前几天打比赛遇到的矩阵快速幂,当时不会做(矩阵快速幂已经忘干净了),现在补上,在加深一下对矩阵快速幂的理解;
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6470
构造矩阵:
f[n] =2*f [n-2]+f [n-1]+n^3; (n+1)^3=n^3+3*n^2+3*n+1;
所以 【 f[n-2], f[n-1], n^3, n^2, n, 1 】*A= 【 f[n-1], f[n], (n+1)^3, (n+1)^2, n+1, 1 】
所以转移矩阵为 A[6][6]={0 ,2 ,0 ,0 ,0 ,0 ,
1 ,1 ,0 ,0 ,0 ,0 ,
0 ,1 ,1 ,0 ,0 ,0 ,
0 ,3 ,3 ,1 ,0 ,0 ,
0 ,3 ,3 ,2 ,1 ,0 ,
0 ,1 ,1 ,1 ,1 ,1 ,};
AC代码:
#include<cstring>
#include<cstdio>
#include<string>
#include<algorithm>
#include<math.h>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const double pi=acos(-1.0);
const ll mod=123456789;
const int N=2e5+10;
ll tmp[6][6];
ll sum[16];
void multi(ll a[][6],ll b[][6])
{
memset(tmp,0,sizeof(tmp));
for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
{
for(int k=0;k<6;k++)
{
tmp[i][j]+=(a[i][k]*b[k][j])%mod;
tmp[i][j]=(tmp[i][j]+mod)%mod;
}
}
}
for(int i=0;i<6;i++)
{
for(int j=0;j<6;j++)
a[i][j]=tmp[i][j];
}
}
ll res[6][6];
void poww(ll a[][6],ll n)
{
ll m=n;
n-=3;
memset(res,0,sizeof(res));
for(int i=0;i<6;i++)
res[i][i]=1;
while(n)
{
if(n&1) multi(res,a);
multi(a,a);
n/=2;
}
ll ans[6][6];
memset(ans,0,sizeof(ans));
ans[0][0]=2;ans[0][1]=31;
ans[0][2]=27;ans[0][3]=9;
ans[0][4]=3;ans[0][5]=1;
multi(ans,res);
printf("%lld\n",ans[0][1]%mod);
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
ll n;
scanf("%lld",&n);
ll a[6][6]={0,2,0,0,0,0,
1,1,0,0,0,0,
0,1,1,0,0,0,
0,3,3,1,0,0,
0,3,3,2,1,0,
0,1,1,1,1,1,};
poww(a,n);
}
return 0;
}