FZU 1683 纪念SlingShot

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思路: 矩阵快速幂

分析:

1 题目给定f(n) = 3*f(n-1)+2*f(n-2)+7*f(n-3) , f(0) = 1 , f(1) = 3 , f(2) = 5 ，给定n求f(0)+...+f(n) %2009

2 矩阵快速幂的水题，我们构造出这样的矩阵，然后利用矩阵快速幂即可

   3 2 7 0      f(n-1)      f(n)

   1 0 0 0 *   f(n-2) =   f(n-1)

   0 1 0 0      f(n-3)      f(n-2)

   3 2 7 1      sum        sum'

代码:

/************************************************
* By: chenguolin *
* Date: 2013-08-27 *
* Address: http://blog.csdn.net/chenguolinblog *
************************************************/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

typedef __int64 int64;
const int MOD = 2009;
const int N = 4;

int64 n;
struct Matrix{
int64 mat

;
Matrix operator*(const Matrix &m)const{
Matrix tmp;
for(int i = 0 ; i < N ; i++){
for(int j = 0 ; j < N ; j++){
tmp.mat[i][j] = 0;
for(int k = 0 ; k < N ; k++)
tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD;
tmp.mat[i][j] %= MOD;
}
}
return tmp;
}
};

void init(Matrix &m){
memset(m.mat , 0 , sizeof(m.mat));
m.mat[0][0] = m.mat[3][0] = 3;
m.mat[0][1] = m.mat[3][1] = 2;
m.mat[0][2] = m.mat[3][2] = 7;
m.mat[1][0] = m.mat[3][3] = 1;
m.mat[2][1] = 1;
}

int Pow(Matrix m){
if(n == 0) return 1;
if(n == 1) return 3;
if(n == 2) return 5;
n -= 2;
Matrix ans;
memset(ans.mat , 0 , sizeof(ans.mat));
for(int i = 0 ; i < N ; i++)
ans.mat[i][i] = 1;
while(n){
if(n%2)
ans = ans*m;
n /= 2;
m = m*m;
}
int sum = 0;
sum += ans.mat[3][0]*5%MOD;
sum += ans.mat[3][1]*3%MOD;
sum += ans.mat[3][2]*1%MOD;
sum += ans.mat[3][3]*9%MOD;
return sum%MOD;
}

int main(){
int Case;
int cas = 1;
Matrix m;
init(m);
scanf("%d" , &Case);
while(Case--){
printf("Case %d: " , cas++);
scanf("%I64d" , &n);
printf("%d\n" , Pow(m));
}
return 0;
}