HLG 1126 Final Destination II（快速幂，矩阵乘法）
| 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 0th 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 |
| Hint |
| When n=2,you can go one step once to go to 2th ,or go 2 steps once to 2th,so the answer is 2. |
给一个n代表你想去第n个台阶，求出有多少种不同的走法，最后对1000000007取余。上台阶时你一步可以走1个台阶，或一次走2个台阶，或一次走3个台阶。
递推公式a(n)=a(n-1)+a(n-2)+a(n-3)，当然在这个题中我们确实用了递推公式，但是却不能直接开一个数组，因为n太大数组开不了这个长度，并且循环起来会超时，
所以我们用矩阵来求。
我们举一个例子来模拟一下用矩阵的递推原理的过程
定义stairs2矩阵为 a[n+3] a[n+2] a[n+1] 另一个stairs1矩阵 a[n+2] a[n+1] a[n] 矩阵的系数coeff
7 4 2 4 2 1 1 1 0
0 0 0 0 0 0 1 0 1
0 0 0 0 0 0 1 0 0
由矩阵求法知要想得到stairs2，是用stairs1*coeff ，而在stairs2中，下标为[0][0]的就是我们所求。
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;
#include <iostream>
using namespace std;
#define mod 1000000007
void mul(long long b[][3],long long a[][3])
{
int i,j,k;
long long c[3][3]={0};
for(i=0;i<3;i++) //矩阵的乘法运算,结果用c[][]来存
for(j=0;j<3;j++)
for(k=0;k<3;k++)
c[i][j]=c[i][j]+((b[i][k]%mod)*(a[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; //这一步必须要对c[][]取余后再付给b[][],否则会出现结果错误。
}
void output(long long n,long long b[][3],long long a[][3])
{
while(n) //矩阵的快速幂求法
{
if(n&1)
mul(b,a);
mul(a,a);
n=n/2;
}
cout<<b[0][0]<<endl;
}
int main()
{
int T,k=1;
cin>>T;
while(T--)
{
long long n;
cin>>n;
long long stairs[3][3]={4,2,1,0,0,0,0,0,0};
long long coeff[3][3]={1,1,0,1,0,1,1,0,0}; //矩阵的系数
cout<<"Case "<<k++<<":"<<endl;
if(n==0||n==1)
cout<<"1"<<endl;
else if(n==2)
cout<<"2"<<endl;
else
output(n-3,stairs,coeff);
}
return 0;
}