HDU 1005 Number Sequence 矩阵乘法 Fib数列

原题： http://acm./showproblem.php?pid=1005

题目：

Number Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 128931 Accepted Submission(s): 31368

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.

Output
For each test case, print the value of f(n) on a single line.

Sample Input
1 1 3
1 2 10
0 0 0

Sample Output
2
5

思路：

按规律求出第n项。
由矩阵乘法我们可以知道：

所以对于fib数列我们可以用矩阵来求，由于矩阵可以左乘右乘，所以我们可以用快速幂来优化。

代码：

#include<iostream>
#include"string.h"
#include<stdio.h>
using namespace std;

const int bc=2;
const int mod = 7;
struct matrix
{
    int x[bc][bc];
};



matrix mutimatrix(matrix a,matrix b)
{
    matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    for(int i=0;i<bc;i++)    //答案的行
    {
        for(int j=0;j<bc;j++)   //答案的列
        {
            for(int k=0;k<bc;k++)
            {
                temp.x[i][j]+=a.x[i][k]*b.x[k][j];
                temp.x[i][j]%=mod;
            }
        }
    }
    return temp;
}

matrix powmatrix(matrix a,int b)
{
    matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    //初始化矩阵为单位阵
    for(int i=0;i<bc;i++)
        temp.x[i][i]=1;
    while(b)
    {
        if(b%2==1)
            temp=mutimatrix(temp,a);
        a=mutimatrix(a,a);
        b=b/2;
    }
    return temp;
}

int main()
{
    int a,b,n;
    //freopen("in.txt","r",stdin);
    while(scanf("%d %d %d",&a,&b,&n)!=EOF)
    {
        if(a==0&&b==0&&n==0) break;
        matrix st;
        //因为每次都是a个F(n-1)和b个F(n-2)相加，所以这里如此初始化
        //Fib数列这里就直接单位阵
        st.x[0][0]=a;
        st.x[0][1]=1;
        st.x[1][0]=b;
        st.x[1][1]=0;
        st=powmatrix(st,n-1);
        printf("%d\n",(st.x[0][1]+st.x[1][1])%mod);
    }
    return 0;
}
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