HDU 5950 Recursive sequence(矩阵快速幂）

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 29    Accepted Submission(s): 19

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and i4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right. 

 

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b < 231 as described above.

 

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

 

Sample Input

23 1 24 1 10

 

Sample Output

85369
Hint
In the first case, the third number is 85 = 2*1十2十3^4. In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.
 

题意：

已知递推公式为F(n)=F(n-1)+F(n-2)+n^4，F(1)=a，F(2)=b，给出n，求出F（n）。

思路：

一道矩阵快速幂的裸题，难点在于由已知的递推式构造矩阵，大家应该大一都学了线性代数，构造矩阵也就不难了，具体的步骤就不说了。

代码：

#include<iostream>#include<cstring>using namespace std;#define mod 2147493647#define LL long longstruct matrix{    LL m[7][7];}mat;int t,n,f1,f2;void init(){    memset(mat.m,0,sizeof(mat));///此处也可以用一个初始化的二维数组通过来赋值，避免一个一个麻烦的赋值。    mat.m[0][1]=1,mat.m[1][0]=2,mat.m[1][1]=1,mat.m[1][2]=1,mat.m[1][3]=4,mat.m[1][4]=6,mat.m[1][5]=4,mat.m[1][6]=1;    mat.m[2][2]=1,mat.m[2][3]=4,mat.m[2][4]=6,mat.m[2][5]=4,mat.m[2][6]=1;    mat.m[3][3]=1,mat.m[3][4]=3,mat.m[3][5]=3,mat.m[3][6]=1;    mat.m[4][4]=1,mat.m[4][5]=2,mat.m[4][6]=1;    mat.m[5][5]=mat.m[5][6]=mat.m[6][6]=1;}matrix mul(matrix a,matrix b){    matrix c;    memset(c.m,0,sizeof(c.m));    for(int i=0; i<7; i++)        for(int j=0; j<7; j++)            for(int k=0; k<7; k++)                c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;///注意要使用long long因为此处可能会越界。    return c;}int slove(){    matrix ans;    memset(ans.m,0,sizeof(ans.m));    ans.m[0][0]=f1,ans.m[1][0]=f2,ans.m[2][0]=16,ans.m[3][0]=8,ans.m[4][0]=4,ans.m[5][0]=2,ans.m[6][0]=1;    n--;    while(n){        if(n&1) ans=mul(mat,ans);///顺序不能改变，矩阵乘法不满足交换侓。        n>>=1;        mat=mul(mat,mat);    }    return ans.m[0][0];}int main(){    cin>>t;    while(t--){        cin>>n>>f1>>f2;        init();        cout<<slove()<<endl;    }}

参考博客：

http://blog.csdn.net/spring371327/article/details/52973534