hrbust Final Destination II

找出递推公式为an=an-1+an-2+an-3;

用到的是快速幂与矩阵乘法。

快速幂：求a的b次方时可直接用：

int res=1;

while（b）｛

if(b&1)//判断b是否为奇数；

res=a*res;

a=a*a;

b/=2;

}

return res;

Final Destination II

Time Limit: 1000 MS

Memory Limit: 65536 K

Total Submit: 103(26 users)

Total Accepted: 27(19 users)

Special Judge: No

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^th step. 

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

#include<stdio.h>
#include<string.h>
#define mod 1000000007
long long stairs[3][3];
void multy(long long a[][3],long long b[][3])//矩阵的乘法
{
    long long i, j, k;
    long long c[3][3];
    for(i = 0; i < 3; i++)
        for(j = 0; j < 3; j++){
            c[i][j] = 0;
            for(k = 0; k < 3; k++){
                c[i][j] += ((a[i][k]%mod)*(b[k][j]%mod))%mod;
            }
        }
    for(i = 0; i < 3; i++)//将结果返回给b[][];
        for(j = 0; j < 3; j++)
            b[i][j] = c[i][j]%mod;
}
long long ultim(long long n,long long a[][3],long long b[][3])
{
    while(n)//矩阵的快速幂求法
    {
        if(n&1)
            multy(a,b);
        multy(a,a);
        n/=2;
    }
    return b[0][0];
}
int main()
{
    long long n,k=1,t;
    scanf("%lld",&t);
    while(t--)
    {
        scanf("%lld",&n);
        memset(stairs,0,sizeof(stairs));
        stairs[0][0]=4;stairs[1][0]=2;stairs[2][0]=1;
        long long coeff[3][3]={1,1,1,1,0,0,0,1,0};
        //multy(coeff,coeff);
        printf("Case %lld:\n",k++);
        if(n==0)
            printf("1\n");
        else if(n == 1)
            printf("1\n");
        else if(n == 2)
            printf("2\n");
        else
            printf("%lld\n",ultim(n-3,coeff,stairs));
    }
    return 0;
}