其实矩阵构造就是对公式的化简，最后运用矩阵快速幂求值
下面来看一题
Everybody knows Fibonacci numbers, now we are talking about the Tribonacci numbers: T[0] = T[1] = T[2] = 1; T[n] = T[n - 1] + T[n - 2] + T[n - 3] (n >= 3)
Given a and b, you are asked to calculate the sum from the ath Fibonacci number to the bth Fibonacci number, mod 1,000,000,007, that is (T[a] + T[a + 1] + ... + T[b]) % 1,000,000,007.
There are multiple cases (no more than 100).
Each case contain two non-negative integers a b(a <= b and a, b < 1,000,000,000)
For each case, output the sum % 1,000,000,007.
T[n] = T[n - 1] + T[n - 2] + T[n - 3]
左边累加，右边累加我们发现
sum[n] = sum[n - 1] + sum[n - 2] + sum[n - 3]
1 1 1
1 0 0
0 1 0
sum[n-1] 0 0
sum[n-2] 0 0
sum[n-3] 0 0
代码：
#include<iostream>
#include<cstdio>
using namespace std;
#define M 1000000007
struct mat
{
int ans[4][4];
};
mat I,MID;
mat cal(mat a,mat b)
{
mat c;
int i,j,k;
for(i=0;i<4;i++)
for(j=0;j<4;j++)
{
c.ans[i][j]=0;
for(k=0;k<4;k++)
c.ans[i][j]+=a.ans[i][k]*b.ans[k][j];
c.ans[i][j]%=M;
}
return c;
}
mat atl(int n)
{
mat res=MID;
mat mid=I;
while(n){
if(n&1)
res=cal(res,mid);
mid=cal(mid,mid);
n>>=1;}
return res;
}
int main()
{
int k,i,j;
int a,b;
while(scanf("%d%d",&a,&b)!=-1)
{
I.ans[0][0]=1;
I.ans[0][1]=1;
I.ans[0][2]=1;
MID.ans[0][0]=1;
MID.ans[0][1]=1;
MID.ans[0][2]=1;
MID.ans[1][0]=1;
MID.ans[2][1]=1;
mat l=atl(b-2);
mat g=atl(a-3);
printf("%d\n",l.ans[0][0]-g.ans[0][0]);
}
return 0;
}