hdu 3306 Another kind of Fibonacci-白红宇

**Another kind of Fibonacci**Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2005    Accepted Submission(s): 787Problem DescriptionAs we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)2 +A(1)2+……+A(n)2.InputThere are several test cases.Each test case will contain three integers , N, X , Y .N : 2<= N <= 231 – 1X : 2<= X <= 231– 1Y : 2<= Y <= 231 – 1OutputFor each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.Sample Input2 1 1 3 2 3 Sample Output6196

题目大意：就是给你三个数 n 和x,y，

f(n) = x*f(n-1) + y*f(n-2);

然后求s(n) = f(0)^2 + f(1)^2 +….+f(n)^2，就是这样了;

解题思路：还是矩阵乘法取余；

直接上代码：

/*2015 - 8 - 14 下午Author: ITAK今天非常非常的不顺心啊。。。。今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。*/#include 
     
      #include 
      
       using namespace std;const int maxn = 4;const int mod = 10007;typedef long long LL;typedef struct{    LL  m[maxn][maxn];}Matrix;Matrix P = {
    1,1,0,0,            0,0,0,0,            0,0,0,0,            0,1,0,0           };Matrix I = {
    1,0,0,0,            0,1,0,0,            0,0,1,0,            0,0,0,1           };Matrix matrix_mul(Matrix a, Matrix b){    int i, j, k;    Matrix c;    for(i=0; i
       
        >= 1;        b = matrix_mul(b, b);    }    return ans;}int main(){    LL n, x, y;    while(~scanf("%lld%lld%lld",&n,&x,&y))    {        Matrix tmp;        x %= mod;        y %= mod;        P.m[1][1] = (x*x) % mod;        P.m[1][2] = (2*x*y) % mod;        P.m[1][3] = (y*y) % mod;        P.m[2][1] = x;        P.m[2][2] = y;        tmp = quick_mod(n);        LL ans = (tmp.m[0][0]+tmp.m[0][1]+tmp.m[0][2]+tmp.m[0][3])%mod;        cout<
        
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