nyist 301 递推求值（矩阵快速幂）

题目地址：http://acm.nyist.net/JudgeOnline/problem.php?pid=301

思路：注意矩阵乘法不满足交换律，所以要注意初始矩阵在右边，要进行快速幂的矩阵在左边

左边的矩阵为    右边的矩阵为

b a c f2 0 0

1 0 0 f1 0 0

0 0 1  1 0 0

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <cstring>
#include <climits>
#include <cmath>
#include <cctype>
const int inf = 0x3f3f3f3f;//1061109567
typedef long long ll;
const int maxn = 40000;
const int mod = 1000007;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
struct node
{
ll a[3][3];
};
node matrixmul(node a,node b)
{
node c;
for(int i=0; i<3; i++)
{
for(int j=0; j<3; j++)
{
c.a[i][j] = 0;
for(int k=0; k<3; k++)
{
c.a[i][j] += a.a[i][k] * b.a[k][j];
c.a[i][j] %= mod;
}
}
}
return c;
}
node quickmod(node s,int k)
{
node ans;
memset(ans.a,0,sizeof(s.a));
ans.a[0][0]=ans.a[1][1]=ans.a[2][2]=1;//初始化为单位矩阵，值不变
while(k)
{
if(k & 1)
ans = matrixmul(ans,s);
k = k >> 1;
s = matrixmul(s,s);
}
return ans;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
int f1,f2,d,e,f,n;
scanf("%d%d%d%d%d%d",&f1,&f2,&d,&e,&f,&n);
if(n == 1)
{
printf("%d\n",(f1+mod)%mod);
continue;
}
if(n == 2)
{
printf("%d\n",(f2+mod)%mod);
continue;
}
node sa;
memset(sa.a,0,sizeof(sa.a));
sa.a[0][0]=e;
sa.a[0][1]=d;
sa.a[0][2]=f;
sa.a[1][0]=1;
sa.a[2][2]=1;
sa = quickmod(sa,n-2);
node sb;
memset(sb.a,0,sizeof(sb.a));
sb.a[0][0]=f2;
sb.a[1][0]=f1;
sb.a[2][0]=1;
sb = matrixmul(sa,sb);//注意顺序
ll ans = (sb.a[0][0]+mod)%mod;
printf("%I64d\n",ans);
}
return 0;
}

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