HDU 1005 矩阵快速幂

Number Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 169273    Accepted Submission(s): 41762

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

 

Author

CHEN, Shunbao

 

Source

ZJCPC2004

题意：给你A B 求f(n) 

题解：构造矩阵 矩阵快速幂

#include<bits/stdc++.h>
#define ll long long
#define mod 7
using namespace std;
int A,B,n;
struct node
{
    int a[3][4];
}exm,ans;
struct node matrix_mulit(struct node aa,struct node bb)
{
    struct node there;
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            there.a[i][j]=0;
            for(int k=0;k<2;k++)
            {
                there.a[i][j]=(there.a[i][j]+(aa.a[i][k]*bb.a[k][j]%mod))%mod;
            }
        }
    }
    return there;
}
int matrix_quick(int gg)
{
    exm.a[0][0]=A;exm.a[0][1]=B;
    exm.a[1][0]=1;exm.a[1][1]=0;
    ans.a[0][0]=1; ans.a[0][1]=0;
    ans.a[1][0]=1; ans.a[1][1]=0;
    while(gg)
    {
        if(gg&1)
            ans=matrix_mulit(exm,ans);
        exm=matrix_mulit(exm,exm);
        gg>>=1;
    }
    return ans.a[0][0];
}
int main()
{
    while(scanf("%d %d %d",&A,&B,&n)!=EOF)
    {
    if(A==0&&B==0&&n==0)
        break;
    if(n==1||n==2)
        printf("1
");
    else
        printf("%d
",matrix_quick(n-2));
    }
    return 0;
}