HLG 1126 Final Destination II 矩阵乘法+快速求幂

Description

JiaoZhu likes going on adventure! One day, he walks into a big castle, and there is an unique stairway. JiaoZhu finds a board ,it says “The one who want to go upstairs only can go three steps the most once, meaning that you can go 1 or 2 or 3 steps once!”. Now, we have a problem, can you tell me the number of ways to go to the destination? If you can’t ,death is the only choice。

In the beginning, you are in the 0^thstep. 

Input

First input a integer T(T<50), represent the number of case.

Each case ,the input will consist only a positive integer n (0<=n<=1000000000), represent the nth steps you want to go to..

Output

Order the sample output format to output.

Line 1,print the Case k.

Line 2,print one integer represent the number of ways to go to nth steps.(MOD 1000000007)

Sample Input

2

1

2

Sample Output

Case 1:

1

Case 2:

2

代码：

#include<stdio.h>
          void multiply(__int64 x[][4],__int64 y[][4])
         {    __int64 i,k,j,s[4][4];
               for(i=1;i<=3;i++)
                  for(j=1;j<=3;j++)
                       {   s[i][j]=0;
                           for(k=1;k<=3;k++)
                          s[i][j]=s[i][j]+x[i][k]*y[k][j];  }
                  for(i=1;i<=3;i++)
                        for(j=1;j<=3;j++)
                            y[i][j]=s[i][j]%1000000007;
         }
         int main()
         {  __int64 i,n,e[4][4],a[32][4][4],b[4][4];
             int p,t;
             e[1][1]=e[1][2]=e[1][3]=e[2][1]=e[3][2]=1;
             e[2][2]=e[2][3]=e[3][1]=e[3][3]=0;
             for(i=0;i<31;i++)//为快速求幂作准备
            { a[i][1][1]=e[1][1]; a[i][1][2]=e[1][2]; a[i][1][3]=e[1][3];
               a[i][2][1]=e[2][1]; a[i][2][2]=e[2][2]; a[i][2][3]=e[2][3];
               a[i][3][1]=e[3][1];a[i][3][2]=e[3][2];  a[i][3][3]=e[3][3];
              multiply(e,e);  }
               p=0;
             scanf("%d",&t);
                while(t--)
               {   p++;
                     scanf("%I64d",&n);
                    {   b[1][1]=4;b[2][2]=2;b[3][3]=1;
                         b[1][2]=b[1][3]=b[2][1]=b[2][3]=b[3][1]=b[3][2]=0;
                  for(i=0;i<31;i++)
                    if((n+1)&(1<<i))    //用快速求幂(按n的二进制数的每为是否为1求幂)原矩阵的n次方 
                       multiply(a[i],b);
                printf("Case %d:\n%d\n",p,b[1][3]);    }
           }
         return 0;
      }