;int pow_mod(int x, int n){  int res = 1;  while(n){    if(n & 1) res = res * x % mod;    x = x * x % mod;    n >>= 1;  }  return res;}int main(){  int T;  scanf("%d",&T);  while(T--){      scanf("%d",&n);      printf("%d\n", (pow_mod(4, n - 1) + pow_mod(2, n - 1)) % mod);  }  return 0;}