【HDU5950】Recursive sequence（矩阵快速幂）

#include <cstdio>
#include <cstring>
#include <algorithm>
#define ll long long
#include <iostream>
using namespace std;
const ll mod=2147493647;
struct Mat{
	int r,c;
	ll a[10][10];
	Mat(int _r,int _c){
		r=_r;c=_c;
		memset(a,0,sizeof a);
	}
	Mat operator *(const Mat &b)const{
		Mat c(r,b.c);
		for(int i=0;i<r;i++)
		for(int j=0;j<b.c;j++)
		for(int k=0;k<b.r;k++){
			c.a[i][j]=(c.a[i][j]+a[i][k]*b.a[k][j]%mod)%mod;
		}
		return c;
	}
}A(7,7),B(7,1);

Mat qpow(Mat a,int b){
	Mat c(a.r,a.c);
	for(int i=0;i<a.r;i++)c.a[i][i]=1;
	while(b){
		if(b&1)c=c*a;
		b>>=1;
		a=a*a;
	}
	return c;
}
int main() {
	int at[10][10]={{1,2,1,4,6,4,1},
				  {1,0,0,0,0,0,0},
				  {0,0,1,4,6,4,1},
				  {0,0,0,1,3,3,1},
				  {0,0,0,0,1,2,1},
				  {0,0,0,0,0,1,1},
				  {0,0,0,0,0,0,1}};
	for(int i=0;i<7;i++)for(int j=0;j<7;j++)A.a[i][j]=at[i][j];
	int t,n,a,b;
	cin>>t;
	while(t--){
		scanf("%d%d%d",&n,&a,&b);
		B.a[0][0]=b;B.a[1][0]=a;
		B.a[6][0]=1;
		for(int i=5;i>1;i--)B.a[i][0]=B.a[i+1][0]*2;
		if(n==1){
			printf("%d\n",a);
		}else if(n==2){
			printf("%d\n",b);
		}else{
			Mat C=qpow(A,n-2)*B;
			printf("%lld\n",C.a[0][0]);
		}
	}
	return 0;
}