HDU 5950（矩阵快速幂）

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, andi4

. 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：a[i]=2*a[i-2]+a[i-1]+i^4，给出n,a1,a2，求an

思路：a[i]=2*a[i-2]+a[i-1]+i^4=2*a[i-2]+a[i-1]+((i-1)+1)^4=2*a[i-2]+a[i-1]+(i-1)^4+4*(i-1)^3+6*(i-1)^2+4*(i-1)+1

构造7*7的矩阵然后快速幂

设矩阵：

1,1,0,0,0,0,0

2,0,0,0,0,0,0

1,0,1,0,0,0,0

4,0,4,1,0,0,0

6,0,6,3,1,0,0

4,0,4,3,2,1,0

1,0,1,1,1,1,1为 martix-1

设 矩阵( a[i-1],a[i-2],(i-1)^4,(i-1)^3,(i-1)^2,(i-1),1 ) 为martix-2.

则 martix-2 * martix-1 = ( a[i],a[i-1],i^4,i^3,i^2,i,1 )

所以ac代码：

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <cmath>using namespace std;typedef long long LL;const LL MOD=2147493647;struct martix{    LL a[7][7];}x;void output(martix a){    for(int i=0;i<7;i++)    {        for(int j=0;j<7;j++)        {            printf("%lld ",a.a[i][j]);        }        puts("");    }    puts("");}martix mul(martix a,martix b){    martix c;    for(int i=0;i<7;i++)for(int j=0;j<7;j++)c.a[i][j]=0;    for(int i=0;i<7;i++)    {        for(int j=0;j<7;j++)        {            for(int k=0;k<7;k++)            {                c.a[i][j]+=a.a[i][k]*b.a[k][j];                c.a[i][j]%=MOD;            }        }    }//    puts("C************");//    output(a);output(b);//    output(c);    return c;}martix quick(LL n){    martix e;//    printf("n=%lld********************\n",n);    for(int i=0;i<7;i++)for(int j=0;j<7;j++)if(i==j)e.a[i][j]=1;else e.a[i][j]=0;    while(n)    {        if(n&1)e=mul(e,x);        x=mul(x,x);        n>>=1;//        printf("n=%lld********************\n",n);    }    return e;}void init(){    x.a[0][0]=1,x.a[0][1]=1,x.a[0][2]=0,x.a[0][3]=0,x.a[0][4]=0,x.a[0][5]=0,x.a[0][6]=0;    x.a[1][0]=2,x.a[1][1]=0,x.a[1][2]=0,x.a[1][3]=0,x.a[1][4]=0,x.a[1][5]=0,x.a[1][6]=0;    x.a[2][0]=1,x.a[2][1]=0,x.a[2][2]=1,x.a[2][3]=0,x.a[2][4]=0,x.a[2][5]=0,x.a[2][6]=0;    x.a[3][0]=4,x.a[3][1]=0,x.a[3][2]=4,x.a[3][3]=1,x.a[3][4]=0,x.a[3][5]=0,x.a[3][6]=0;    x.a[4][0]=6,x.a[4][1]=0,x.a[4][2]=6,x.a[4][3]=3,x.a[4][4]=1,x.a[4][5]=0,x.a[4][6]=0;    x.a[5][0]=4,x.a[5][1]=0,x.a[5][2]=4,x.a[5][3]=3,x.a[5][4]=2,x.a[5][5]=1,x.a[5][6]=0;    x.a[6][0]=1,x.a[6][1]=0,x.a[6][2]=1,x.a[6][3]=1,x.a[6][4]=1,x.a[6][5]=1,x.a[6][6]=1;}LL n,a1,a2;int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%lld%lld%lld",&n,&a1,&a2);        if(n==1)printf("%lld\n",a1);        else if(n==2)printf("%lld\n",a2);        else        {            init();//            puts("X*************");//            output(x);            x=quick(n-2);//            puts("Y*************");//            output(x);            LL b[7]={a2,a1,16,8,4,2,1};            LL ans=0;            for(int i=0;i<7;i++)            {                ans=(ans+b[i]*x.a[i][0])%MOD;            }            printf("%lld\n",ans);        }    }    return 0;}

大概就这样了。。。。。。

本人蒟蒻，如有错误，还望指正