HDU 1005 Number Sequence

先上题目

**Problem Description

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Input

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.**

题目类似斐波那契，相同的A,B两个数相乘了n次，可以联想到矩阵快速幂，那就先构造矩阵

| f(n) |=|A B| *| f(n-1)|

| f(n-1) | |1 0| | f(n-2)|

然后矩阵相乘取模

AC代码

#include<bits/stdc++.h>
typedef long long LL;
#define Max(a,b)  ((a)>(b)?(a):(b))
#define Min(a,b)  ((a)<(b)?(a):(b))
#define INF 0x3f3f3f3f
using namespace std;
struct Matrix{
int a[2][2];
}Num1,Num2,Num3;

void __inti__(){//初始化矩阵
Num2.a[0][0]=1;//f(n-1)
Num2.a[0][1]=0;
Num2.a[1][0]=1;//f(n-2)
Num2.a[1][1]=0;
}

Matrix Matrixpow(Matrix num1,Matrix num2,int mod)//矩阵相乘
{
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
Num3.a[i][j]=0;//存放结果矩阵初始化
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
Num3.a[i][j]=(Num3.a[i][j]+(num1.a[i][k]%mod)*(num2.a[k][j]%mod))%mod;
return Num3;
}

void Matrixquickpow(int n,int mod)//快速幂
{
Matrix num1=Num1,num2=Num2;
while(n)
{
if(n&1)
{
num2=Matrixpow(num1,num2,mod);
}
n=n/2;
num1=Matrixpow(num1,num1,mod);//A,B所在矩阵相乘
}
printf("%d\n",num2.a[0][0]);//输出结果
}

int main()
{
// clock_t time_start,time_end;
//  time_start=clock();
int A,B,mod=7;
LL n;
while(scanf("%d%d%lld",&A,&B,&n))
{
if(A==0&&B==0&&n==0)
break;
if(n==1||n==2)//f(1)=f(2)=1
printf("1\n");
else
{
__inti__();
Num1.a[0][0]=A;
Num1.a[0][1]=B;
Num1.a[1][0]=1;
Num1.a[1][1]=0;
Matrixquickpow(n-2,mod);
}
}
//time_end=clock();
// cout<<"Run time:"<<(time_end-time_start)<<"ms"<<endl;
return 0;
}